MEASURE THEORY - BGU Math | Homepageyadina/teaching/MT/MTnotes.pdf · This lecture has 12 exercises.62 Lecture 7. Lebesgue-Stieltjes Theory 63 ... Measure Theory Ariel Yadin Lecture

MEASURE THEORY

ARIEL YADIN

Course: 201.1.0081 Fall 2014-15

Lecture notes updated: January 22, 2015 (partial solutions)

Contents

Lecture 1. Introduction 4

1.1. Measuring things 4

1.2. Elementary measure 5

This lecture has 6 exercises. 11

Lecture 2. Jordan measure 12

2.1. Jordan measure 12


Lecture 3. Lebesgue outer measure 25

3.1. From finite to countable 25


Lecture 4. Lebesgue measure 30

4.1. Definition of Lebesgue measure 30

4.2. Lebesgue measure as a measure 37


Lecture 5. Abstract measures 44

5.1. σ-algebras 44

5.2. Measures 47

5.3. Fatou’s Lemma and continuity 50


Lecture 6. Outer measures 531

2

6.1. Outer measures 53

6.2. Measurability 54

6.3. Pre-measures 57


Lecture 7. Lebesgue-Stieltjes Theory 63

7.1. Lebesgue-Stieltjes measure 63

7.2. Regularity 68

7.3. Non-measureable sets 69

7.4. Cantor set 69


Lecture 8. Functions of measure spaces 71

8.1. Products 71

8.2. Measurable functions 73

8.3. Simple functions 78


Lecture 9. Integration: positive functions 81

9.1. Integration of simple functions 81

9.2. Integration of positive functions 83


Lecture 10. Integration: general functions 90

10.1. Real valued functions 90

10.2. Complex valued functions 91

10.3. Convergence 93

10.4. Riemann vs. Lebesgue integration 97


Lecture 11. Product measures 102

11.1. Sections 105

11.2. Product integrals 107

3

11.3. The Fubini-Tonelli Theorem 114


Lecture 12. Change of variables 121

12.1. Linear transformations of Lebesgue measure 121

12.2. Change of variables formula 124


Lecture 13. Lebesgue-Radon-Nykodim 129

13.1. Signed measures 129

13.2. The Lebesgue-Radon-Nikodym Theorem 136


Lecture 14. Convergence 147

14.1. Modes of convergence 147

14.2. Uniform and almost uniform convergence 153


Lecture 15. Differentiation 158

15.1. Hardy-Littlewood Maximal Theorem 158

15.2. The Lebesgue Differentiation Theorem 161

15.3. Differentiation and Radon-Nykodim derivative 162


Lecture 16. The Riesz Representation Theorem 165

16.1. Compactly supported functions 165

16.2. Linear functionals 168

16.3. The Riesz Representation Theorem 169


Total number of exercises: 177 174

4

Measure Theory

Ariel Yadin

Lecture 1: Introduction

1.1. Measuring things

Already the ancient Greeks developed a theory of how to measure length, area, and

volume and area of 1, 2 and 3 dimensional objects. In this setting (i.e. in Rd for d ≤ 3)

it stands to reason that the “size” or “measure” of an object must satisfy some basic

axioms:

• If m(A) is the measure of a set A, it should be the same for any reflection,

translation or rotation of A. That is, m(A) = m(A+ x) = m(UA) where U is a

rotation matrix.

For example, [0, 1]2 should heave the same measure as [−2,−1]2 which should

have the same measure as{

(x, y) : |x|+ |y| ≤ 1√2

}, the diamond of side-length

1.

• If A can be broken into disjoint pieces, the sum of their measures should be the

measure of A. That is, m(A ]B) = m(A) +m(B).

Example: [0, 1] × [0, 2] should have measure that is the sum of the measures

of [0, 1]2 and [0, 1]× (1, 2].

X We use ] to denote disjoint union; that is, A ] B is not only notation for a set,

but this notation claims that A ∩ B = ∅. The small + sign remind us of the additive

property above.

This is already quite fruitful. If the unit square (0, 1]2 is of measure 1, then:

• We can determine the measure of any square of rational side length.

m((0, n]2) =n∑

j,k=1

m((j − 1, j]× (k − 1, k]) = n2

5

and

m((0, 1]2) =

n∑j,k=1

m(( j−1n , jn ]× (k−1

n , kn ]) = n2m((0, 1n ]2).

• We can measure any rectangle of rational side length: decompose it into squares.

• We can measure right-angle triangles: disjoint union of two is a rectangle.

• We can then measure any triangle, by bounding it in a rectangle and subtracting

the excess right-angle triangles.

• With triangles we can measure any polygon.

X What about measuring a disc?

1.2. Elementary measure

Let us first formally define the above.

• Definition 1.1 (Boxes). Consider Rd for some d ≥ 1.

• I ⊂ R is an interval if I is one of [a, b], (a, b), (a, b], [a, b) for some −∞ < a ≤ b <∞. Note that we allow the singleton {a} = [a, a] as an interval, and the empty set

∅ = (a, a). The length or measure of such I is defined to be |I| = m(I) = b−a.

• B ⊂ Rd is a box if B = I1 × · · · × Id where Ij are intervals. The volume or

measure of such a box B is defined to be |B| = m(B) = |I1| · · · |Id|.• E ⊂ Rd is an elementary set if E = B1 ∪ · · · ∪ Bn for some finite number of

boxes.

• E0 = E0(Rd) denotes the set of elementary sets in Rd.

� Exercise 1.1. Show that E0 is closed under finite unions, finite intersections,

set-difference, symmetric difference and translations. That is, show that if E,F are

elementary sets then so are:

• E ∪ F and E ∩ F ,

• E \ F := {x ∈ E : x 6∈ F} = E ∩ F c,• E4F := (E \ F ) ∪ (F \ E),

6

• E + x := {y + x : y ∈ E}.

� Exercise 1.2. Show that if E is an elementary set then there exist B1, . . . , Bn

pairwise disjoint boxes such that E = B1 ] · · · ]Bn.

It would be tempting to define the measure of E = B1]· · ·]Bn asm(B1)+· · ·+m(Bn).

But we are not guarantied that the decomposition is a unique one.

• Proposition 1.2 (Discretisation Formula). Let I be an interval. Then,

m(I) = limn→∞

1

n#(I ∩ 1

nZ).

Consequently, if B ⊂ Rd is a box then,

m(B) = limn→∞

1

nd#(B ∩ 1

nZd),

and if E = B1 ] · · · ]Bn ⊂ Rd is an elementary set then

m(B1) + · · ·+m(Bn) = limn→∞

1

nd#(E ∩ 1

nZd).

Proof. Let a < b and I be an interval with endpoints a, b. For n large enough, note that

#(I ∩ 1nZ) = #

{z ∈ Z : z

n ∈ I}≤ # {z ∈ Z : na ≤ z ≤ nb}

≤ # {z ∈ Z : bnac ≤ z ≤ dnbe} ≤ nb+ 1− (na− 1) + 1 = n(b− a) + 3.

Similarly,

#(I ∩ 1nZ) ≥ # {z ∈ Z : na < z < nb} ≥ n(b− a)− 1.

Dividing by n and taking n→∞ we obtain the formula for intervals.

Now, if B = I1 × · · · × Id then

B ∩ 1nZ

d = (I1 ∩ 1nZ)× · · · × (Id ∩ 1

nZ),

7

which implies that

#(B ∩ 1nZ

d) =

d∏j=1

#(Ij ∩ 1nZ).

Dividing by nd and taking n→∞ gives the formula for boxes.

The final assertion is a consequence of the fact that if A ∩B = ∅ then

(A ]B) ∩ 1nZ

d = (A ∩ 1nZ

d) ] (B ∩ 1nZ

d).

ut

• Definition 1.3. If E = B1 ] · · · ] Bn is an elementary set we define the measure of

E to be

m(E) = m(B1) + · · ·+m(Bn) = limn→∞

1

nd#(E ∩ 1

nZ),

which is well defined by Proposition ??.

� Exercise 1.3. Give an example of a set A ⊂ [0, 1] such that the limit

limn→∞

1

n#(A ∩ 1

nZ)

does not exist.

limn→∞

1

n#(A ∩ 1

nZ) and limn→∞

1

n#((A+ x) ∩ 1

nZ)

both exist but are not equal.

♣ Solution to ex:1.3. :(

For the first example: Set

A ={

2m−12n : Z 3 n,m ≥ 1 , 2m− 1 ≤ 2n

}.

Since for any z ≥ 1 we can write z = (2m − 1)2k for some k ≥ 0 and m ≥ 1, we have

that

#(A ∩ 2−nZ) ≥ #{z

2n : 1 ≤ z ≤ 2n}

= 2n.

8

So

lim supn→∞

1n#(A ∩ 1

nZ) ≥ lim supn→∞

2−n#(A ∩ 2−nZ) = 1.

On the other hand, if x ∈ A ∩ 3−kZ then for some m,n ≥ 1 and z ∈ Z, x = 2m−12n = z

3k

which implies that 2nz = 3k(2m − 1). The left hand side is even while the right hand

side is odd, which is a contradiction. So A ∩ 3−kZ = ∅ for all k, and we conclude that

lim infn→∞

1n#(A ∩ 1

nZ) ≤ lim infn→∞

3−n#(A ∩ 3−nZ) = 0.

For the second example take A = Q ∩ [0, 1] and x = π. :) X

This last exercise shows that we do not want to use the discretization formula to

define the measure of general sets, but we can use it for more than just elementary sets.

� Exercise 1.4. Show that the measure of elementary sets has the following prop-

erties.

• m : E0 → [0,∞).

• (Additivity) m(E ] F ) = m(E) +m(F ).

• m(∅) = 0.

• If B is a box, then m(B) = |B|.• (Monotonicity) If E ⊂ F then m(E) ≤ m(F ).

• (Subadditivity) m(E ∪ F ) ≤ m(E) +m(F ).

• (Translation invariance) For any x ∈ Rd, m(E + x) = m(E).


The discretization formula guaranties that m is non-negative and additive, m(∅) = 0

and also that if B is a box then m(B) = |B|.For monotonicity recall that F \ E is elementary, and F = (F ∩ E) ] (F \ E). Since

E ⊂ F we have that F ∩ E = E. So m(F ) = m(E) +m(F \ E) ≥ m(E).

Subadditivity follows from E ∪ F = E ] (F \ E) so

m(E ∪ F ) = m(E) +m(F \ E) ≤ m(E) +m(F ).

9

Translation invariance is another consequence of the discretization formula: For any

interval I with endpoints a < b, we have that zn−x ∈ I if and only if z ∈ nI+nx. Thus,

if zn − x ∈ I then z ∈ [bna + nxc, dnb + nxe] ∩ Z and if z ∈ [dna + nxe, bnb + nxc] ∩ Z

then zn − x ∈ I. Since

#([bna+nxc, dnb+nxe]∩Z) ≤ n(b−a)+3 and #([dna+nxe, bnb+nxc]∩Z) ≥ n(b−a)−1,

we have ∣∣#{z ∈ Z : zn − x ∈ I

}− n(b− a)

∣∣ ≤ 3.

Dividing by n and taking a limit we get that

m(I + x) = limn→∞

1n#{z ∈ Z : z

n − x ∈ I}

= b− a = m(I).

If B = I1×· · ·×Id is a box in Rd, then for any x ∈ Rd, B+x = (I1+x1)×· · ·×(Id+xd),

so m(B + x) =∏dj=1m(Ij + xj) =

∏dj=1m(I) = m(B).

Finally, if E = B1]· · ·]Bk is an elementary set, then E+x = (B1+x)]· · ·](Bk+x).

So m(E + x) = m(E). :) X

� Exercise 1.5. Show that if m′ : E0 → [0,∞) is additive and translation invariant,

i.e. m(E]F ) = m(E)+m(F ) and m(E+x) = m(E), then there exists a constant c ≥ 0

such that m′ = cm.


If E = B1 ] · · · ] Bk is an elementary set then m′(E) = m′(B1) + · · ·+m′(Bk). So we

only need to show that m′(B) = c|B| for any box B.

First, note that since

[0, 1) =

n⊎j=1

1n [j − 1, j) and [0, nk ) =

n⊎j=1

1k [j − 1, j),

additivity and translation invariance guaranty that

m′([0, 1n)d) = 1

ndm′([0, 1)d) and m′([0, q)d) = qd ·m′([0, 1))

10

for any q ∈ Q (make sure to fill in the details here). By translation invariance, for any

aj ≤ bj ∈ Q we then have

m′([a1, b1)× · · · × [ad, bd)) =

d∏j=1

(bj − aj) ·m′([0, 1)d).

Set C = m′([0, 1)d).

Let B = I1 × · · · × Id. Assume that the endpoints of Ij are aj < bj .

Fix ε > 0 such that for all j we have bj − aj > 2ε. Let a−j , a+j , b−j , b

+j be rational

numbers such that a+j ∈ (aj , aj + ε), a−j ∈ (aj − ε, aj), b+j ∈ (bj , bj + ε), b−j ∈ (bj − ε, bj).

Let I+j = [a−j , b

+j ) and I−j = [a+

j , b−j ). Let B± = I±1 × · · · × I±d , so B− ⊂ B ⊂ B+.

Since m′ is additive and non-negative, it is also monotone. Monotonicity tells us that

m′(B−) ≤ m′(B) ≤ m′(B+).

Because a±j , b±j ∈ Q we get that

m′(B+) =

d∏j=1

(b+j − a−j ) · C ≤d∏j=1

(bj − aj + 2ε) · C,

m′(B−) =d∏j=1

(b−j − a+j ) · C ≥

d∏j=1

(bj − aj − 2ε) · C.

Setting M = maxj(bj − aj) we have that

m′(B+)−d∏j−1

(bj − aj) · C ≤ C · 2ε · dMd−1,

d∏j−1

(bj − aj) · C −m′(B−) ≤ C · 2ε · dMd−1.

Taking ε→ 0 we get that

m′(B) =d∏j=1

(bj − aj) · C = m′([0, 1)d) ·m(B).

:) X

� Exercise 1.6. Let E ⊂ Rd and F ⊂ Rk be elementary sets. Show that E × F is

elementary (in Rd+k) and that m(E × F ) = m(E) ·m(F ).

11

Number of exercises in lecture: 6

Total number of exercises until here: 6

12

Measure Theory

Ariel Yadin

Lecture 2: Jordan measure

2.1. Jordan measure

We have seen that the measure of elementary sets is a good way to measure length,

area and volume for squares and rectangles, and anything that can be composed of

finite unions of such. How about measuring a triangle? Or a circle? We saw that the

discretisation formula also has its limitations. However, Jordan measure is essentially

the way to use the discretisation formula.

Jordan measure is essentially approximating general shapes by elementary ones.

• Definition 2.1 (Jordan measure). Let A ⊂ Rd be a bounded set (i.e. A ⊂ B(0, r) :=

{x : |x| ≤ r}). Define:

• The Jordan inner measure of A

J∗(A) := supE03E⊂A

m(A).

• The Jordan outer measure of A

J∗(A) := infE03F⊃A

m(F ).

• If A is such that J∗(A) = J∗(A) we say that A is Jordan measurable. We then

define the Jordan measure of A as this common value, m(A) = J∗(A) = J∗(A).

� Exercise 2.1. Show that J∗, J∗ are monotone. Show that J∗ is subadditive.


Let A ⊂ C. Let (En)n, (Fn)n be sequences of elementary sets such that En ⊂ A,C ⊂ Fn

13

and m(En)→ J∗(A),m(Fn)→ J∗(C). Note that for every n,

J∗(C) ≥ m(En)→ J∗(A) and J∗(A) ≤ m(Fn)→ J∗(C).

Also, if (Gn)n is a sequence of elementary sets such that A ⊂ Gn and m(Gn)→ J∗(A),

then A ∪ C ⊂ Gn ∪ Fn so

J∗(A ∪ C) ≤ m(Gn ∪ Fn) ≤ m(Gn) +m(Fn)→ J∗(A) + J∗(C).

:) X

Note that every elementary set E is Jordan measurable and m(E) = J∗(E) = J∗(E).

Jordan sets are those which are “almost elementary”.

� Exercise 2.2. Show that the following are equivalent for bounded A ⊂ Rd.

• A is Jordan measurable.

• For every ε > 0 there exist elementary sets E ⊂ A ⊂ F , E,F ∈ E0, such that

m(F \ E) < ε.

• For every ε > 0 there exists an elementary set E0 3 E ⊂ A such that J∗(A\E) <

ε.


For any ε > 0 there exist elementary sets E ⊂ A ⊂ F such that m(E) > J∗(A)− ε and

m(F ) < J∗(A) + ε. So if A is Jordan measurable then m(F ) < J∗(A) + ε = J∗(A) + ε <

m(E) + 2ε. Since E ⊂ F we have that m(F ) = m(E) +m(F \ E), so m(F \ E) < 2ε.

Now assume that there exist E,F ∈ E0, E ⊂ A ⊂ F such that m(F \ E) < ε. Then,

A \ E ⊂ F \ E, so J∗(A \ E) ≤ m(F \ E) < ε.

Finally assume that for every ε > 0 there exists an elementary set Eε ⊂ A such that

J∗(A \ Eε) < ε. Since Eε ⊂ A we have that m(Eε) ≤ J∗(A). Using the subadditivity of

J∗ we get

J∗(A) ≤ J∗(A \ Eε) +m(Eε) < ε+ J∗(A).

14

Taking ε → 0 and recalling that J∗(A) ≤ J∗(A) by definition, we have that J∗(A) =

J∗(A) and aso A is jordan measurable. :) X

� Exercise 2.3. Show that for Jordan measurable sets A,C the following holds.

• A ∪ C,A ∩ C,A \ C,A4C are all Jordan measurable.

• m(A) ≥ 0.

• (Additivity) If A ∩ C = ∅ then m(A ] C) = m(A) +m(C).

• (Monotonicity) If C ⊂ A then m(C) ≤ m(A).

• (Subadditivity) m(A ∪ C) ≤ m(A) +m(C).

• (Translation invariance) m(A+ x) = m(A).

� Exercise 2.4. Show that a bounded set A is Jordan measurable if and only if for

every ε > 0 there exists an elementary set E such that A ⊂ E and J∗(E \A) < ε.


Let B be a box containing the bounded set A.

Suppose that there exists an elementary set F ⊂ B \A such that J∗((B \A)\F ) < ε.

Then, with E = B \ F we have that A ⊂ E and J∗(E \A) = J∗((B \A) \ F ) < ε.

Thus we obtain that A is Jordan measurable iff B \ A is Jordan measurable iff for

every ε > 0 there exists an elementary set F , F ⊂ B \A, such that J∗((B \A) \ F ) < ε

iff for every ε > 0 there exists an elementary set E, A ⊂ E, such that J∗(E \A) < ε. :)

X

� Exercise 2.5. [Tao, ex. 1.1.7] Let B ⊂ Rd be a closed box, and let f : B → R be a

continuous function. Show that

• The graph {(x, f(x)) : x ∈ B} is Jordan measurable with Jordan measure 0.

15

• The volume under the graph {(x, t) : x ∈ B, 0 ≤ t ≤ f(x)} is Jordan measur-

able.


f is uniformly continuous in B; that is, for every ε > 0 there exists δ > 0 such that for

any ||x− y|| < δ we have |f(x)− f(y)| < ε.

Write B =⋃Nn=1Bδ(xn) where (Bδ(xn))n are disjoint boxes of diameter less than δ,

with xn the center of Bδ. Uniform continuity of f gives that for any x ∈ B there exists (a

unique) n such that x ∈ Bδ(xn), and so (x, f(x)) ∈ Bδ(xn)×(f(xn)−ε, f(xn)+ε) =: Qδ,n.

Thus, {(x, f(x)) : x ∈ B} ⊂ ⋃Qδ,n. Since m(Qδ,n) = m(Bδ(xn)) · 2ε, we have that

J∗({(x, f(x)) : x ∈ B}) ≤∑n

m(Qδ,n) ≤ 2ε∑n

m(Bδ(xn)) = 2εm(B).

Taking ε→ 0 gives the first assertion.

For the second assertion, let V = {(x, t) : x ∈ B, 0 ≤ t ≤ f(x)}. Fix ε, δ > 0 and

B =⋃nBδ(x) as above. For every n set

mn = infx∈Bδ(xn)

f(x) Mn = supx∈Bδ(xn)

f(x).

Note that |Mn −mn| ≤ ε by uniform continuity. Now, set Kn = bmnε c. For 0 ≤ k ≤Kn − 1 set Qn,k = Bδ(xn)× ε[k, k + 1) and Qn = Bδ(xn)× [εKn,Mn].

For any n, if 0 ≤ k < Kn and (x, t) ∈ Qn,k then x ∈ Bδ(xn) and t < εKn ≤ mn ≤ f(x).

So ⋃n

Kn−1⋃k=0

Qn,k ⊂ V.

On the other hand, if x ∈ B and 0 ≤ t ≤ f(x) then there exists n such that x ∈ Bδ(xn)

and so f(x) ∈ [mn,Mn]. Thus, either t < εKn in which case (x, t) ∈ Qn,k for some

0 ≤ k < Kn or εKn ≤ t ≤ f(x) ≤Mn in which case (x, t) ∈ Qn. Thus,

⋃n

Kn−1⋃k=0

Qn,k ⊂ V ⊂⋃n

Kn−1⋃k=0

Qn,k⊎⋃

n

Qn.

16

Now, let Λ := m(⋃n

⋃Kn−1k=0 Qn,k). Then,

J∗(V ) ≤ Λ +∑n

m(Qn) = Λ +∑n

m(Bδ(xn)) · (Mn − εKn)

≤ Λ +∑n

m(Bδ(xn)) · (Mn −mn + ε) ≤ Λ +m(B) · 2ε

≤ J∗(V ) +m(B) · 2ε.

Taking ε→ 0 completes the proof. :) X

� Exercise 2.6. [Tao, p. 31, ex. 1.2.6] Show that it is not true that

J∗(E) = supE⊃U open

J∗(U).


Take E = [0, 1] \ Q. If U ⊂ E is an open set, then U = ∅, because E does not contain

any open intervals. Thus, supU⊂E,U openm∗(U) = 0. However, if E ⊂ ⋃n In where (In)n

are disjoint intervals, then it must be that [0, 1] ⊂ ⋃n In. So∑

nm(In) ≥ m([0, 1]) = 1.

Thus, m∗(E) = 1. :) X

� Exercise 2.7. [Tao, p. 12, ex. 1.1.13] Prove the discretisation formula

m(A) = limN→∞

1Nd#

(A ∩ 1

NZd)

for all Jordan measurable subsets A ⊂ Rd.


We know the discretisation formula holds for all elementary sets. For simplicity of the

presentation, let us denote IN (A) = 1Nd#

(A ∩ 1

NZd).

17

Let A be Jordan measurable, and let ε > 0. Let E ⊂ A ⊂ F be elementary sets such

that m(F ) < m(A) + ε and m(E) > m(A)− ε. Note that

E ∩ 1NZd ⊂ A ∩ 1

NZd ⊂ F ∩ 1NZd,

so IN (E) ≤ IN (A) ≤ IN (F ). Taking N →∞,

m(A) < m(E) + ε = ε+ limN→∞

IN (E) ≤ ε+ limN→∞

IN (A)

≤ ε+ limN→∞

IN (F ) = ε+m(F ) < 2ε+m(A).

Taking ε→ 0 we get the formula. :) X

� Exercise 2.8. [Tao, p. 12, ex. 1.1.14] A dyadic cube of scale 2−n is a box of the

form:

Dn(z1, . . . , zd) :=[z12n ,

z1+12n

)× · · · ×

[zd2n ,

zd+12n

),

For n ∈ N and (z1, . . . , zd) ∈ Zd. Note that these are half-open half-closed.

Let E∗(A,n) be the number of dyadic cubes of scale 2−n that are contained in A, and

let E∗(A,n) be the number of dyadic cubes of scale 2−n that intersect A.

Show that a bounded set A is Jordan measurable if and only if

limn→∞

2−dn(E∗(A,n)− E∗(A,n)) = 0.

Show that Jordan measurable sets admit

m(A) = limn→∞

2−dnE∗(A,n) = limn→∞

2−dnE∗(A,n).


If limn→∞ 2−dn(E∗(A,n) − E∗(A,n)) = 0: Let S∗(A,n) be the set of dyadic cubes of

scale 2−n that are contained in A, and let S∗(A,n) be the set of dyadic cubes of scale

2−n that intersect A. Then,⋃S∗(A,n) ⊂ A ⊂

⋃S∗(A,n),

18

and these are elementary sets. Since dyadic cubes of the same scale are always disjoint,

we get that

2−dnE∗(A,n) = m(⋃S∗(A,n)) ≤ J∗(A) ≤ J∗(A) ≤ m(

⋃S∗(A,n)) = 2−dnE∗(A,n).

Thus, taking n→∞ we get that J∗(A) = J∗(A). Moreover, we have that

m(A) = limn→∞

2−dnE∗(A,n) = limn→∞

2−dnE∗(A,n).

If A is Jordan measurable: Let B = I1 × · · · × Id be a box where Ij is the interval

with endpoints aj < bj , define

• (B)n := [a1 − 2−n, b1 + 2−n] × · · · × [ad − 2−n, bd + 2−n], the enlargement of B

by 2−n in each coordinate direction.

• (B)n := (a1 + 2−n, b1 − 2−n)× · · · × (ad + 2−n, bd − 2−n), shrinking B by 2−n in

each coordinate direction.

Under these definitions, if a dyadic cube Dn(z) of scale 2−n intersects B then it is

contained in (B)n; if Dn(z) is not contained in B then Dn(z) does not intersect (B)n.

Finally note that

m((B)n)−m((B)n) =d∏j=1

(bj − aj + 2 · 2−n)−d∏j=1

(bj − aj − 2 · 2−n)

=∑

S⊂{1,...,d}

∏j∈S

(bj − aj) ·(

(2 · 2−n)d−|S| − (−2 · 2−n)d−|S|)

≤ 2d · 4 · 2−n ·m(B).

Fix ε > 0 and let E ⊂ A ⊂ F be elementary sets such that m(F ) ≤ m(A) + ε and

m(E) ≥ m(A) − ε. Suppose that E =⊎nk=1Ek and F =

⊎mj=1 Fj where En, Fj are

boxes.

Let Dn(z) be a dyadic cube of scale 2−n such that Dn(z) intersects A but is not

contained in A. Then, there exists j such that Dn(z) intersects some Fj , so Dn(z) is

contained in (Fj)n. Also, Dn(z) is not contained in Ek for all k, so for all k we get that

Dn(z) does not intersect (Ek)n. Thus, if we define

(E)n :=n⊎k=1

(Ek)n and (F )n :=m⋃j=1

(Fj)n

19

then

m((E)n) =n∑k=1

m((Ek)n) ≥n∑k=1

m(Ek) ·(

1− 2d · 22−n)

= m(E) ·(

1− 2d · 22−n),

m((F )n) ≤m∑j=1

m((Fj)n) ≤

m∑j=1

m(Fj) ·(

1 + 2d · 22−n)

= m(F ) ·(

1 + 2d · 22−n),

and since for any Jordan measurable set J we have that 2−dnE∗(J) ≤ m(J) ≤ 2−dnE∗(J),

2−dn(E∗(A,n)− E∗(A,n)) ≤ 2−dn(E∗((F )n, n)− E∗((E)n, n)) ≤ m((F )n)−m((E)n)

≤ m(F )−m(E) + 2d · 22−n · (m(F ) +m(E))

≤ ε+ 2d · 22−n · (2m(A) + ε).

Taking n→∞, we have that for all ε > 0,

0 ≤ limn→∞

2−dn(E∗(A,n)− E∗(A,n)) ≤ ε,

so this limit must be 0. :) X

� Exercise 2.9. [Tao, p. 13, ex. 1.1.18] Show that for any bounded set A:

• J∗(A) = J∗(A).

• J∗(A◦) = J∗(A).

• A is Jordan measurable if and only if J∗(∂A) = 0.


First, A ⊂ A so J∗(A) ≤ J∗(A), and we only need to prove J∗(A) ≤ J∗(A). Let ε > 0.

Then choose disjoint boxes (Bn)Nn=1 such that A ⊂ ⋃nBn and∑

nm(Bn) ≤ J∗(A) + ε.

Note that

A ⊂⋃n

Bn ⊂⋃n

Bn,

and since Bn are also boxes,

J∗(A) ≤∑n

m(Bn) =∑n

m(Bn) ≤ J∗(A) + ε.

20

Taking ε→ 0 completes the proof of the first item.

For the second item, since A◦ ⊂ A we only need to show that J∗(A) ≤ J∗(A◦). Fix

ε > 0 and let (Bn)Nn=1 be a finite number of disjoint boxes such that⋃nBn ⊂ A and∑

nm(Bn) ≥ J∗(A)− ε. Then, since B◦n are disjoint boxes, and since m(B◦n) = m(Bn),

from⋃nB◦n ⊂ A◦ we deduce that

J∗(A◦) ≥∑n

m(B◦n) =∑n

m(Bn) ≥ J∗(A)− ε.

Taking ε→ 0 completes the second item.

For the final item: First, note that (∂A)◦ = ∅. If Dn(z) ⊂ ∂A for some dyadic cube

Dn(z), then (∂A)◦ ⊃ (Dn(z))◦ 6= ∅, a contradiction. So ∂A contains no dyadic cubes,

which is to say that E∗(∂A, n) = 0 for all n. Thus, ∂A is Jordan measurable with

J(∂A) = 0 if and only if 2−dnE∗(∂A, n)→ 0.

Note that if Dn(z) is a dyadic cube, then it intersects A and is not contained in A, if

and only if it intersects ∂A. That is, E∗(∂A, n) = E∗(A,n)− E∗(A,n).

Now, by a previous exercise, A is Jordan measurable, if and only if 2−dn(E∗(A,n) −E∗(A,n))→ 0 if and only if 2−dnE∗(∂A, n)→ 0, if and only if ∂A is Jordan measurable

with J(∂A) = 0, if and only if J∗(∂A) = 0.

The last step follows from the fact that if J∗(∂A) = 0 then J∗(∂A) ≤ J∗(∂A) = 0

and so ∂A is Jordan measurable with J(∂A) = 0. :) X

� Exercise 2.10. Show that any triangle ABC in R2 is Jordan measurable.

Show that any compact convex polygon in R2 is Jordan measurable.

� Exercise 2.11. Show that the ball B(x, r) ={x ∈ Rd : |x| < r

}is Jordan mea-

surable with measure m(B(x, r)) = cdrd where cd > 0 is a constant depending only on

the dimension d.

21

� Exercise 2.12. Show that if A ⊂ Rd, C ⊂ Rk are Jordan measurable, then A×Cis Jordan measurable and m(A× C) = m(A) ·m(C).


Fix ε > 0. Let E ⊂ A ⊂ F,E′ ⊂ C ⊂ F ′ such that E,E′, F, F ′ ∈ E0 and m(F \ E) < ε,

m(F ′ \E′) < ε. Note that E×E′ ⊂ A×C ⊂ F ×F ′ and E×E′, F ×F ′ are elementary

sets. Thus, since we assumed that A,C are Jordan measurable,

J∗(A× C) ≤ m(F × F ′) = m(F ) ·m(F ′) < (m(E) + ε) · (m(E′) + ε) ≤ (m(A) + ε) · (m(C) + ε).

Also, since m(A) = m(A \ E) + m(E) ≤ m(F \ E) + m(E) and m(C) = m(C \ E′) +

m(E′) ≤ m(F ′ \ E′) +m(E′),

J∗(A× C) ≥ m(E × E′) ≥ (m(A)− ε) · (m(C)− ε).

Taking ε→ 0 we have that J∗(A× C) = J∗(A× C) = m(A) ·m(C). :) X

� Exercise 2.13. Show that if J∗(A) = 0 then A is Jordan measurable.

Show that if m(A) = 0 for Jordan measurable A, then any C ⊂ A is Jordan measur-

able.

� Exercise 2.14. give an example of a sequence (An)n of Jordan measurable sets

such that A :=⋃nAn is bounded but not Jordan measurable.

22

••• Theorem 2.2. Let L : Rd → Rd be an invertible linear transformation. If A is

Jordan measurable, then L(A) is also, and m(L(A)) = | detL|m(A).

Proof. If E is an elementary set, then L(E) is Jordan measurable. Indeed, recall that

any invertible matrix L can be written as a product of elementary operation matrices:

L = T1 · · ·Tn where each Tj is either multiplication of a row by a scalar c, addition of

one row to another row, or swapping of two rows. That is, Tj is in one of the following

families of matrices:

Mc,j =

1 0 ··· 0...

. . ....

··· c ···...

. . ....

0 ··· 1

Ri,j =

1 0 ··· 0...

. . ....

···1 0··· 1 ···...

. . ....

0 ··· 1

Si,j =

1 0 ··· 0...

. . ....

··· 0 1 ······ 1 0 ···

.... . .

...0 ··· 1

That is Mc,j is obtain by multiplying the j-th row of the identity matrix by c; Ri,j is

obtained by adding the i-th row in the identity matrix to the j-th row; Si,j is obtained

by swapping the i-th and j-th rows of the identity matrix.

It is immediate to compute that detMc,j = c, |detRi,j | = | detSi,j | = 1. Also, if

B = I1 × · · · × Id we have that Mc,j(B) = I1 × · · · × cIj × · · · × Id and for i < j,

Si,j(B) = I1 × · · · Ij × · · · × Ii × · · · × Id. Thus, we have that for any box B, and any

linear map L ∈ {Mc,j , Si,j}, m(L(B)) = | detL| · m(B). If E = B1 ] · · · ] Bk is an

elementary set, where (Bj)kj=1 are boxes, then L(E) = L(B1) ] · · · ] L(Bk), so L(E) is

an elementary set and

m(L(E)) =

k∑j=1

m(L(Bj)) =

k∑j=1

|detL| ·m(Bj) = |detL| ·m(E).

Finally, if A is Jordan measurable, the for any ε > 0 choose elementary sets E ⊂ A ⊂ Fsuch that m(F ) < m(E) + ε. So, L(E) ⊂ L(A) ⊂ L(F ) and

J∗(L(A)) ≤ m(L(F )) = |detL| ·m(F ) ≤ |detL| ·m(E) + |detL| · ε

≤ |detL| · J(A) + |detL| · ε ≤ |detL| ·m(F ) + |detL| · ε

≤ |detL| ·m(E) + |detL| · 2ε = m(L(E)) + | detL| · 2ε ≤ J∗(L(A)) + | detL| · 2ε.

Taking ε → 0 we have that J∗(L(A)) = J∗(L(A)) = |detL| · J(A). This proves the

theorem for the case that L ∈ {Mc,j , Si,j}.

23

Now for a somewhat more cumbersome computation:

Ri,j(B) ={

(x1, . . . , xi, . . . , xi + xj , . . . , xd) ∈ Rd : ∀ k , xk ∈ Ik},

which is the Cartesian product of the parallelepiped {(x, x+ y) : x ∈ Ii, y ∈ Ij} with a

box in Rd−2. Indeed, if we apply S = Si,1Sj,2 to this set,

Ri,j(B) ={

(x1, . . . , xi, . . . , xi + xj , . . . , xd) ∈ Rd : ∀ k , xk ∈ Ik}

= S{

(xi, xi + xj , x3, . . . , xi−1, x1, xi+1, . . . , xj−1, x2, xj+1, . . . , xd) ∈ Rd : ∀ k , xk ∈ Ik}

= S({(x, x+ y) : x ∈ Ii, y ∈ Ij} × I3 × · · · Ii−1 × I1 × Ii+1 × · · · × Ij−1 × Ij × Ij+1 × · · · × Id).

Since S preserves Jordan measure (and measurability), it suffices to show the Jordan

measurability of P := {(x, x+ y) : x ∈ Ii, y ∈ Ij} and compute its Jordan measure.

By translating P we may assume without loss of generality that the endpoints of Ii are

0 < a and the endpoints of Ij are 0, b. We may also write P = T1 ]B ] T2 where T1, T2

are right-angle triangles, with orthogonal sides parallel to the axes, and B is a box, as

in Figure 1. In a previous exercise it was shown that triangles are Jordan measurable.

A translation of T2 by the vector −(0, b) shows that T1 ] T2 has the Jordan measure of

the box Ii× Ii. Thus, P is Jordan measurable and has Jordan measure J(P ) = |Ii| · |Ij |,which consequently is J(P ) = m(Ii × Ij).

T1

T2

B

P

T2 − (0, b)

T1

Figure 1. Computation of the Jordan measure of the parallelepiped P .

24

� Exercise 2.15. Let P := {(x, x+ y) : x ∈ I, y ∈ J} where I, J are intervals. Show

that P is Jordan measurable and that J(P ) = |I| · |J |.

We conclude that Ri,j(B) is Jordan measurable, and that

J(Ri,j(B)) = J(S(P ×B)) = J(P ×B) = J2(P ) · Jd−2(B) = |Ii| · |Ij | ·∏k 6=i,j

|Ik| = m(B),

where P = {(x, x+ y) : x ∈ Ii, y ∈ Ij} and B = I3 × · · · Ii−1 × I1 × Ii+1 × · · · × Ij−1 ×Ij × Ij+1 × · · · × Id ⊂ Rd−2. Since |detRi,j | = 1 we have just proved that for any box

B, the set Ri,j(B) is Jordan measurable and J(Ri,j(B)) = | detRi,j | ·m(B).

Just as for Mc,j and Si,j we now use the fact that Ri,j(A]B) = Ri,j(A)]Ri,j(B) to

prove the theorem for any L ∈ {Mc,j , Si,j , Ri,j}.Finally, if L is a general invertible linear map, then L = L1 · · ·Ln where Lk ∈

{Mc,j , Si,j , Ri,j} for all k. So for any A that is Jordan measurable, also Ln(A) is Jordan

measurable, and thus Ln−1Ln(A) is as well, and so on to get that L(A) = L1 · · ·Ln(A)

is Jordan measurable. We also compute the measure by

J(L(A)) = J(L1(L2 · · ·Ln(A))) = |detL1| · J(L2 · · ·Ln(A)) =

= · · · = |detL1| · · · |detLn| · J(A) = |detL| · J(A).

ut



25

Measure Theory

Ariel Yadin

Lecture 3: Lebesgue outer measure

3.1. From finite to countable

Recall that in order to measure sets from outside we used the outer measure by

approximating with finitely many elementary sets.

� Exercise 3.1. Show that

J∗(A) = infA⊂B1∪···∪Bn

m(B1) + · · ·+m(Bn),

where B1, . . . , Bn are always boxes.

Lebesgue’s idea is that there is no reason to stop with finite, and not countable

collections. (Mathematicians are not afraid of infinity anymore...)

• Definition 3.1. Let A ⊂ Rd be a set. The Lebesgue outer measure of A is defined

to be

m∗(A) := inf{∑

n

|Bn| : (Bn)n is a sequence of boxes , A ⊂⋃n

Bn}.

X Note that we may have m∗(A) =∞.

� Exercise 3.2. Show that in general m∗(A) ≤ J∗(A) for bounded sets A.

Show that for Q := Q ∩ [0, 1] we have that

J∗(Q) = 1 and m∗(Q) = 0.

26


If A is bounded then the infimum for Lebesgue outer measure is on a larger set than

the infimum for Jordan outer measure.

For Q, we have that Q = [0, 1], so J∗(Q) = J∗([0, 1]) = m([0, 1]) = 1 and if Q =

{q1, q2, . . .} is an enumeration of Q then m∗(Q) ≤∑n | {qn} | = 0. :) X

� Exercise 3.3. Give an example of an unbounded set with Lebesgue outer measure

0.

• Proposition 3.2. Properties of Lebesgue outer measure:

• m∗(∅) = 0.

• (Monotonicity) If A ⊂ C then m∗(A) ≤ m∗(C).

• (Countable subadditivity) If (An)n is a sequence of sets then m∗(⋃nAn) ≤∑

nm∗(An).

Proof. First, ∅ is a box of volume 0; another proof is by noting that ∅ ⊂ [0, 1n ]d for all

n, so m∗(∅) ≤ infnm([0, 1n ]d) = 0.

For A ⊂ C we have that the infimum used to obtain m∗(A) is over a larger set that

the one used to obtain the infimum for m∗(C).

Let (An)n be a sequence of sets. Fix ε > 0. For every n let (Bn,k)k be a sequence of

boxes such that An ⊂⋃k Bn,k and∑

k

|Bn,k| ≤ m∗(An) + ε · 2−n.

Consider the sequence of boxes (Bn,k)n,k. We have that⋃n

An ⊂⋃n,k

Bn,k,

27

and so

m∗(⋃n

An) ≤∑n,k

|Bn,k| ≤∑n

m∗(An) +∑n

ε · 2−n =∑n

m∗(An) + 2ε.

Taking ε→ 0 we have countable subadditivity. ut

It is now natural to ask for the additivity property: if A ∩ C = ∅ is it true that

m∗(A ] C) = m∗(A) + m∗(C)? As it turns out, this is not always the case, a counter

example will be given in the future. However, in some cases, when the sets are separated

enough, we have additivity.

• Proposition 3.3. If A,C are such that dist(A,C) > 0 then m∗(A ] C) = m∗(A) +

m∗(C).

Specifically, if A,C are closed disjoint sets and A is compact then dist(A,C) > 0.

Proof. By sub-additivity it suffices to prove m∗(A) +m∗(C) ≤ m∗(A]C). We may also

assume w.l.o.g. that m∗(A]C) <∞ and by monotonicity that m∗(A) <∞,m∗(C) <∞.

Fix ε > 0. Let (Bn)n be a sequence of boxes such that A]C ⊂ ⋃nBn and∑

n |Bn| ≤m∗(A ] C) + ε.

Let r = dist(A,C) > 0. Note that for any n, we may replace the box Bn by a finite

number of disjoint boxes Bn,1, . . . , Bn,k such that the diameter of any Bn,j is less than r.

So we obtain a sequence (B′n)n such that A]C ⊂ ⋃nB′n and

∑n |B′n| ≤ m∗(A]C) + ε,

and such that any box B′n cannot intersect both A and C.

Thus, we have that N = NA ] NC where NA = {n : B′n ∩A 6= ∅} and NC =

{n : B′n ∩ C 6= ∅}.We now have that

A ⊂⋃

n∈NAB′n and C ⊂

⋃n∈NC

B′n,

and so

m∗(A) +m∗(C) ≤∑n∈NA

|B′n|+∑n∈NC

|B′n| ≤∑n

|B′n| ≤ m∗(A ] C) + ε.

Taking ε→ 0 completes the proof.

28

As for the case where A,C are closed and A is compact, note that dist(x,C) is a

continuous function of x (because C is closed) and so achieves a minimum on A when

A is compact. ut

� Exercise 3.4. Give an example of two closed sets A,C such that A ∩ C = ∅ but

dist(A,C) = 0.

The next proposition shows that the notation “m” is an appropriate one, as an ex-

tension of elementary measure.

• Proposition 3.4. For any elementary set E we have that m∗(E) = m(E).

Proof. If suffices to prove that J∗(E) ≤ m∗(E).

Case I. E is closed. Since E is bounded, it is compact, and the Heine-Borel Theorem

tells us that every open cover of E has a finite sub-cover.

Fix any ε > 0. Let (Bn)n be a sequence of boxes such that E ⊂ ⋃nBn and∑

n |Bn| ≤m∗(E) + ε.

X The boxes (Bn)n do not form an open cover – they need not be open.

For every n let B′n be an open box containing Bn ⊂ B′n such that |B′n| ≤ |Bn|+ ε2−n.

(Exercise: show this is possible.) Then,∑

n |B′n| ≤ m∗(E) + 2ε and (B′n)n are an open

cover of E. Thus, there is a finite sub-cover

E ⊂N⋃n=1

B′n,

for some large enough N . Thus,

J∗(E) ≤N∑n=1

|B′n| ≤∑n

|B′n| ≤ m∗(E) + 2ε.

Taking ε→ 0 completes the proof for closed E.

Case II. E is a general elementary set. Write E = B1 ] · · · ] Bn where (Bj)nj=1 are

disjoint boxes. These need not be closed boxes.

29

Fix ε > 0. For every 1 ≤ j ≤ n let B′j ⊂ Bj be a closed box such that |B′j | ≥ |Bj |− εn .

(Exercise: Show this is always possible.) Then, E′ = B′1 ] · · · ] B′n is a finite union of

disjoint closed boxes, and thus a closed elementary set. Also,

m(E′) =

n∑j=1

|B′j | ≥n∑j=1

|Bj | − ε = m(E)− ε.

So by Case I,

m(E) ≤ m(E′) + ε = m∗(E′) + ε ≤ m∗(E) + ε.

Taking ε→ 0 completes the proof. ut

� Exercise 3.5. Show that for any bounded set A we have

J∗(A) ≤ m∗(A) ≤ J∗(A).

Show that if A is Jordan measurable then m(A) = m∗(A).


We have already seen m∗(A) ≤ J∗(A). If E is an elementary set such that E ⊂ A, then

m(E) = m∗(E) ≤ m∗(A) by monotonicity. Taking supremum over all such elementary

sets contained in A, we get that J∗(A) ≤ m∗(A). This proves the first assertion.

If A is Jordan measurable, then m(A) = J∗(A) ≤ m∗(A) ≤ J∗(A) = m(A). :) X



30

Measure Theory

Ariel Yadin

Lecture 4: Lebesgue measure

4.1. Definition of Lebesgue measure

Recall that A is Jordan measurable if and only if for every ε > 0 there exists an

elementary set E such that A ⊂ E and J∗(E \ A) < ε. This motivates the following.

Later we will see that there is another way to approach the issue of Lebesgue measure,

and outer measures in general.

• Definition 4.1 (Lebesgue measure). A set A ⊂ Rd is said to be Lebesgue measur-

able if for every ε > 0 there exists an open set U such that A ⊂ U and m∗(U \A) < ε.

for Lebesgue measurable A we denote m(A) := m∗(A) and refer to this as the

Lebesgue measure of A.

X Lebesgue measurable sets are sets that are “almost open”.

? Why do open sets enter the picture?

• Definition 4.2. We say that boxes (Bn)n are almost disjoint if any two boxes can

only intersect at their boundary; that is, for every n 6= m, B◦n ∩B◦m = ∅.

• Lemma 4.3. Any open set U ⊂ Rd can be written as a countable union of almost

disjoint closed boxes.

Proof. For any z ∈ Zd define the dyadic cube at scale n ≥ 0:

Dn(z) =[z12n ,

z1+12n

]× · · · ×

[zd2n ,

zd+12n

].

This is the d-dimensional cube of side-lengths 2−n and corner z.

The following is the crucial property to verify: If Dn(z)◦ ∩Dm(z′)◦ 6= ∅ then one is

contained in the other.

31

Specifically, if Dn(z) ( Dm(z′) then n > m.

Now, set

Γ ={

(z, n) : z ∈ Zd, n ≥ 0, Dn(z) ⊂ U}.

These are all dyadic cubes whose closure is contained in U . Also, set

Λ ={

(z, n) ∈ Γ : 6 ∃ (z′, n′) ∈ Γ , Dn(z) ⊂ Dn′(z′)}.

These are all dyadic cubes contained in U that are maximal with respect to inclusion.

We claim that

U =⋃

(z,n)∈Λ

Dn(z).

One inclusion is obvious, since all closures of dyadic cubes indexed by Λ are contained

in U . For the other inclusion, let x ∈ U . Then, since U is open (this is the only place

we use that U is an open set!), there is a small ball B(x, ε) ⊂ U . However, for large

enough n (so that√d2−n < ε), if x ∈ Dn(z) then Dn(z) ⊂ B(x, ε) ⊂ U (because Dn(z)

has diameter√d2−n). Since for every n there exists some dyadic cube of scale n that

contains x (because Rd =⋃z∈Zd Dn(z) for every fixed n), we get that for some large

enough n there exists z ∈ Zd with (z, n) ∈ Γ and x ∈ Dn(z). Thus, there must be

(z, n) ∈ Λ such that x ∈ Dn(z) (by just taking the cube of minimal scale containing x).

Since this holds for all x ∈ U we get that

U ⊂⋃

(z,n)∈Λ

Dn(z).

This is of course a countable union. Also, since Λ is the indices of maximal dyadic

cubes, any two cannot intersect except for the boundary; that is, they are almost disjoint.

ut

This construction has a remarkable consequence.

� Exercise 4.1. Show that if U =⋃nBn where (Bn)n are almost disjoint boxes

then m∗(U) =∑

n |Bn| = J∗(U).

32

Figure 2. Part of the dyadic decomposition of a set. If the set is open,

one can continue to capture all points in the set by taking smaller and

smaller squares.


m∗(U) ≤∑n |Bn| so it suffices to show that∑

n |Bn| ≤ J∗(U). Note that∑m

n=1 |Bn| ≤J∗(U) for all m because B1 ∪ · · · ∪Bm ⊂ U . Taking m→∞ we have

∑n |Bn| ≤ J∗(U).

:) X

� Exercise 4.2. Show that if (Bn)n are almost disjoint and (B′n)n are almost disjoint,

and if⋃nBn =

⋃nB′n then ∑

n

|Bn| =∑n

|B′n|.

• Proposition 4.4 (Outer regularity, Lebesgue measure). For any set A we have

m∗(A) = infA⊂U open

m∗(U).

Proof. One direction m∗(A) ≤ m∗(U) for any open U ⊃ A, is just monotonicity.

33

For the other direction, if m∗(A) =∞ there is nothing to prove. Assume m∗(A) <∞.

Fix ε > 0. Let (Bn)n be boxes such that A ⊂ ⋃nBn and∑

n |Bn| ≤ m∗(A) + ε. For

every n let B′n be an open box such that Bn ⊂ B′n and |B′n| ≤ |Bn| + ε2−n. Thus, the

set U =⋃nB′n is an open set containing A and

m∗(U) ≤∑n

|B′n| ≤ m∗(A) + 2ε.

Taking infimum over the left hand side and ε→ 0 completes the proof. ut

� Exercise 4.3. Show that it is false that

m∗(A) = supA⊃U open

m∗(U).

• Proposition 4.5 (Lebesgue measurable sets). Examples of Lebesgue measurable sets:

• If U is open it is Lebesgue measurable.

• If m∗(A) = 0 then A is Lebesgue measurable.

• ∅ is Lebesgue measurable.

• If (An)n is a sequence of Lebesgue measurable sets, then⋃nAn is Lebesgue

measurable.

• Any closed set F is Lebesgue measurable.

• If A is Lebesgue measurable then so is Ac = Rd \A.

• If (An)n is a sequence of Lebesgue measurable sets, then⋂nAn is Lebesgue

measurable.

Proof. For open sets this is obvious by definition.

If m∗(A) = 0 then by outer regularity for any ε > 0 there exists an open set U ⊃ A

such that m∗(U \A) ≤ m∗(U) < m∗(A) + ε = ε. So A is Lebesgue measurable.

∅ has m∗(∅) ≤ J∗(∅) = 0.

34

Countable unions. Now, if (An)n are all Lebesgue measurable, then: Fix ε > 0.

For every n there exists an open set Un ⊃ An such that m∗(Un \ An) < ε2−n. Thus,

U :=⋃n Un is an open set containing

⋃nAn such that by subadditivity,

m∗(U \⋃n

An) ≤∑n

µ∗(Un \An) ≤ ε.

Since this holds for all ε > 0 we get that⋃nAn is Lebesgue measurable.

Closed sets. Let F be a closed set. Note that F =⋃n(F ∩ B[0, n]) where B[0, n]

is the closed ball of radius n. So it suffices to prove that F is Lebesgue measurable

for closed and bounded F (because then F ∩ B[0, n] is Lebesgue measurable for all n).

Heine-Borel guaranties then that F is compact.

Fix ε > 0. Let U be an open set such that F ⊂ U and m∗(U) ≤ m∗(F ) + ε. The

set U \ F is open, so we can write U \ F =⋃nBn where (Bn)n are almost disjoint and

closed boxes. For any m > 0, the set Cm := B1 ∪ · · · ∪ Bm is closed and disjoint from

F . Also, F is compact, so we have additivity:

m∗(F ) +m∗(Cm) = m∗(F ∪ Cm) ≤ m∗(F ∪ (U \ F )) = m∗(U) ≤ m∗(F ) + ε.

Since F is bounded, we have that m∗(F ) <∞, so we conclude that for all m > 0,

m∑n=1

|Bn| = m∗(B1 ∪ · · · ∪Bm) ≤ ε,

where the equality is because (Bn)n are almost disjoint. Thus, taking m→∞,

m∗(U \ F ) =∑n

|Bn| ≤ ε.

This holds for all ε > 0, so F is Lebesgue measurable.

Complements. Now, if A is Lebesgue measurable: For every n let Un ⊃ A be an

open set such that m∗(Un \ A) < 2−n. Let Fn = U cn and let F =⋃n Fn. Since Fn are

closed they are Lebesgue measurable, and thus F is Lebesgue measurable as a countable

union. Note that m∗(Ac \ Fn) = m∗(Un \A) < 2−n for all n. Since Ac \ F ⊂ Ac \ Fn,

m∗(Ac \ F ) ≤ infnm∗(Ac \ Fn) = 0.

Thus, Ac = F ∪ (Ac \ F ) which is the union of two Lebesgue measurable sets, and thus

Lebesgue measurable itself.

35

Countable intersections. If (An)n are all Lebesgue measurable, then so are (Acn)n

and thus also ⋂n

An =

(⋃n

Acn

)c.

ut

� Exercise 4.4. Show that the following are equivalent.

(1) A is Lebesgue measurable.

(2) For every ε > 0 there exists an open set U ⊃ A such that m∗(U \A) < ε.

(3) For every ε > 0 there exists an open set U such that m∗(U4A) < ε.

(4) For every ε > 0 there exists a closed set F such that m∗(A4F ) < ε.

(5) For every ε > 0 there exists a closed set F ⊂ A such that m∗(A \ F ) < ε.

(6) For every ε > 0 there exists a Lebesgue measurable set C such that m∗(A4C) <

ε.


(1) ⇐⇒ (2) is the definition.

(2) ⇒ (3) follows by taking the same open set U ⊃ A.

(3) ⇒ (2): Fix ε > 0 and let U be such that m∗(U4A) < ε. Let V ⊃ U4A be an

open set such that m∗(V ) < 2ε, by outer regularity. Note that A ⊂ U ∪(A\U) ⊂ U ∪V ,

which is an open set, and

m∗((U ∪ V ) \A) ≤ m∗(U \A) +m∗(V \A) ≤ m∗(U4A) +m∗(V ) < 3ε.

So (1) ⇐⇒ (2) ⇐⇒ (3).

(1) ⇒ (4): A is Lebesgue measurable, so also Ac is. Fix ε > 0 and let U be an open

set such that m∗(U4Ac) < ε. Let F = U c. So m∗(F4A) = m∗(U4Ac) < ε, and F is

closed.

(4) ⇒ (5): Fix ε > 0 and let F be a closed set such that m∗(A4F ) < ε. By outer

regularity let U ⊃ A4F be an open set such that m∗(U) < 2ε. Then F ∩U c is a closed

36

set and F ∩ U c ⊂ F ∩A ⊂ A, and

m∗(A \ F ∩ U c) ≤ m∗(A \ F ) +m∗(A \ U c) ≤ m∗(A4F ) +m∗(U) < 3ε.

(5) ⇒ (6): Fix ε > 0 and let F ⊂ A be a closed set such that m∗(A \ F ) < ε. The

C = F is Lebesgue measurable, and m∗(A4C) = m∗(A \ F ) < ε.

(6) ⇒ (3): Fix ε > 0 and let C be Lebesgue measurable such that m∗(A4C) < ε.

Let U ⊃ C be an open set such that m∗(U4C) = m∗(U \ C) < ε. Note that A4U ⊂A4C⋃C4U , so

m∗(A4U) ≤ m∗(A4C) +m∗(C4U) < 2ε.

:) X

� Exercise 4.5. Show that if A is Jordan measurable then it is also Lebesgue mea-

surable.

• Definition 4.6. A family of subsets F is a σ-algebra if

• ∅ ∈ F ;

• F is closed under complements, i.e. A ∈ F implies Ac ∈ F ;

• F is closed under countable unions, i.e. (An)n ⊂ F implies⋃nAn ∈ F .

� Exercise 4.6. Show that if F is a σ-algebra it is closed under countable intersec-

tions, set difference and symmetric difference.

� Exercise 4.7. Show that the family of Lebesgue measurable sets form a σ-algebra.

37

4.2. Lebesgue measure as a measure

• Proposition 4.7 (Measure axioms). Let L be the family of all Lebesgue measurable

sets. Recall that for A ∈ L we define m(A) = m∗(A). We the have that m : F → [0,∞]

with the following properties:

• m(∅) = 0;

• (Countable additivity) If (An)n ⊂ L are pairwise disjoint Lebesgue measurable

sets then

m(⊎n

An) =∑n

m(An).

Proof. We only need to prove the assertion regarding additivity.

Case I. (An)n are all compact. Since m∗ is additive on disjoint compact sets, for any

N ,N∑n=1

m(An) = m(

N⊎n=1

An) ≤ m(⊎n

An) ≤∑n

m(An),

where the last two inequalities are monotonicity and countable subadditivity. Taking

N →∞ we get the compact case.

Case II. (An)n are all bounded. Fix ε > 0. For every n let Fn be a closed set with

Fn ⊂ An and m∗(An \ Fn) < ε2−n. Since Fn ⊂ An is bounded, Fn is compact. By

subadditivity, m(An) ≤ m(Fn) + ε2−n, so∑n

m(An) ≤∑n

m(Fn) + ε = m(⊎n

Fn) + ε ≤ m(⊎n

An) + ε.

Taking ε→ 0 completes the bounded case.

Case III. Now for the general case. Let Ck = B(0, k) \ B(0, k − 1) for all n, and

set Dn,k = An ∩ Ck. So (Dn,k)n,k is a sequence of pairwise disjoint bounded Lebesgue

measurable sets. Thus, for every n,

m(An) = m(⊎k

Dn,k) =∑k

m(Dn,k),

and

m(⊎n

An) = m(⊎n,k

Dn,k) =∑n,k

m(Dn,k) =∑n

m(An).

ut

38

� Exercise 4.8. Show that if A is Lebesgue measurable then

m(A) = supA⊃K compact

m(K).


Monotonicity implies that it suffices to prove that m(A) ≤ supA⊃K compact m(K).

First suppose that A is bounded, so m(A) <∞.

Fix ε > 0. Since A is Lebesgue measurable there exists a closed set K ⊂ A such that

µ∗(A \K) < ε. Since K ⊂ A and A is bounded, we have that K is bounded as well, so

K is compact because it is bounded and closed. So K is a compact set such that K ⊂ Aand µ(K) = µ(A)−µ(A \K) ≥ µ(A)− ε. This holds for all ε > 0 establishing the claim

in the case that A is bounded.

For the case where A is unbounded, consider An = A∩(B(0, n)\B(0, n−1)) for all n.

These are all bounded, so for every ε > 0 and every n there exists a compact set Kn ⊂ Ansuch that m(Kn) ≥ m(An)−ε2−n. Also, (An)n are disjoint so m(A) =

∑nm(An). Thus,

if K ′n =⊎nj=1Kj , then K ′n is compact, K ′n ⊂ A and

m(K ′n) =n∑j=1

m(Kj) ≥n∑j=1

m(Aj)− ε→ m(A)− ε.

This implies that there exists n large enough so that m(K ′n) ≥ m(A)− 2ε.

Conclusion, for any ε > 0 there exists a compact K ⊂ A such that m(K) ≥ m(A)−ε.This proves the unbounded case. :) X

Finally, let us end with the following criterion for Lebesgue measurability.

• Proposition 4.8 (Charatheodory criterion). The following are equivalent.

• A is Lebesgue measurable.

• For every box B we have |B| = m∗(B ∩A) +m∗(B \A).

• For every elementary set E we have m(E) = m∗(E ∩A) +m∗(E \A).

39

Proof. If A is Lebesgue measurable and E is elementary, then E = (E∩A)](E\A) which

are both Lebesgue measurable sets, so by additivity, m(E) = m∗(E ∩A) +m∗(E \A).

Suppose the second bullet holds. Let E be any elementary set, and write E =⊎nj=1Bj

for disjoint boxes (Bj)nj=1. Then by subadditivity,

m∗(E ∩A) +m∗(E \A) ≤n∑j=1

m∗(Bj ∩A) +m∗(Bj \A) =n∑j=1

|Bj | = m(E).

Since m(E) ≤ m∗(E ∩A) +m∗(E \A) for any set E just by subadditivity, we conclude

that equality holds for all elementary sets, proving the third bullet.

Now assume the third bullet. We want to show that A is Lebesgue measurable.

Assume first that m∗(A) < ∞. Fix ε > 0. Let (Bn)n be disjoint boxes such that∑n |Bn| ≤ m∗(A) + ε and A ⊂ ⋃nBn (this exists by the definition of m∗). For every n

let B′n ⊃ Bn be an open box such that |B′n| ≤ |Bn|+ ε2−n. Note that U =⋃nB′n is an

open set and U ⊃ A. For every n we have that

m∗(B′n ∩A) +m∗(B′n \A) = |B′n| ≤ |Bn|+ ε2−n.

By subadditivity,

m∗(A) +m∗(U \A) ≤∑n

m∗(B′n ∩A) +m∗(B′n \A) ≤∑n

|Bn|+ ε ≤ m∗(A) + 2ε.

Thus, m∗(U \A) ≤ 2ε. Since this holds for all ε > 0 this completes the proof for A with

m∗(A) <∞.

Now assume that m∗(A) = ∞. Let E be any elementary set, and let B be a box.

Since E ∩B is elementary, and since E \ (A∩B) ⊆ (E \B)∪ ((E ∩B)\A), we have that

m∗(E ∩A ∩B) +m∗(E \ (A ∩B)) ≤ m∗(E ∩B ∩A) +m∗((E ∩B) \A) +m(E \B)

= m(E ∩B) +m(E \B) = m(E).

Thus, for any elementary E we have the Caratheodory criterion for A ∩B,

m(E) = m∗(E ∩ (A ∩B)) +m∗(E \ (A ∩B)).

Since B is bounded, we have m∗(A∩B) <∞ and by the previous part A∩B is Lebesgue

measurable. Since A =⋃n(A ∩ Bn) where Bn are boxes of side length n around 0, we

40

have that A is Lebesgue measurable as a countable union of Lebesgue measurable sets.

ut

� Exercise 4.9. For a bounded set A define the Lebesgue inner measure by

m∗(A) = m(E)−m∗(E \A),

for a Lebesgue mesurable set E such that A ⊂ E and m(E) <∞.

Show that this definition does not depend on the choice of the set E.

Show that m∗(A) ≤ m∗(A) and equality holds if and only if A is Lebesgue measur-

able.


Let E,F be Lebesgue measurable sets such that A ⊂ F ⊂ E and m(E) <∞.

Fix ε > 0. Let (Bn)n be a sequence of boxes such that E \A ⊂ ⋃nAn and∑

n |Bn| ≤m∗(E \A) + ε. Note that

F \A ⊂ (E \A) \ (E \ F ) ⊂⋃n

(Bn \ (E \ F )) and E \ F ⊂⋃n

(Bn ∩ (E \ F )).

So,

m(E \ F ) +m∗(F \A) ≤∑n

m∗(Bn ∩ (E \ F )c) +m∗(Bn ∩ (E \ F ))

=∑n

|Bn| ≤ m∗(E \A) + ε.

Taking ε→ 0 and rearranging we have that

m(E)−m∗(E \A) ≤ m(F )−m∗(F \A).

Since E \A ⊂ (E \ F ) ∪ (F \A) we have

m(E)−m∗(E \A) ≥ m(E)−m(E) +m(F )−m∗(F \A).

So we have show that

m(E)−m∗(E \A) = m(F )−m(F \A)

41

whenever A ⊂ F ⊂ E.

For general Lebesgue measurable sets E,F such that A ⊂ F and A ⊂ E and m(E) <

∞,m(F ) <∞, we have that A ⊂ E ∩ F , so

m(E)−m∗(E \A) = m(E ∩ F )−m∗((E ∩ F ) \A) = m(F )−m∗(F \A).

This shows that m∗ is well defined.

Now, if A is Lebesgue measurable then for any Lebesgue measurable set E ⊃ A with

m(E) <∞ we have that m(E \ A) = m(E)−m(A). So m∗(A) = m(E)−m(E \ A) =

m(A) = m∗(A).

On the other hand, assume that m∗(A) = m∗(A). Fix ε > 0 and let U ⊃ A be an

open set such that A ⊂ U and m(U) ≤ m∗(A) + ε. Note that

m(U)− ε ≤ m∗(A) = m∗(A) = m(U)−m∗(U \A),

so m∗(U \A) ≤ ε. :) X

� Exercise 4.10. Reminder: A Gδ set is a set A =⋂n Un where (Un)n are all open

(i.e. a countable intersection of open sets). A Fσ set is a set A =⋃n Fn where (Fn)n

are all closed (i.e. a countable union of closed sets).

A null set is a set with Lebesgue (outer) measure 0.

Show that the following are equivalent.

• A is Lebesgue measurable.

• A = G \N is where G is Gδ and N is null.

• A = F ∪N where F is Fσ and N is null.

� Exercise 4.11. Show that if A is Lebesgue measurable then A+x is also Lebesgue

measurable and m(A+ x) = m(A).

42

� Exercise 4.12. Let L be the family of Lebesgue measurable sets. Suppose that

m′ : L → [0,∞] admits the following properties:

• m′(∅) = 0;

• If (An)n ⊂ L are pairwise disjoint Lebesgue measurable sets then m′(⊎nAn) =∑

nm′(An);

• m′(A+ x) = m′(A) for all A ∈ L, x ∈ Rd;

• m′([0, 1]d) = 1.

Show that m′ is Lebesgue measure.


We already know that m′(E) = m(E) for any Jordan measurable set E, and specifically

for boxes.

It is simple to prove that m′ is sub-additive and monotone.

Step I. We show that for A ∈ L, if m(A) = 0 then m′(A) = 0: Let A be a Lebesgue

measurable set of 0 measure, m(A) = 0. For any ε > 0, let (Bn)n be a sequence of

boxes such that A ⊂ ⋃nBn and such that∑

n |Bn| ≤ ε. Then, m′(A) ≤∑nm′(Bn) =∑

n |Bn| ≤ ε, and taking ε→ 0, we have that m′(A) = 0.

Step II. We show that if U is open of finite Lebesgue measure, then m′(U) = m(U):

Let U be an open set such that m(U) < ∞. Write U =⋃nBn where (Bn)n are

almost disjoint closed boxes. Let B′n = (Bn)◦, so (B′n)n are pairwise disjoint. Let

F =⋃nBn \

⊎nB′n. Since

m(F ) = m(U)−m(⊎n

B′n) =∑n

|Bn| −∑n

|B′n| = 0,

we have that m′(F ) = 0. Note that U = F ]⊎nB′n, so

m′(U) = m′(F ) +∑n

m′(B′n) =∑n

|B′n| = m(U).

43

Step III. We show that for any A ∈ L, the inequality m′(A) ≤ m(A) holds: Let A

be a Lebesgue measurable set with m(A) <∞. Fix ε > 0 and let U be an open set such

that A ⊂ U and m(U) −m(A) = m(U \ A) < ε. Then, m′(A) ≤ m′(U) ≤ m(A) + ε.

Taking ε → 0 we have that for any A ∈ L, the inequality m′(A) ≤ m(A) holds. (The

case where m(A) =∞ is immediate.)

Step IV. We show that for any bounded set A ∈ L we havem′(A) = m(A): Let A ∈ Lbe a bounded set. Let B be a box bounding A ⊂ B. We have m′(B\A) = m′(B)−m′(A)

by additivity of m′. Also,

m′(A) ≤ m(A) = m(B)−m(B \A) ≤ m′(B)−m′(B \A) = m′(A).

Step V. We show that for any A ∈ L we have m′(A) = m(A): Let A ∈ L. Write

A =⊎An where An = Bn \Bn−1, and Bn = [−n, n]d. Then by additivity of both m,m′

and since An are all bounded,

m′(A) =∑n

m′(An) =∑n

m(An) = m(A).

:) X



44

Measure Theory

Ariel Yadin

Lecture 5: Abstract measures

We now review the construction of Lebesgue measure, isolating the main abstract

properties, in order to generalize it to other settings.

5.1. σ-algebras

We saw in the discussion of Lebesgue measure that the family of Lebesgue measurable

sets form a special structure called a σ-algebra.

• Definition 5.1 (σ-algebra). Let X be any set. We denote by 2X = P(X) =

{A : A ⊂ X} the set of all subsets of X.

A family F ⊂ 2X is called a σ-algebra (on X) if:

• ∅ ∈ F ;

• F is closed under complements, i.e. A ∈ F implies X \A ∈ F ;

• F is closed under countable unions, i.e. if (An)n is a sequence in F then⋃nAn ∈

F .

� Exercise 5.1. Show that if F is a σ-algebra on X then:

• F is closed under countable intersections, i.e. if (An)n is a sequence in F then⋂nAn ∈ F .

• X ∈ F .

• F is closed under finite unions and finite intersections.

• F is closed under set differences.

• F is closed under symmetric differences.

45

� Exercise 5.2. Suppose F ⊂ 2X is a family of subsets satisfying the following:

• ∅ ∈ F ;

• F is closed under complements;

• F is closed under countable intersections.

Show that F is a σ-algebra.

� Exercise 5.3. Show that if (Fα)α∈I is a collection of σ-algebras on X, then⋂α Fα

is also a σ-algebra on X.

• Proposition 5.2 (σ-algebra generated by subsets). Let K be a collection of subsets

of X.

There exists a σ-algebra, denoted σ(K) such that K ⊂ σ(K) and for every other σ-

algebra F such that K ⊂ F we have that σ(K) ⊂ F .

That is, σ(K) is the smallest σ-algebra containing K.

We call σ(K) the σ-algebra generated by K.

Proof. Define

σ(K) :=⋂{F : F is a σ-algebra on X , K ⊂ F} .

This is a σ-algebra with the required properties. ut

� Exercise 5.4. Show that if K ⊂ L then σ(K) ⊂ σ(L). Also, if K ⊂ F and F is a

σ-algebra, then σ(K) ⊂ F .

46

• Definition 5.3 (Borel σ-algebra). Given a topological space X, the Borel σ-algebra

is the σ-algebra generated by the open sets. It is denoted B(X).

Specifically in the case X = Rd we have that

B = Bd = B(Rd) = σ(U : U is an open set ).

X A Borel-measurable set, i.e. a set in B(X), is called a Borel set.

� Exercise 5.5. Prove that

B = Bd = B(Rd) = σ(B : B is an open box ) = σ(B′ : B′ is a closed box ).


If B is an open box then B ∈ B so σ(B : B is an opend box ) ⊂ B.

If U is an open set then it is a countable union of closed boxes, so U ∈ σ(B :

B is a closed box ), which shows that B ⊂ σ(B′ : B′ is a closed box ).

If B′ = [a1, b1]× · · · × [ad, bd] is a closed box, then for

Bn = (a1 − 1n , b1 + 1

n)× · · · × (ad − 1n , bd + 1

n),

we have thatB′ =⋂nBn andBn are all open boxes. Thus, B′ ∈ σ(B : B is an open box ),

which implies that σ(B′ : B′ is a closed box ) ⊂ σ(B : B is an open box ). :) X

47

� Exercise 5.6. Show that

B(R) = σ((a, b) : −∞ < a < b <∞) = σ([a, b] : −∞ < a < b <∞)

= σ((a, b] : −∞ < a < b <∞) = σ([a, b) : −∞ < a < b <∞)

= σ((a,∞) : −∞ < a <∞) = σ([a,∞) : −∞ < a <∞)

= σ((−∞, a) : −∞ < a <∞) = σ((−∞, a] : −∞ < a <∞).

5.2. Measures

• Definition 5.4. A pair (X,F) where F is a σ-algebra on X is call a measurable

space. Elements of F are called measurable sets.

Given a measurable space (X,F), a function µ : F → [0,∞] is called a measure (on

(X,F)) if

• µ(∅) = 0;

• (Additivity) For all sequences (An)n ⊂ F of pairwise disjoint sets in F , we have

that

µ(⊎n

An) =∑n

µ(An).

(X,F , µ) is called a measure space.

X A measure space (X,F , µ) is called finite if µ(X) < ∞. It is called σ-finite if

X =⋃nAn where An ∈ F and µ(An) <∞ for all n.

� Exercise 5.7. Show that any measure is finitely additive; that is, if (X,F , µ) is a

measure space then for any disjoint sets A,B ∈ F we have µ(A ]B) = µ(A) + µ(B).

48

Example 5.5. The counting measure: Take F = 2X and µ(A) = |A|.If f : X → [0,∞) is a function then

µ(A) := sup(an)n⊂A

∑n

f(an)

is a measure on (X, 2X).

If X is uncountable and

F = {A ⊂ X : A is countable, or Ac is countable } ,

then (X,F) is a measurable space and

µ(A) =

1 Ac is countable

0 A is countable

is a measure on (X,F). 454

� Exercise 5.8. Give an example of a set X such that

µ(A) :=

∞ |A| =∞,

0 |A| <∞,

is not a measure on (X, 2X).

• Proposition 5.6 (Basic properties of measures). Let (X,F , µ) be a measure space.

Then:

• (Monotonicity) For A ⊂ B ∈ F we have µ(A) ≤ µ(B).

• (Subadditivity) If (An)n ⊂ F then µ(⋃nAn) ≤∑n µ(An).

Proof. Monotonicity follows from B = A ] (B \ A) and finite additivity, so µ(B) =

µ(A) + µ(B \A) ≥ µ(A).

49

For subadditivity, let (An)n ⊂ F and set

Bn = An \ (

n−1⋃j=1

Aj).

So ⊎n

Bn =⋃n

An and Bn ⊂ An.

Thus,

µ(⋃n

An) = µ(⊎n

Bn) =∑n

µ(Bn) ≤∑n

µ(An).

ut

• Definition 5.7. For a measure space (X,F , µ) a set N ∈ F is called a null set (or

µ-null set) if µ(N) = 0.

A measure space (X,F , µ) such that for all null sets N ∈ F we have that any A ⊂ Nis also measurable (i.e. in F) is called complete.

� Exercise 5.9. Let (X,F , µ) be a measure space. Let N be the set of all µ-null

sets. Define

F := {A ∪ F : A ∈ F , F ⊂ N ∈ N} .

Show that F is a σ-algebra.

Show that if we define µ : F → [0,∞] by

µ(A ∪ F ) = µ(A) ∀ A ∈ F , F ⊂ N ∈ N ,

then µ is a well defined complete measure on F ; moreover, µ∣∣F = µ and if ν is a complete

measure on F such that ν∣∣F = µ then ν = µ.

� Exercise 5.10. Show that if µ1, . . . , µn are measures on (X,F) the for any non-

negative numbers a1, . . . , an the function µ :=∑

j ajµj is also a measure on (X,F).

50

5.3. Fatou’s Lemma and continuity

• Proposition 5.8 (Monotone convergence). Let (X,F , µ) be a measure space.

If A1 ⊂ A2 ⊂ · · · is an increasing sequence in F then

limn→∞

µ(An) = µ(⋃n

An).

If A1 ⊃ A2 ⊃ · · · is an decreasing sequence in F then under the condition that there

exists n such that µ(An) <∞ we have that

limn→∞

µ(An) = µ(⋂n

An).

Proof. For the increasing sequence case define Bn = An\An−1. Then, (Bn)n are pairwise

disjoint, so

limn→∞

µ(An) = limn→∞

µ(n⊎j=1

Bj) = limn→∞

n∑j=1

µ(Bj)

=∑n

µ(Bn) = µ(⊎n

Bn) = µ(⋃n

An).

For the decreasing sequence case, since µ(Ak) <∞ we define Bn = Ak\An (so Bn = ∅if n ≤ k). Note that µ(Bn) + µ(An) = µ(Ak) for n ≥ k and since µ(An) ≤ µ(Ak) < ∞we have that µ(Bn) = µ(Ak)−µ(An). Note that (Bn)n≥k is an increasing sequence such

that⋃n≥k Bn = Ak \

⋂n≥k An. Since

⋂n≥k An ⊂ Ak we have

µ(Ak) = µ(⋂n≥k

An) + µ(⋃n≥k

Bn) = µ(⋂n

An) + limn→∞

µ(Bn)

= µ(⋂n

An) + limn→∞

(µ(Ak)− µ(An)).

Since µ(Ak) <∞ we may subtract it from both sides to get the proposition. ut

Let (An)n be a sequence of subsets of X. Define

lim supAn =⋂n

⋃k≥n

Ak and lim inf An =⋃n

⋂k≥n

Ak.

51

lim supAn is the set of all x ∈ X that appear in infinitely many of the subsets An.

Similarly, lim inf An is the set of all x ∈ X that appear in all but finitely many of the

subsets An.

We say that the sequence (An)n converges if lim inf An = lim supAn, and denote this

common set by limAn = lim supAn = lim inf An in this case.

� Exercise 5.11. Show that lim inf An ⊆ lim supAn.

� Exercise 5.12. Show that if (An)n is an increasing sequence then limAn =⋃nAn

and if (An)n is a decreasing sequence then limAn =⋂nAn.

• Lemma 5.9 (Fatou’s Lemma). Let (X,F , µ) be a measure space.

If (An)n is a sequence in F then

µ(lim inf An) ≤ lim infn→∞

µ(An).

If in addition µ(⋃nAn) <∞ then

µ(lim supAn) ≥ lim supn→∞

µ(An).

Proof. For every n we have that⋂k≥nAk ⊂ An, so

µ(⋂k≥n

Ak) ≤ infk≥n

µ(Ak).

Note that Bn :=⋂k≥nAk for an increasing sequence so

lim infn→∞

µ(An) = limn→∞

infk≥n

µ(Ak) ≥ limn→∞

µ(Bn) = µ(⋃n

Bn) = µ(lim inf An).

For A :=⋃nAn

A \ lim supAn = A \⋂n

⋃k≥n

Ak =⋃n

⋂k≥n

(A \Ak) = lim inf(A \An).

52

Thus, if we have µ(A) < ∞ then µ(A \ lim supAn) = µ(A) − µ(lim supAn) and µ(A \An) = µ(A)− µ(An), so

µ(A)−µ(lim supAn) = µ(A\lim supAn) ≤ lim infn→∞

(µ(A)−µ(An)) = µ(A)−lim supn→∞

µ(An),

which completes the proof by subtracting µ(A) from both sides. ut

� Exercise 5.13. Show that if limAn exists and µ(⋃nAn) <∞ then

µ(limAn) = limn→∞

µ(An).



53

Measure Theory

Ariel Yadin

Lecture 6: Outer measures

6.1. Outer measures

• Definition 6.1. Let X be any set. An outer measure on X is a function µ∗ : 2X →[0,∞] such that

• µ∗(∅) = 0;

• µ∗(A) ≤ µ∗(B) for all A ⊂ B;

• µ∗(⋃nAn) ≤∑n µ∗(An).

� Exercise 6.1. Show that Lebesgue outer measure m∗ is an outer measure (on

Rd).

Analogously to the way we defined Lebesgue outer measure, we have:

� Exercise 6.2. Let E ⊂ 2X such that ∅ ∈ E , and let ρ : E → [0,∞] such that

ρ(∅) = 0. Define

µ∗(A) := inf

{∑n

ρ(En) : A ⊂⋃n

En , ∀ n , En ∈ E},

where inf ∅ =∞. Show that µ∗ is an outer measure.


Since ∅ ⊂ ⋃n ∅ and ∅ ∈ E we have that µ∗(∅) = 0.

54

If A ⊂ B then any sequence (En)n participating in the infimum for µ∗(B) also par-

ticipates in the infimum for A. So µ∗(A) ≤ µ∗(B).

For a sequence (An)n: If µ∗(An) = ∞ for some n then there is nothing to prove. So

assume that µ∗(An) <∞ for all n.

Fix ε > 0 and for every n let (En,k)k ⊂ E be such that∑

k ρ(En,k) ≤ µ∗(An) + ε2−n

and An ⊂⋃k En,k. Then,

⋃nAn ⊂

⋃n,k En,k and

µ∗(⋃n

An) ≤∑n,k

ρ(En,k) ≤∑n

µ∗(An) + ε.

Taking ε→ 0 completes the proof. :) X

Compare this to the case that E is the collection of boxes in Rd.

6.2. Measurability

Recall the Caratheodory’s criterion for Lebesgue measurability: A ⊂ Rd is Lebesgue

measurable if and only if for every elementary set E we have m(E) = m∗(E ∩ A) +

m∗(E \A). This motivates the following definition.

• Definition 6.2 (Measurable sets). Let µ∗ be an outer measure on X. A subset A ⊂ Xis called µ∗-measurable (or simply measurable) if for every subset E ⊂ X we have

µ∗(E) = µ∗(E ∩A) + µ∗(E \A).

� Exercise 6.3. Show that A is µ∗-measurable if and only if for every E ⊂ X such

that µ∗(E) <∞ we have

µ∗(E ∩A) + µ∗(E ∩Ac) ≤ µ∗(E).

� Exercise 6.4. Let m be Lebesgue measure in Rd. We have seen that m∗ is an

outer measure. Show that A is Lebesgue measurable if and only if it is m∗-measurable.

55


If A is m∗-measurable, then we have already seen that A is Lebesgue measurable, since

it is enough to require that m(E) = m∗(E ∩ A) +m∗(E ∩ Ac) for every elementary set

E.

Assume that A is Lebesgue measurable. Then |B| = m∗(B ∩ A) + m∗(B ∩ Ac) for

every box B.

Let E ⊂ X be any subset such that m∗(E) < ∞. Fix ε > 0. Let (Bn)n be pairwise

disjoint boxes such that E ⊂ ⊎nBn and∑

n |Bn| ≤ m∗(E) + ε. E ∩ A ⊂ ⊎n(Bn ∩ A)

and E ∩Ac ⊂ ⊎n(Bn ∩Ac), so

m∗(E ∩A) +m∗(E ∩Ac) ≤∑n

m(Bn ∩A) +m(Bn ∩Ac) =∑n

|Bn| ≤ m∗(E) + ε.

Taking ε→ 0 shows that A is m∗-measurable. :) X

••• Theorem 6.3 (Charatheodory’s Theorem). Let µ∗ be an outer measure on X and

let F be the collection of all µ∗-measurable subsets. Then, F is a σ-algebra, and µ∗

restricted to F is a complete measure on (X,F).

Proof. First, ∅ ∈ F since µ∗(E) = µ∗(E ∩X) + µ∗(E ∩ ∅) for all E ⊂ X.

Also, F is closed under complements, because µ∗(E) = µ∗(E ∩ A) + µ∗(E ∩ Ac) is a

symmetric equation in A,Ac.

Now, if A,B ∈ F then A ∪ B = (A ∩ B) ] (A ∩ Bc) ] (Ac ∩ B), so for any E with

µ∗(E) <∞,

µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac)

= µ∗(E ∩A ∩B) + µ∗(E ∩A ∩Bc) + µ∗(E ∩Ac ∩B) + µ∗(E ∩Ac ∩Bc)

≥ µ∗(E ∩ (A ∪B)) + µ∗(E ∩ (A ∪B)c).

Thus A ∪B ∈ F as well.

This shows that F is closed under finite unions.

56

Note that if A,B ∈ F , A ∩B = ∅ then

µ∗(A ]B) = µ∗((A ]B) ∩A) + µ∗((A ]B) ∩Ac) = µ∗(A) + µ∗(B).

So µ∗ is finitely additive on F .

Now, if (An)n are pairwise disjoint sets in F then for all n,

µ∗(E ∩n⊎j=1

Aj) = µ∗(E ∩n⊎j=1

Aj ∩An) + µ∗(E ∩n⊎j=1

Aj ∩Acn)

= µ∗(E ∩An) + µ∗(E ∩n−1⊎j=1

Aj) = · · · =n∑j=1

µ∗(E ∩Aj).

Thus, for any n,

µ∗(E) = µ∗(E ∩n⊎j=1

An) + µ∗(E \n⊎j=1

An) ≥n∑j=1

µ∗(E ∩Aj) + µ∗(E \⊎n

An).

Taking n→∞ we have that for A =⊎nAn,

µ∗(E) ≥∑n

µ∗(E ∩An) + µ∗(E ∩Ac) ≥ µ∗(E ∩A) + µ∗(E ∩Ac).

This implies that A ∈ F and that the above inequalities are all equalities, so

µ∗(E ∩A) =∑n

µ∗(E ∩An).

Thus, taking E = A we get that µ∗ is countable additive on F and that F is closed

under countable disjoint unions.

So we are only left with showing that F is closed under countable unions.

Now if (An)n is any sequence in F (not necessarily disjoint), then set Bn = An \⋃n−1j=1 Aj . So (Bn)n are pairwise disjoint. Because F is closed under complements and

finite unions we have that Bn ∈ F for all n. Thus,⋃nAn =

⊎nBn ∈ F , which proves

that F is closed under countable unions.

Finally, to show that (X,F , µ∗∣∣F ) is indeed complete: If A admits µ∗(A) = 0 and

then by monotonicity, for any E with µ∗(E) <∞,

µ∗(E ∩A) + µ∗(E ∩Ac) ≤ µ∗(E ∩Ac) ≤ µ∗(E).

So A ∈ F . ut

57

� Exercise 6.5. [Folland p.32 ex.17] Let µ∗ be an outer measure on X and let (An)n

be a sequence of pairwise disjoint µ∗-measurable subsets. Show that for any A ⊂ X,

µ∗(A ∩ (⊎n

An)) =∑n

µ∗(A ∩An).

6.3. Pre-measures

It is also interesting to what degree this resulting measure is unique.

• Definition 6.4. An algebra is a collection of subsets A ⊂ 2X that is closed under

finite unions and complements.

� Exercise 6.6. Let B be a box in Rd. Show that

A = {E ⊂ B : E is elementary }

is an algebra over B.

� Exercise 6.7. Show that an algebra A is closed under finite intersections, set

differences, symmetric differences.

Show that ∅ ∈ A.

� Exercise 6.8. Let A be an algebra such that for all sequences (An)n ⊂ A that are

pairwise disjoint we have that⊎nAn ∈ A.

Show that A is a σ-algebra.

58

• Definition 6.5 (Pre-measure). Given an algebra A on X, a pre-measure is a func-

tion µ0 : A → [0,∞] such that µ0(∅) = 0 and for any sequence (An)n ⊂ A that are

pairwise disjoint and admit⊎nAn ∈ A we have that µ0(

⊎nAn) =

∑n µ0(An).

• Lemma 6.6. Let µ0 be a pre-measure on A over X. Define

µ∗(A) := inf

{∑n

µ0(En) : A ⊂⋃n

En , ∀ n , En ∈ A}.

Then µ∗ is an outer measure such that µ∗∣∣A = µ0 and every set in A is µ∗-measurable.

Proof. We have already seen that such a definition gives rise to an outer measure.

For any A ∈ A let (En)n be a sequence in A such that A ⊂ ⋃nEn. Let Bn =

A ∩ (En \⋃n−1j=1 Ej). Then, Bn ∈ A for all n and En \ Bn ∈ A for all n. Thus,

µ0(En) = µ0(Bn) + µ0(En \Bn) ≥ µ0(Bn) for all n, and also because⊎nBn = A ∈ A,

µ0(A) =∑n

µ0(Bn) ≤∑n

µ0(En).

Taking infimum over all such sequences (En)n ⊂ A we obtain that µ0(A) ≤ µ∗(A).

Since µ∗(A) ≤ µ0(A) by definition, we get that µ0(A) = µ∗(A). This completes the first

assertion.

For the second assertion, let A ∈ A and E ⊂ X such that µ∗(E) <∞. Fix ε > 0 and

let (En)n ⊂ A be such that E ⊂ ⋃nEn and∑

n µ0(En) ≤ µ∗(E) + ε. For all n we have

µ0(En ∩A) + µ0(En ∩Ac) = µ0(En). Thus,

µ∗(E) ≥ −ε+∑n

µ0(En ∩A) +∑n

µ0(En ∩Ac) = −ε+∑n

µ∗(En ∩A) +∑n

µ∗(En ∩Ac)

≥ −ε+ µ∗(E ∩A) + µ∗(E ∩Ac).

So A is µ∗-measurable. ut

••• Theorem 6.7 (Charatheodory’s Extension Theorem). Let µ0 be a pre-measure on

an algebra A over X. Let F = σ(A). Then, there exists a measure µ on (X,F) such

that µ∣∣A = µ0.

59

Also, if ν is a measure on (X,F) such that ν∣∣A = µ0 then ν(A) ≤ µ(A) for all A ∈ F ,

and ν(A) = µ(A) for all A ∈ F with µ(A) <∞.

Moreover, if X =⋃nAn for An ∈ A, µ0(An) <∞ (i.e. µ0 is σ-finite), then ν = µ.

Proof. Lemma 6.6 tells us that µ0 extends to a measure µ on the µ∗-measurable sets,

which form a σ-algebra that contains A, and thus contains F = σ(A). Here

µ∗(A) := inf

{∑n

µ0(En) : A ⊂⋃n

En , ∀ n , En ∈ A}.

Now, if A ∈ F , and (En)n ⊂ A is such that A ⊂ ⋃nEn, then

ν(A) ≤∑n

ν(En) =∑n

µ0(En).

Taking infimum over all such sequences we get that ν(A) ≤ µ∗(A) = µ(A) for all A ∈ F .

If µ(A) = µ∗(A) < ∞ then for any ε > 0 we may choose (En)n such that A ⊂⋃nEn and

∑n µ0(En) ≤ µ∗(A) + ε. Thus, µ(

⋃nEn) ≤ µ(A) + ε which implies that

µ(⋃nEn \A) ≤ ε. Hence,

µ(A) ≤ µ(⋃n

En) = limn→∞

µ0(n⋃j=1

Ej) = ν(⋃n

En)

≤ ν(A) + ν(⋃n

En \A) ≤ ν(A) + µ(⋃n

En \A) ≤ ν(A) + ε.

Thus, taking ε→ 0 implies that ν(A) = µ(A).

Finally, if µ0 is σ-finite, then we may write X =⊎Xn for Xn ∈ A pairwise disjoint

and µ0(Xn) < ∞. So for any A ∈ F we have A =⊎n(A ∩Xn). Since µ(A ∩Xn) < ∞

for all n we get that

µ(A) =∑n

µ(A ∩Xn) =∑n

ν(A ∩Xn) = ν(A).

ut

� Exercise 6.9. [Folland, p.32, ex.18] Let A ⊂ 2X be an algebra. Let Aσ be all count-

able unions of sets in A; let Aσδ be all countable intersections of sets in Aσ. Let µ0 be

a pre-measure on A and let µ∗ be the induced outer measure.

60

(1) Show that for any E ⊂ X and any ε > 0 there exists A ∈ Aσ such that

µ∗(A) ≤ µ∗(E) + ε and E ⊂ A.

(2) Show that if µ∗(E) < ∞, then E is µ∗-measurable if and only if there exists

B ∈ Aσδ with E ⊂ B and µ∗(B \ E) = 0.

(3) Show that if µ0 is σ-finite then the assumption in (b) above that µ∗(E) <∞ is

superfluous.


• Fix ε > 0. Let (An)n be a sequence in A such that E ⊂ A :=⋃nAn and∑

n µ0(An) ≤ µ∗(E) + ε. Then since µ∗(A) ≤ ∑n µ∗(An) =

∑n µ0(An) we are

done.

• (⇐) Let (An,k)n,k be sets in A such that B =⋂k

⋃nAn,k ∈ Aσδ and such that

E ⊂ B and µ∗(B \ E) = 0.

Let F ⊂ X. Then, since An,k are all µ∗-measurable, then also B is µ∗-

measurable. So,

µ∗(F ∩ E) + µ∗(F ∩ Ec) ≤ µ∗(F ∩B) + µ∗(F ∩ Ec ∩Bc) + µ∗(F ∩ Ec ∩B)

≤ µ∗(F ∩B) + µ∗(F ∩Bc) + µ∗(B \ E) = µ∗(F ).

This holds for any F so E is µ∗-measurable. (*) Note: this direction did not

require µ∗(E) <∞.

(⇒) If E is µ∗-measurable, then for any n > 0 there exists An ∈ Aσ such that

E ⊂ An and µ∗(An) ≤ µ∗(E) + 1n . Thus, for B =

⋂nAn ∈ Aσδ, since E ⊂ B,

µ∗(B) = µ∗(B ∩ E) + µ∗(B ∩ Ec) = µ∗(E) + µ∗(B \ E).

Also, since B ⊂ An for all n,

µ∗(B) ≤ infnµ∗(An) ≤ µ∗(E).

Thus, µ∗(B \ E) ≤ 0.

61

• The only direction we need to show is (⇒) because the other direction holds

without the finite measure assumption.

If X =⊎n Fn such that µ0(Fn) <∞ for all n, then for any A ⊂ X, since Fn

are all µ∗-measurable,

µ∗(A) =∑n

µ∗(A ∩ Fn).

(This was shown in a previous exercise.)

For every k we have that E ∩ Fk is measurable and has finite outer measure.

So let Bk ⊂ Fk be a Aσδ set such that E ∩ Fk ⊂ Bk and µ∗(Bk \ (E ∩ Fk)) = 0.

Let B =⊎k Bk. We have that E ⊂ B and B ∩ Fk ∩ Ec = Bk \ (E ∩ Fk). So,

µ∗(B ∩ Ec) =∑k

µ∗(B ∩ Fk ∩ Ec) =∑k

µ∗(Bk \ (E ∩ Fk)) = 0.

Thus we are left with showing that B ∈ Aσδ.For every k write

Bk =⋂n

⋃j

Akn,j ,

where Akn,j ∈ A and Akn,j ⊂ Fk for all n, j, k. Writing Cn :=⊎k

⋃j A

kn,j we have

that Cn ∈ Aσ. We have that Bk =⋂n(Cn ∩ Fk) = Fk ∩

⋂nCn, so

B =⊎k

Bk =⊎k

Fk ∩⋂n

Cn =⋂n

Cn ∈ Aσδ.

:) X

� Exercise 6.10. [Folland p.32, ex.19] Let µ∗ be an outer measure induced from a

pre-measure µ0 on X such that µ0(X) < ∞. Define µ∗(A) = µ0(X) − µ∗(Ac). Show

that µ∗(A) ≤ µ∗(A) and that A is measurable if and only if µ∗(A) = µ∗(A).

� Exercise 6.11. [Folland p.32, ex.23] Let A be the collection of finite unions of sets

of the form (a, b] ∩Q, for all −∞ ≤ a < b ≤ ∞. Show that A is an algebra over Q.

62

What is the σ-algebra generated by A?

Define µ0(∅) = 0 and µ0(A) =∞ for ∅ 6= A ∈ A. Show that µ0 is a pre-measure, and

that µ0 can be extended to a measure on 2Q in more than one way.

� Exercise 6.12. [Folland p.33] Let (X,F , µ) be a finite measure space. Let µ∗ be

the outer measure induced by µ. Let Y 6∈ F be such that µ∗(Y ) = µ∗(X).

Show that if A,B ∈ F are such that A ∩ Y = B ∩ Y , then µ(A) = µ(B).

Define

FY := {A ∩ Y : A ∈ F} .

Show that FY is a σ-algebra on Y .

Define ν on (Y,FY ) by ν(A ∩ Y ) = µ(A), which is well defined by the above. Prove

that ν is a measure.



63

Measure Theory

Ariel Yadin

Lecture 7: Lebesgue-Stieltjes Theory

7.1. Lebesgue-Stieltjes measure

Define A1 = A(R) to be the collection of finite disjoint unions of intervals of the form

(a, b], (a,∞), (−∞, b], ∅ for −∞ < a < b <∞.

� Exercise 7.1. Show that A1 is an algebra on R and that σ(A1) = B1 (the Borel

σ-algebra).

• Lemma 7.1. Let F : R → R be a non-decreasing function that is right-continuous

(continue a droite). Define µ0(∅) = 0 and for −∞ ≤ a < b < ∞, µ0((a, b]) = F (b) −F (a). Also, define µ((a,∞)) = F (∞) − F (a). (Here by F (−∞) and F (∞) we mean

the limits at −∞ and ∞ respectively.) For any A ∈ A1 that is a finite disjoint union of

such intervals, A = I1 ] · · · ] In, define µ0(A) = µ0(I1) + · · ·+ µ0(In).

Then, µ0 is a well defined pre-measure on A1.

Proof. Let I denote the collection of all intervals of the form (a, b], (a,∞), (−∞, b], ∅ for

−∞ < a < b <∞.

Well defined. To show that µ0 is well defined: Note that if I, J ∈ I, then also

I ∩J ∈ I.Now, if (a, b] = (a1, b1]] · · ·] (an, bn], then without loss of generality a = a1 <

b1 = a2 < b2 = · · · = an < bn = b. So,

F (b)− F (a) =n∑j=1

F (bj)− F (aj).

Also, a similar identity holds when writing (−∞, b] or (a,∞) as finite disjoint unions.

Thus, if A = I1 ] · · · ] In = J1 ] · · · ] Jk then for all i we have that Ii = (Ii ∩ J1)] · · · ]

64

(Ii ∩ Jk), so

µ0(Ii) =k∑j=1

µ0(Ii ∩ Jj),

and we get thatn∑i=1

µ0(Ii) =∑i,j

µ0(Ii ∩ Jj) =k∑j=1

µ0(Jj).

This shows that µ0 is well defined.

Finite additivity. This also proves that µ0 is finitely additive. Indeed, if A1, . . . , An

are pairwise disjoint sets in A1, then we may write Aj =⊎njk=1 Ij,k for Ij,k ∈ I, and then

µ0(

n⊎j=1

Aj) = µ0(

n⊎j=1

nj⊎k=1

Ij,k) =

n∑j=1

nj∑k=1

µ0(Ij,k) =

n∑j=1

µ0(Aj).

We will use the finite additivity of µ0 in what follows.

Pre-measure. To show that it is a pre-measure we need to consider (An)n ⊂ A1

that are pairwise disjoint and⊎nAn = A ∈ A1. Since every An is a finite disjoint

union of intervals, we may assume without loss of generality that An is an interval for

all n. Since A = I1 ] · · · ] Ik is also a finite disjoint union of intervals, we have that⊎n(An ∩ Ij) = Ij which is a countable disjoint union of intervals whose union is an

interval. So using the additivity of µ0, we may assume without loss of generality that

A is an interval. Also, we may re-order (An)n so that An = (an, bn] and an < bn = an+1

for all n. Set Bn =⊎nj=1Aj .

A \Bn ∈ A1, so µ0(A) = µ0(Bn) + µ0(A \Bn). Thus,

µ0(A) = µ0(Bn) + µ0(A \Bn) ≥n∑j=1

µ0(Aj).

Taking n → ∞ we get that µ0(A) ≥ ∑n µ0(An). So we only need to prove the other

inequality.

Start with the case A = (a, b] with −∞ < a < b <∞. Fix ε > 0. F was assumed to

be right-continuous. So we may choose δ > 0 and δn > 0 such that for all n,

F (a+ δ) < F (a) + ε and F (bn + δn) < F (bn) + ε2−n.

We have the open cover [a+ δ, b] ⊂ ⋃n(an, bn+ δn), and since [a+ δ, b] is compact, there

is a finite sub-cover. That is, there exists m > 0 such that

65

• [a+ δ, b] ⊂ ⋃mk=1(ak, bk + δk),

• and for all k, we have bk + δk ∈ (ak+1, bk+1 + δk+1).

In this case we get that

µ0(A) = F (b)− F (a) ≤ F (b)− F (a+ δ) + ε ≤ F (bm + δm)− F (a1) + ε

= F (bm + δm)− F (am) +m−1∑j=1

F (aj+1)− F (aj) + ε

≤ F (bm + δm)− F (am) +

m−1∑j=1

F (bj + δj)− F (aj) + ε

=m−1∑j=1

F (bj + δj)− F (bj) + F (bj)− F (aj) + ε

≤m∑j=1

µ0(Aj) +

m∑j=1

ε2−j + ε

≤∑n

µ0(An) +∑n

ε2−n + ε ≤∑n

µ0(An) + 2ε.

Taking ε→ 0 completes the case where A = (a, b].

If A = (−∞, b] then for any large enough r > 0 we have that (−r, b] =⊎n(An∩(−r, b]).

Since µ0(An ∩ (−r, b]) ≤ µ0(An ∩ (−r, b]) + µ0(An ∩ (−r, b]c) = µ0(An),

F (b)− F (−r) = µ0((−r, b]) =∑n

µ0(An ∩ (−r, b]) =∑n

µ0(An),

and taking r →∞ we get that

µ0(A) = F (b)− F (−∞) ≤∑n

µ0(An).

If A = (a,∞), then similarly, for any large enough r > 0 we have that (a, r] =⊎n(An ∩ (a, r]), so

F (r)− F (a) =∑n

µ0(An ∩ (a, r]) ≤∑n

µ0(An),

and taking r →∞ completes the proof. ut

??? THEOREM 7.2 (Lebesgue-Stieltjes measure). If F : R→ R is a right-continuous

non-decreasing function, then there exists a unique measure µF on (R,B) such that for

66

all a < b,

µF ((a, b]) = F (b)− F (a).

In fact, µF = µG if and only if F −G is constant.

Moreover, one may recover F from µF : If µ is a measure on (R,B) that is finite on

all bounded sets in B, if we define

F (x) =

µ((0, x]) x > 0

0 x = 0

−µ((x, 0]) x < 0

then F is non-decreasing and right-continuous, and µF = µ.

Proof. Lemma 7.1 and Charathodory’s Extension Theorem tell us that we may extend

the pre-measure µ0((a, b]) = F (b) − F (a) uniquely to (R,B). Also, if µF = µG then

F (b)− F (a) = G(b)−G(a) for all a < b which implies that F = G+ F (0)−G(0).

If µ is a measure on (R,B) that is finite on all bounded sets in B, one may check

that by the definition of F in the theorem, µ((a, b]) = F (b) − F (a) for all a < b. So F

is non-decreasing (since µ is non-negative). F is right continuous from the continuity

properties of measures: For any x and εn ↘ 0

F (x+ εn)− F (x) = µ((x, x+ εn])→ µ(⋂n

(x, x+ εn]) = µ(∅) = 0.

ut

X The measure µF is called the Lebesgue-Stieltjes measure associated to F .

� Exercise 7.2. Show that µF satisfies the following properties.

• µF ({a}) = F (a)− F (a−).

• µF ((a, b)) = F (b−)− F (a).

• µF ([a, b]) = F (b)− F (a−).

• µF ([a, b)) = F (b−)− F (a−).

67

� Exercise 7.3. Show that for F (x) = x the associated measure µF is Lebesgue

measure on R.


Note that µF is a measure with µ((0, 1]) = 1 and µ((a, b] + x) = F (b+ x)− F (a+ x) =

b− a = µ((a, b]). So µ is the extension of Jordan measure on elementary sets, and must

be Lebesgue measure. :) X

� Exercise 7.4. Let A ⊂ R be Lebesgue measurable. Suppose that m(A) > 0. Show

that for any 0 < ε < 1 there exists an open interval I such that m(A∩I) ≥ (1−ε)m(I).


First suppose the 0 < m(A) <∞. Define

β = sup{m(A∩I)m(I) : I is an open interval

}.

Of course β ≤ 1. We need to show that β = 1.

Fix ε > 0 and let (In)n be disjoint intervals (not necessarily open or closed) such that

A ⊂ ⊎n In and∑

n λ(In) ≤ m(A) + ε.

Recall that Lebesgue measure of all points is 0. So m(A ∩ I) is the same for open,

closed, half-open-closed intervals, and similarly for m(I). Thus, m(A ∩ In) ≤ βm(In)

for all n, and we obtain

m(A) =∑n

m(A ∩ In) ≤ β∑n

m(In) ≤ β(m(A) + ε).

Taking ε→ 0 we get that β = 1.

68

In the case that m(A) = ∞, let N be large enough so that m(A ∩ (−N,N)) > 0.

Since m(A ∩ (−N,N)) ≤ 2N <∞, by the previous part, for any 0 < ε < 1 there exists

an open interval I such that

m(A ∩ I) ≥ m(A ∩ (−N,N) ∩ I) > (1− ε)m(I).

:) X

7.2. Regularity

We have already seen outer and inner regularity for Lebesgue measure. This is a more

general fact, for Lebesgue-Stieltjes measures.

• Proposition 7.3 (Outer and inner regularity). For any Lebesgue measurable set A

we have

µF (A) = infA⊂U open

µF (U) = supA⊃K compact

µF (K).

Proof. If µ(A) = ∞ then any U ⊃ A also admits µ(U) = ∞ so outer regularity is

simple in this case. So assume that µ(A) < ∞. By the definition of the outer measure

from the pre-measure µ0, for every ε > 0 there exists intervals ((an, bn])n such that

A ⊂ ⋃n(an, bn] and∑

n µ((an, bn]) ≤ µ(A) + ε. (Note that even if A is unbounded

this is possible with only bounded intervals.) For every n there exists δn > 0 such that

F (bn + δn) < F (bn) + ε2−n. Thus, taking U =⋃n(an, bn + δn), which is an open set

containing A, we get that

µ(U) ≤∑n

µ((an, bn + δn]) =∑n

F (bn + δn)− F (an)

≤∑n

F (bn)− F (an) + ε =∑n

µ((an, bn]) + ε ≤ µ(A) + 2ε.

Taking infimum of the left hand side, and ε→ 0 we have outer regularity.

For inner regularity, first assume that A is bounded. Consider B = A \A. Fix ε > 0.

There exists an open set U such that B ⊂ U and µ(U) < µ(B) + ε. Let K = A \ U .

This is a closed bounded set, so it is compact. Also, K ⊂ A and

µ(B) + µ(A) = µ(A) ≤ µ(K) + µ(U) ≤ µ(K) + µ(B) + ε,

69

so µ(K) ≥ µ(A) − ε. Taking supremum of such K ⊂ A and ε → 0 we have inner

regularity for bounded A.

If A is unbounded: Let ε > 0. For every r > 0 there exists a compact Kr ⊂ A∩(−r, r)such that µ(Kr) ≥ µ(A ∩ (−r, r)) − ε. Note that µ(A ∩ (−r, r)) → µ(A) by monotone

convergence. So

supA⊃K compact

µ(K) ≥ limr→∞

µ(A ∩ (−r, r))− ε = µ(A)− ε.

Taking ε→ 0 completes the proof. ut

7.3. Non-measureable sets

In this section we will show that the Lebesgue measurable sets L are not all of 2R;

that is, there exist non-Lebesgue-measurable sets (and specifically non-Borel sets).

� Exercise 7.5. Let Q = Q∩ [0, 1). Define an equivalence relation on R by r ∼ r′ if

r − r′ ∈ Q. Let R be a set of representatives for the equivalence classes of this relation

(chosen by the axiom of choice!).

For every q ∈ Q let Rq = {r + q (mod 1) : r ∈ R ∩ [0, 1)}.Show that if R ∈ L then also Rq ∈ L for all q ∈ Q and λ(Rq) = λ(R) where λ is

Lebesgue measure.

Show that this leads to a contradiction, so R 6∈ L.

7.4. Cantor set

The Cantor set C ⊂ [0, 1] is defined as follows. Start with C0 = [0, 1]. Given Cn

define Cn+1 by “removing the middle third of each interval composing Cn”; formally:

Cn+1 = 13Cn

⋃(23 + 1

3Cn).

Then take

C =⋂n

Cn.

70

� Exercise 7.6. Show that C is Lebesgue measurable and has Lebesgue measure 0.

(*) Show that C has cardinality of [0, 1].


Note that for every n the set Cn is the union of two measurable sets so is measurable

by induction. Thus, C is measurable as an intersection of measurable sets.

Let us calculate the measure of Cn. Since Cn ⊂ [0, 1], we have that 13Cn ⊂ [0, 1

3 ] and

23 + 1

3Cn ⊂ [23 , 1]. This implies that these sets are disjoint, and

λ(Cn+1) = λ(13Cn) + λ(2

3 + 13Cn) = 1

3λ(Cn) + 13λ(Cn) = 2

3λ(Cn).

Thus, λ(Cn) = (2/3)n and

λ(C) ≤ infnλ(Cn) = 0.

:) X



71

Measure Theory

Ariel Yadin

Lecture 8: Functions of measure spaces

8.1. Products

Let (X,F), (Y,G) be two measurable spaces. What is the natural measurable struc-

ture on X ×Y ? Naturally, any set A×B for A ∈ F , B ∈ G should be measurable in the

product.

� Exercise 8.1. Let (X,F), (Y,G) be two measurable spaces. Let πX : X × Y → X

and πY : X × Y → Y be the natural projections. Show that

σ(A×B : A ∈ F , B ∈ G) = σ(π−1X (A), π−1

Y (B) : A ∈ F , B ∈ G).


We only need to show that every generator in one σ-algebra is in the other.

If A ∈ F , B ∈ G then

π−1X (A) = A× Y and π−1

Y (B) = X × Y,

which are both in the left σ-algebra.

On the other hand, A×B = π−1X (A) ∩ π−1

Y (B) which is in the right σ-algebra. :) X

� Exercise 8.2. Generalize the previous exercise: If ((Xn,Fn))n is a sequence of

measurable spaces, then

σ(A1 × · · · ×An × · · · : An ∈ Fn ∀ n) = σ(π−1n (An) : An ∈ Fn),

72

where πn : X1 × · · · ×Xn × · · · → Xn is the natural projection.

X We use the notations

⊗n

Xn = X1 × · · · ×Xn × · · · and

n⊗j=1

X1 × · · · ×Xn

and⊗n

Fn = σ(⊗nAn : An ∈ Fn ∀ n) and

n⊗j=1

Fj = σ(⊗nj=1Aj : Aj ∈ Fj ∀ j).

These are called product σ-algebras. Also,⊗n

(Xn,Fn) = (⊗nXn,⊗nFn)

and similarly for finite products. These latter spaces are called product (measure)

spaces.

� Exercise 8.3. Show that B(Rd) = ⊗dj=1B(R).

� Exercise 8.4. Show that if Fn = σ(En) for some sets En, then⊗n

Fn = σ(π−1n (En) : En ∈ En).


Let

σ1 =⊗n

Fn = σ(π−1n (An) : An ∈ Fn)

by a previous exercise. Let σ2 = σ(π−1n (En) : En ∈ En).

73

It is immediate that σ2 ⊂ σ1.

So we only need to show that σ1 ⊂ σ2. For this it suffices to prove that for any n and

any An ∈ Fn we have that π−1n (An) ∈ σ2.

Fix n and set Gn ={A ⊂ Xn : π−1

n (A) ∈ σ2

}. Then: π−1

n (∅) = ∅ ∈ σ2 so ∅ ∈ Gn.

Also, if A ∈ Gn then π−1n (Ac) = π−1

n (A)c ∈ σ2, so Ac ∈ Gn as well; i.e. , G is closed under

complements. If (Ak)k ⊂ Gn is a sequence in Gn then π−1n (⋃k Ak) =

⋃k π−1n (Ak) ∈ σ2;

so Gn is closed under countable unions as well. In conclusion Gn is a σ-algebra. By

definition of σ2, we have that En ⊂ Gn, so Fn = σ(En) ⊂ Gn.

Since this holds for all n, we conclude that for any n and any An ∈ Fn we have

π−1n (An) ∈ σ2. :) X

Of course now one would like to construct a product measure space from two (or more)

measure spaces. For this, it is convenient to first go through the theory of measurable

functions.

8.2. Measurable functions

• Definition 8.1. Let (X,F), (Y,G) be measurable spaces. A function f : X → Y is

called (F ,G)-measurable, or just measurable, if f−1(A) ∈ F for all A ∈ G.

That is, f pulls back measurable sets to measurable sets.

Sometimes we denote this by f : (X,F)→ (Y,G).

• Proposition 8.2. Let (X,F), (Y,G) be measurable spaces. Let f : X → Y . Suppose

that G = σ(K).

f is measurable if and only if for every K ∈ K, f−1(K) ∈ F .

Proof. One direction is trivial.

For the other direction, verify that{A ⊂ Y : f−1(A) ∈ F

}is a σ-algebra containing

K, and so it contains G as well. ut

� Exercise 8.5. Show that if X,Y are topological spaces and f : X → Y is a con-

tinuous function, then f is Borel measurable (i.e. (B(X),B(Y ))-measurable).

74

X For a function f : Rd → C we say that f is Borel if f is (Bd,B(C))-measurable

and f is Lebesgue if f is (L,B(C))-measurable. Note that Borel implies Lebesgue, but

the converse is not necessarily true.

� Exercise 8.6. Show that for measurable spaces (X,F), (Y,G), (Z,H) and measur-

able functions f : (X,F)→ (Y,G) and g : (Y,G)→ (Z,H), we have that g ◦ f : X → Z

is (F ,H)-measurable.

� Exercise 8.7. Let (X,F), (Y,G), (Z,H) be measurable spaces, and πX : X ×Y →X,πY : X × Y → Y the natural projections. Show that f : Z → X × Y is measurable if

and only if πX ◦ f and πY ◦ f are measurable.

� Exercise 8.8. Let (X,F), (Y,G), (Z,H) be measurable spaces. Let f : X →Y, g : X → Z be measurable functions. Show that F : X → Y × Z defined by

F (x) = (f(x), g(x)) is a measurable function.


If πY : Y×Z → Y, πZ : Y×Z → Z are the natural projections, then πY ◦F = f, πZ◦F = g

which are measurable. :) X

� Exercise 8.9. Show that f : (X,F) → C is measurable if and only if Ref, Imf

are measurable.

75

� Exercise 8.10. Show that if f, g : (X,F)→ C are measurable then f + g and fg

are also.


Let F (x) = (f(x), g(x)) and ψ(x, y) = x + y, φ(x, y) = xy. Then ψ, φ are continuous,

and F is measurable. So f + g = ψ ◦ F and fg = φ ◦ F are measurable. :) X

� Exercise 8.11. [Folland p.48] Show that for f : (X,F) → R, if for all q ∈ Q we

have that f−1(−∞, q] ∈ F then f is measurable.

� Exercise 8.12. Show that any monotone function f : R→ R is Borel.

X We will also like to have functions with values ±∞. One must be careful when

adding ∞−∞ and when multiplying 0 · ∞, but as a topological space we can consider

[−∞,∞]. The Borel σ-algebra is defined in the same way as for R: it is the σ-algebra

generated by intervals.

• Proposition 8.3. Let (fn)n be a sequence of measurable functions fn : (X,F) →[−∞,∞]. Then,

supnfn inf

nfn lim inf

nfn lim sup

nfn

are all measurable.

If the limit limn→∞ fn(x) = f(x) exists for all x then f is measurable.

76

Proof. For any a ≤ ∞,

(supnfn)−1(−∞, a] =

⋂n

f−1n (−∞, a] ∈ F .

Since (−∞, a] generate B([−∞,∞]) we get that supn fn is measurable.

Similarly,

(infnfn)−1(−∞, a] =

⋃n

f−1n (−∞, a] ∈ F .

A bit more cumbersome to check, but not too difficult is

(lim inf fn)−1(−∞, a] =⋂n

⋃k≥n

f−1k (−∞, a],

and

(lim sup fn)−1(−∞, a] =⋃n

⋂k≥n

f−1k (−∞, a].

Which gives the measurability of lim inf fn and lim sup fn.

Finally, if f = lim fn exists then f = lim sup fn = lim inf fn and so is measurable. ut

� Exercise 8.13. Show that for f : (X,F)→ [−∞,∞],

f+ = max {0, f} and f− = max {0,−f}

are measurable if f is.

� Exercise 8.14. Show that for measurable f : (X,F)→ C the functions |f | : X → R

and sgnf : X → R defined by |f |(x) = |f(x)| and

sgnf(x) =

f(x)|f(x)| f(x) 6= 0,

0 f(x) = 0,

are measurable.

Show that arg f is measurable.

77

� Exercise 8.15. [Folland p.48] Let (X,F) be a measurable space. Let f : X →[−∞,∞] and let Y = f−1(R). Show that f is measurable if and only if f−1(∞) ∈F , f−1(−∞) ∈ F and f

∣∣Y

is measurable as a function f∣∣Y

: (Y,FY ) → R, where

FY = {A ∩ Y : A ∈ F}.

� Exercise 8.16. [Folland p.48] Let (X,F) be a measurable space. Let f, g : X →[−∞,∞].

Define fg(x) = f(x)g(x) where ±∞· 0 = 0 · (±∞) = 0 and x · (±∞) = ±∞· x = ±∞for x > 0 and x · (±∞) = ±∞ · x = ∓∞ if x < 0.

Show that if f, g are measurable then fg is measurable.

For a ∈ [−∞,∞] define

(f + g)a(x) =

f(x) + g(x) if f(x) 6= −g(x) = ±∞ or |f(x)| ∧ |g(x)| <∞,

a if f(x) = −g(x) = ±∞

(that is, replacing the problematic ∞−∞ with a).

Show that if f, g are measurable then (f + g)a is measurable.

� Exercise 8.17. Show that if (fn)n is a sequence of measurable functions on (X,F)

then A = {x : limn→∞ fn(x) exists } is a measurable set. Show that for any a ∈ R,

f(x) =

limn→∞ fn(x) if the limit exists,

a otherwise

is measurable.

78

8.3. Simple functions

• Definition 8.4 (Simple functions). Let (X,F) be a measurable space.

The indicator function of a set A ⊂ X, denoted 1A is the function

1A : X → R 1A(x) :=

1 x ∈ A

0 x 6∈ A.

A simple function is a measurable function f : X → [0,∞) such that f(X) is a

finite set.

� Exercise 8.18. Let (X,F) be a measurable space, and let A ⊂ X. Show that 1A

is measurable if and only if A ∈ F .

� Exercise 8.19. Let (X,F) be a measurable space. Show that for any simple func-

tion f we can write f =∑n

j=1 aj1Aj where 0 < a1 < a2 < · · · < an and Aj ∈ F for all

j. Show that in fact, we have the standard representation

f =∑

0<a∈f(X)

a1f−1(a).


Since f(X) is finite, and since f ≥ 0, we may write f(X) = {a1 < a2 < · · · < an} for

a1 > 0 (if 0 6∈ f(X)) or f(X) = {0 = a0 < a1 < · · · < an} (in the case that 0 ∈ f(X)).

Since f is measurable, Aj := f−1(aj) ∈ F for all j. Finally, since X = f−1(f(X)) =⊎j f−1(aj) =

⊎j Aj , for every x ∈ X there exists a unique j(x) such that x ∈ Aj(x).

79

X

[0,∞)

2−n

2−n(k + 1)

2−nk

An,k

ϕn−1

ϕn

Figure 3. Approximating functions with simple functions

Note that f(x) = aj(x) = aj(x)1Aj(x)(x). Since (Aj)j are disjoint, if 1Aj (x) 6= 0 then

1Ai(x) = 0 for all i 6= j. Altogether this gives

f(x) =∑

a∈f(X)

a1{f(a)=x} =∑

0<a∈f(X)

a1f−1(a)(x).

:) X

The following is a very useful building block in the theory of measurable functions.

• Proposition 8.5. Let (X,F) be a measurable space. Let f : X → C be a Borel

function.

• If f ≥ 0 then there exists a monotone sequence of simple functions 0 ≤ ϕ1 ≤ϕ2 ≤ · · · ≤ ϕn ≤ · · · such that ϕn ↗ f . In fact, the convergence is uniform on

any set for which f is bounded.

• If f is real-valued then f = f+−f− and f+, f− are non-negative Borel functions.

• In general, f = Ref + iImf where Ref, Imf are real valued and Borel. So

f = f1 − f2 + i(f3 − f4) where all fj are non-negative Borel functions.

80

Proof. We only prove the first bullet. The other bullets have actually been proven in

previous exercises.

If f ≥ 0 then for every n define ϕn := min {n, 2−nb2nfc}.Note that ϕn(X) ⊂ {2−nk : k ∈ N , k ≤ 2nn} which is finite, so ϕn is simple (why

is it measurable?). Also, one may verify that for the sets An,k = f−1[2−nk, 2−n(k + 1))

and En = f−1[n+ 2−n,∞) we have that the standard representation of ϕn is

ϕn =

2nn∑k=0

2−nk1An,k + n1En .

To show that ϕn ≤ ϕn+1, note that 2brc ≤ b2rc for any r ≥ 0, so

ϕn = n ∧ 2−nb2nfc ≤ (n+ 1) ∧ 2−nb2nfc ≤ (n+ 1) ∧ 2−n 12b2 · 2nfc = ϕn+1.

Now, if A is a set on which f is bounded, say supx∈A f(x) = M < ∞, then for all

n > M we have that supx∈A 2−nb2nf(x)c < n so ϕn(x) = 2−nb2nf(x)c for all x ∈ A.

Thus, supx∈A |ϕn(x)− f(x)| ≤ 2−n for all n > M . So ϕn → f uniformly on A. ut

� Exercise 8.20. Let f : (X,F) → [0,∞] be a measurable function. Show that

there exists a sequence of simple functions 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ ϕn ≤ · · · such that

ϕn ↗ f .


Write f = f1A +∞1Ac where 0 · ∞ =∞ · 0 = 0 and Ac = f−1(∞).

f1A is measurable and non-negative, so there is a monotone sequence a simple func-

tions ψn ↗ f1A. Define ϕn = ψn1A + n1Ac . :) X



81

Measure Theory

Ariel Yadin

Lecture 9: Integration: positive functions

9.1. Integration of simple functions

Let (X,F , µ) be a measure space. Let ϕ be a simple function, with standard repre-

sentation ϕ =∑n

j=1 aj1Aj . Define the integral of ϕ with respect to µ as∫ϕdµ :=

n∑j=1

ajµ(Aj).

For A ∈ F define ∫Aϕdµ :=

∫ϕ1Adµ.

• Proposition 9.1. Let ϕ,ψ be simple functions on a measure space (X,F , µ). Then:

• For all r > 0 we have∫rϕdµ = r

∫ϕdµ.

• If ϕ =∑n

j=1 aj1Aj for pairwise disjoint (Aj)nj=1, then

∫ϕdµ =

∑nj=1 ajµ(Aj).

•∫

(ϕ+ ψ)dµ =∫ϕdµ+

∫ψdµ.

• If ϕ ≤ ψ then∫ϕdµ ≤

∫ψdµ.

Proof. For r > 0, rϕ is a simple function such that rϕ(X) = {ra : a ∈ ϕ(X)}. So,

rϕ =∑

0<a∈ϕ(X)

ra1ϕ−1(a)

is the standard representation of rϕ. Thus by definition,∫rϕdµ =

∑0<a∈ϕ(X)

raµ(ϕ−1(a)) = r

∫ϕdµ.

Also, X =⊎a∈ϕ(X) ϕ

−1(a). Thus, if ϕ =∑n

j=1 aj1Aj then Aj =⊎a∈ϕ(X)Aj∩ϕ−1(a).

Note that for x ∈ Aj ∩ ϕ−1(a) we have that a = ϕ(x) = aj . Thus, ajµ(Aj ∩ ϕ−1(a)) =

aµ(Aj ∩ ϕ−1(a)). Also, for B =(]nj=1Aj

)cwe have that aµ(B ∩ ϕ−1(a)) = 0 because

82

either B ∩ ϕ−1(a) = ∅ or for x ∈ B ∩ ϕ−1(a), a = ϕ(x) =∑n

j=1 aj1Aj (x) = 0. Since

X =⊎nj=1Aj

⊎B,

n∑j=1

ajµ(Aj) =n∑j=1

∑a∈ϕ(X)

aµ(Aj ∩ ϕ−1(a)) =∑

a∈ϕ(X)

aµ(ϕ−1(a)) =

∫ϕdµ.

Now, (ϕ + ψ)(X) = {a+ b : a ∈ ϕ(X), b ∈ ψ(X)} so it is indeed a simple function.

Note that for Ea,b = ϕ−1(a) ∩ ψ−1(b), these sets are pairwise disjoint, and

ϕ+ ψ =∑

a∈ϕ(X),b∈ψ(X)

(a+ b)1Ea,b .

Similarly,

ϕ =∑

a∈ϕ(X),b∈ψ(X)

a1Ea,b and ψ =∑

a∈ϕ(X),b∈ψ(X)

b1Ea,b .

So by the above,∫(ϕ+ ψ)dµ =

∑a∈ϕ(X),b∈ψ(X)

(a+ b)µ(Ea,b) =

∫ϕdµ+

∫ψdµ.

Finally, if ϕ ≤ ψ we claim that for all a ∈ ϕ(X), b ∈ ψ(X) we have that aµ(Ea,b) ≤bµ(Ea,b), because either Ea,b = ∅ or if x ∈ Ea,b then a = ϕ(x) ≤ ψ(x) = b. Thus,∫

ϕdµ =∑

a∈ϕ(X),b∈ψ(X)

aµ(Ea,b) ≤∑

a∈ϕ(X),b∈ψ(X)

bµ(Ea,b) =

∫ψdµ.

ut

• Proposition 9.2. Let ϕ be a simple functions on a measure space (X,F , µ). The

map A 7→∫A ϕdµ is a measure on (X,F).

Proof. The function ϕ1∅ is just the zero function, which has 0 integral by definition.

For any E ∈ F ,

ϕ1E =∑

a∈ϕ(X)

a1ϕ−1(a)∩E .

The sets (ϕ−1(a) ∩ E)a∈ϕ(X) are pairwise disjoint so∫Eϕdµ =

∑a∈ϕ(X)

aµ(ϕ−1(a) ∩ E).

83

Now if A =⊎nAn then∫

Aϕdµ =

∑a∈ϕ(X)

aµ(ϕ−1(a) ∩A) =∑

a∈ϕ(X)

a∑n

µ(ϕ−1(a) ∩An) =∑n

∫An

ϕdµ.

ut

9.2. Integration of positive functions

We use L+(X,F) to denote the set of measurable functions f : (X,F)→ [0,∞].

• Definition 9.3. Let (X,F , µ) be a measure space. For f ∈ L+(X,F) define∫fdµ := sup

{∫ϕdµ : 0 ≤ ϕ ≤ f , ϕ is simple

}.

� Exercise 9.1. Show that if f ≤ g for f, g ∈ L+(X,F) then∫fdµ ≤

∫gdµ.

Show that for any c > 0,∫cfdµ = c

∫fdµ.


This simple functions participating in the supremum for f also participate in the supre-

mum for g.

If c > 0 then for any simple function ϕ, ϕ ≤ f if and only if cϕ ≤ cf . Taking

supremums over∫cϕdµ = c

∫ϕdµ yields the result. :) X

It is usually difficult to compute supremums over such big sets (all simple functions

dominated by some f ∈ L+). The next fundamental result will make life much simpler

when wishing to compute integrals of positive functions.

??? THEOREM 9.4 (Monotone Convergence). Let (X,F , µ) be a measure space. If

(fn)n is a sequence in L+(X,F) such that 0 ≤ fn ≤ fn+1 for all n (i.e. a monotone

sequence), then for the limit f(x) = limn→∞ fn(x) (which always exists as f = supn fn),∫fdµ = lim

n→∞fndµ.

84

Proof. Because fn ≤ f for all n we have that∫fn ≤

∫f and so lim

∫fn ≤

∫f . We are

left with proving the other inequality.

Let 0 < ε < 1 and let ϕ be a simple function such that 0 ≤ ϕ ≤ f . Define

An = {x ∈ X : fn(x) ≥ (1− ε)ϕ(x)} = {fn ≥ (1− ε)ϕ} .

Note that An ⊂ An+1 because fn ≤ fn+1. Thus (An)n is an increasing sequence. Since

A 7→∫A ϕdµ is a measure on (X,F), we get that∫

An

ϕdµ→∫ϕdµ

because⋃n

An = {x : ∃ n , fn(x) ≥ (1− ε)ϕ(x)} =

{x : sup

nfn(x) ≥ (1− ε)ϕ(x)

}= {x : f(x) ≥ (1− ε)ϕ(x)} = X.

Since fn ≥ fn1An ≥ (1− ε)ϕ1An , we get that∫fndµ ≥ (1− ε) ·

∫An

ϕdµ→ (1− ε) ·∫ϕdµ.

Taking supremum over ϕ and ε→ 0 we get that

limn→∞

∫fndµ ≥

∫fdµ,

which completes the proof. ut

• Proposition 9.5 (Poor man’s Fubini). If (fn)n is a sequence of functions in L+(X,F)

then ∫ ∑n

fndµ =∑n

∫fndµ.

Proof. If f, g ∈ L+(X,F), then let (ϕn)n, (ψn)n be monotone sequences of simple func-

tions such that ϕn ↗ f and ψn ↗ g. Then, ϕn + ψn ↗ f + g. So, using the Monotone

Convergence Theorem,∫(f + g)dµ = lim

n→∞

∫(ϕn + ψn)dµ =

∫fdµ+

∫gdµ.

85

Now, for a sequence (fn)n in L+(X,F), for any n,∫ n∑k=1

fkdµ =

n∑k=1

∫fkdµ.

Since (fn)n are all non-negative,∑n

k=1 fk ↗∑

n fn, so by Monotone Convergence again,∑n

∫fndµ = lim

n→∞

∫ n∑k=1

fkdµ =

∫ ∑n

fndµ.

ut

X For a measure µ we say that a measurable set A occurs almost everywhere, or a.e.,

if µ(Ac) = 0. sometimes we stress the dependence on the measure by saying µ-a.e.

X For a function f we may sometimes shorthand {f ≤ a} = {x : f(x) ≤ a}, and

similarly for other measurable sets.

• Proposition 9.6. For f ∈ L+(X,F), f = 0 a.e. (that is, µ({x : f(x) 6= 0}) = 0) if

and only if∫fdµ = 0.

Proof. If f is a simple function then∫fdµ =

∑0<a∈f(X) aµ(f−1(a)) and this is 0 if and

only if for every a ∈ f(X), aµ(f−1(a)) = 0, which is if and only if µ({x : f(x) > 0}) =

0.

Now, if f ∈ L+, then let ϕn ↗ f be an approximating monotone sequence of simple

functions. Since 0 ≤ ϕn ≤ f we have that f = 0 implies that ϕn = 0 for all n and so∫f = lim

∫ϕn = 0.

In the other direction, if since {f > 0} =⋃n

{f > 1

n

}we have that if µ(f > 0) > 0

then there exists n such that µ(f > 1n) > 0. For A =

{f > 1

n

},∫

Afdµ ≥ 1

n

∫1Adµ = 1

nµ(A) > 0.

ut

� Exercise 9.2. Show that if (fn)n is a monotone sequence in L+ such that fn ↗ f

a.e., then∫fndµ→

∫fdµ.

86


Let A = {fn ↗ f}. So gn = fn1A is a monotone sequence on L+ that increases to

gn ↗ g := f1A. Also, fn(1− 1A) = fn1Ac , f(1− 1A) = f1Ac ∈ L+ and µ(Ac) = 0, so∫fndµ =

∫gndµ and

∫fdµ =

∫gdµ.

Thus, ∫fdµ =

∫gdµ = lim

n→∞

∫gndµ = lim

n→∞

∫fndµ.

:) X

• Lemma 9.7 (Fatou’s Lemma). Let (X,F , µ) be a measure space. Let (fn)n be a

sequence on L+(X,F). Then,∫lim inf fndµ ≤ lim inf

n→∞

∫fndµ.

Proof. For every j ≥ k we have that infn≥k fn ≤ fj and so∫

infn≥k fndµ ≤ infj≥k∫fjdµ.

The sequence gk := infn≥k fn is a monotone sequence in L+ converging to gk ↗lim inf fn. By Monotone Convergence,∫

lim inf fndµ = limk→∞

∫gkdµ ≤ lim

k→∞infj≥k

∫fjdµ = lim inf

n→∞

∫fndµ.

ut

� Exercise 9.3. Assume that (fn)n is a sequence in L+ such that∫

sup fndµ <∞.

Show that

lim supn→∞

∫fndµ ≤

∫lim sup fndµ.


Define gn = (sup fn) − fn. Then (gn)n ⊂ L+. Since gn, fn are non-negative functions,

87∫fn ≤

∫gn +

∫fn =

∫sup fn < ∞, and we can subtract

∫fn from both sides to get

that∫gn =

∫sup fn −

∫fn for all n. Also, since lim inf gn, lim sup fn are non-negative

functions, and since lim inf gn + lim sup fn = sup fn, we have that∫lim sup fndµ ≤

∫lim inf gndµ+

∫lim sup fndµ =

∫sup fndµ <∞,

and subtracting∫

lim sup fndµ from both sides we have that∫lim inf gndµ =

∫sup fndµ−

∫lim sup fndµ.

However, by Fatou’s Lemma,∫lim inf gndµ ≤ lim inf

∫gndµ =

∫sup fndµ− lim sup

∫fndµ.

Since∫

sup fndµ <∞ we may subtract it from both sides. :) X

� Exercise 9.4. Let f ∈ L+(X,F) for a measure space (X,F , µ). Show that

ν(A) :=∫A fdµ is a measure on (X,F). Show that for any g ∈ L+(X,F),∫

gdν =

∫gfdµ.


ν(∅) = 0 is simple.

If A =⊎nAn then f1A =

∑n f1An and so

ν(A) =

∫Afdµ =

∑n

∫An

fdµ =∑n

ν(An)

(by poor man’s Fubini).

This shows that ν is a measure.

Now, if g is a simple function with standard representation g =∑n

j=1 aj1Aj , then∫gdν =

n∑j=1

ajν(Aj) =

n∑j=1

aj

∫Aj

fdµ =

∫gfdµ.

88

Now for general g ∈ L+, let ϕn ↗ g be a monotone sequence of simple functions

approximating g. So ϕnf ↗ gf . Monotone Convergence guaranties that∫gdν = lim

n→∞

∫ϕndν = lim

n→∞

∫ϕnfdµ =

∫gfdµ.

:) X

� Exercise 9.5. [Folland, p.52, ex.13] Suppose (fn)n is a sequence of non-negative

functions such that fn → f a.e. and∫fn →

∫f <∞.

Show that for any measurable A,∫A fn →

∫A f .

Show that this is not necessarily true if∫f =∞.


Since fn1A → f1A, by Fatou’s Lemma, for any A,∫Af ≤ lim inf

∫Afn.

Since∫f <∞,∫

Af =

∫f −

∫Acf ≥

∫f − lim inf

∫Acfn = lim sup(

∫fn −

∫Acfn) = lim sup

∫Afn.

Hence, ∫Af ≤ lim inf

∫Afn ≤ lim sup

∫Afn ≤

∫Af.

Now for a counter-example when∫f =∞. Set

fn = 1[0,n−1]n2 + 1[n−1,n).

So fn → 1[0,∞) =: f . Also,∫fn = n+ n− n−1 →∞ =

∫f.

However, for A = [0, 1], ∫Afn = n+ 1− n−1 →∞ 6= 1 =

∫Af.

:) X

89

� Exercise 9.6. Let (X,F , µ) be a measure space. Assume that if (fn)n is a sequence

in L+(X,F) such that fn+1 ≤ fn for all n. Assume that∫f1dµ <∞ and that fn ↘ f

a.e.

Show that limn→∞∫fndµ =

∫fdµ.

� Exercise 9.7. Let (X,F , µ) be a measure space and f ∈ L+(X,F) such that∫fdµ <∞.

Show that {f =∞} = f−1(∞) is a null set.

Show that {f > 0} is σ-finite (i.e. can be written as a countable union of finite-

measure sets).


If µ(f =∞) > 0 then

∞ =∞ · µ(f =∞) ≤∫{f=∞}

fdµ ≤∫fdµ <∞,

a contradiction.

Write {f > 0} = {f =∞} ] ⊎∞n=1 {n− 1 < f ≤ n}. We have already seen that

{f =∞} is a null set (and specifically has finite measure). For any n ≥ 1, if µ(n− 1 <

f ≤ n) =∞ then

∞ = (n− 1) · µ(n− 1 < f ≤ n) ≤∫{n−1<f≤n}

fdµ ≤∫fdµ <∞,

a contradiction. So µ(n− 1 < f ≤ n) <∞ for all n. :) X



90

Measure Theory

Ariel Yadin

Lecture 10: Integration: general functions

We have already seen that any measurable function can be decomposed into real and

imaginary parts, and those into positive and negative parts each. So essentially we can

split the task of integrating a functions into integrating four positive functions. The

only thing we have to be careful about is ∞−∞.

10.1. Real valued functions

• Definition 10.1. Let (X,F , µ) be a measure space. Let f : (X,F) → R be a

measurable function.

Consider the integrals I+ =∫f+dµ, I− =

∫f−dµ (f+ = 0 ∨ f, f− = 0 ∨ −f).

If at least one integral I+, I− is finite then we define∫fdµ :=

∫f+dµ−

∫f−dµ

and we say the integral∫fdµ exists. If I+ = I− = ∞ then

∫fdµ does not exist (or is

undefined).

If I+ <∞ and I− <∞ then we say that f is integrable.

As usual we define∫A fdµ =

∫f1Adµ.

� Exercise 10.1. Let (X,F , µ) be a measure space. Let f : (X,F)→ R be a mea-

surable function. Show that f is integrable if and only if∫|f |dµ <∞.

• Proposition 10.2. For a measure space (X,F , µ) the set of integrable real-valued

functions is a vector space, and the integral is a linear functional on this space.

91

Proof. If f, g are integrable then∫|f + αg|dµ ≤

∫(|f |+ |α||g|)dµ ≤

∫|f |dµ+ |α|

∫|g|dµ <∞.

So f + αg is also integrable.

Now, write h = f + g. So h+ − h− = f+ − f− + g+ − g−, which implies that

h+ + f− + g− = h− + f+ + g+. Since these are all positive functions,∫h+ +

∫f− +

∫g− =

∫h− +

∫f+ +

∫g+,

which after rearranging gives∫h =

∫h+ −

∫h− =

∫f+ −

∫f− +

∫g+ −

∫g− =

∫f +

∫g.

If α > 0 then (αf)+ = αf+, (αf)− = αf− and (−αf)+ = αf−, (−αf)− = αf+. So∫αf =

∫αf+ −

∫αf− = α

∫f and

∫(−αf) =

∫αf− −

∫αf+ = −α

∫f.

ut

10.2. Complex valued functions

• Definition 10.3. Let (X,F , µ) be a measure space and let f : (X,F) → C be a

measurable function.

We say that f is integrable if∫|f |dµ <∞ (this makes sense because |f | ∈ L+(X,F)).

We define ∫fdµ =

∫Refdµ+ i

∫Imfdµ.

As usual we define∫A fdµ =

∫f1Adµ.

The set of all complex valued integrable functions is denoted L1(X,F , µ) = L1(X) =

L1(µ).

� Exercise 10.2. f ∈ L1(X,F , µ) if and only if Ref, Imf are both integrable.

Thus, the integral of a complex function is well defined. It is a (complex) linear

functional the vector space L1(X,F , µ).

92


The first assertion follows since |f | ≤ |Ref |+ |Imf | ≤ 2|f |. :) X

� Exercise 10.3. Show that for f ∈ L1,

∣∣ ∫ fdµ∣∣ ≤ ∫ |f |dµ.


If∫f = 0 this is immediate.

Otherwise, if f is real valued then |f | = f+ + f−, so

∣∣ ∫ f∣∣ =

∣∣ ∫ f+ −∫f−∣∣ ≤ ∫ f+ +

∫f− =

∫|f |.

If f is complex valued and∫f 6= 0, then set α =

|∫f |∫f∈ C. So α

∫f =

∫αf ∈ R and

∣∣ ∫ f∣∣ =

∫αf = Re

∫αf =

∫Re(αf) ≤

∫|Re(αf)| ≤

∫|α||f | =

∫|f |.

:) X

� Exercise 10.4. Show that for f ∈ L1 the set {f 6= 0} is σ-finite.

• Proposition 10.4. For f, g ∈ L1(X,F , µ) we have that the following are equivalent.

• f = g a.e.

•∫|f − g|dµ = 0.

• For every A ∈ F ,∫A fdµ =

∫A gdµ.

Proof. If f = g a.e. then |f − g| = 0 a.e. so∫|f − g|dµ = 0.

93

If∫|f − g|dµ = 0 then for any A ∈ F ,∣∣ ∫

Afdµ−

∫Agdµ

∣∣ ≤ ∫ 1A|f − g|dµ ≤∫|f − g|dµ = 0.

If∫A fdµ =

∫A gdµ for all A ∈ F then also

∫A Re(f − g)dµ =

∫A Im(f − g)dµ = 0 for

all A ∈ F . So without loss of generality we may assume that f, g are real valued. Set

A = {f > g} and B = {f < g}. We then have∫|f − g|dµ =

∫A

(f − g)dµ+

∫B

(g − f)dµ = 0.

So |f − g| = 0 a.e., and f = g a.e. ut

10.3. Convergence

� Exercise 10.5. Let (fn)n be a sequence in L1(X,F , µ) such that fn → f uniformly

(i.e. for every ε > 0 there exists n0 such that for all n > n0, supx |fn(x)− f(x)| < ε).

Show that if µ(X) <∞ then∫fndµ→

∫fdµ.

Show that if µ(X) =∞ this is not necessarily true.

??? THEOREM 10.5 (Dominated Convergence Theorem). Let (fn)n be a sequence in

L1(X,F , µ) such that fn → f a.e. and there exists g ∈ L1(X,F , µ) such that |fn| ≤ g

for all n.

Then f ∈ L1(X,F , µ) and∫fndµ→

∫fdµ.

Proof. Let E = {fn → f}. So µ(Ec) = 0 and∫h =

∫E h for all h ∈ L1. Since

|f |1E ≤ | lim fn|1E ≤ g, we have that f ∈ L1.

Suppose first that (fn)n, f are real valued. Then, g − fn ≥ 0 and g + fn ≥ 0 so by

Fatou’s Lemma,∫g +

∫f =

∫lim inf(g + fn) ≤ lim inf

∫(g + fn) =

∫g + lim inf

∫fn,∫

g −∫f =

∫lim inf(g − fn) ≤ lim inf

∫(g − fn) =

∫g − lim sup

∫fn.

94

Subtracting the finite quantity∫g from both sides of both equations we get that

lim sup

∫fn ≤

∫f ≤ lim inf

∫fn,

which implies these are equal. ut

• Proposition 10.6. If (fn)n is a sequence in L1(X,F , µ) such that∑

n

∫|fn|dµ <∞,

then f =∑

n fn is a.e. well defined (the sum converges a.e.), and∫fdµ =

∑n

∫fndµ.

Proof. Poor man’s Fubini tells us that for g =∑

n |fn| we have∫g =

∑n

∫|fn| < ∞,

so g ∈ L1. So N = {g =∞} is a null set, and off this set the sum f =∑

n fn converges

absolutely.

Also, |∑nj=1 fj | ≤ g for every n. This sequence converges a.e. to f so by Dominated

Convergence,

∑n

∫fndµ = lim

n→∞

n∑j=1

∫fjdµ = lim

n→∞

∫ n∑j=1

fjdµ =

∫fdµ.

ut

� Exercise 10.6. [Folland p.59] Let (fn)n, (gn)n, f, g all be functions in L1(X,F , µ)

such that fn → f a.e., gn → g a.e., |fn| ≤ gn a.e., and∫gndµ→

∫gdµ.

Show that∫fndµ→

∫fdµ.

� Exercise 10.7. Let (X,F , µ) be a measure space and let (X, F , µ) be its com-

pletion. Show that if f : X → C is F-measurable, then there exists g : X → C

such that g is F-measurable and there exists a set N ∈ F such that µ(N) = 0 and

{x : g(x) 6= f(x)} ⊂ N , so g1Nc = f1Nc (and so also g = f µ-a.e.)

95


Recall that

F = {A ∪ F : A ∈ F , F ⊂ N for some N ∈ F with µ(N) = 0} .

Suppose that f = 1A∪F for A ∪ F ∈ F , A ∈ F , F ⊂ N ∈ F and µ(N) = 0. Then if we

set g = 1A∪N then g is F measurable, and if g(x) 6= f(x) then x ∈ (A ∪N) \ (A ∪ F ) =

N \ F ⊂ N .

Now suppose f =∑

k ak1Ak∪Fk is a simple function for X =⊎k(Ak ∪ Fk), Ak ∈ F ,

Fk ⊂ Nk ∈ F with µ(Nk) = 0. Then, defining g =∑

k ak1Ak∪Nk we have that if

g(x) 6= f(x) then x ∈ ⋃kNk which has µ-measure 0.

If f ≥ 0 and F-measurable, then let fn ↗ f be a sequence of F-measurable simple

functions converging monotony to f . Then for every n there is a set Nn ∈ F and a

simple F-measurable function gn such that µ(Nn) = 0 and gn1Ncn

= fn1Ncn. Taking

N =⋃nNn we have that µ(N) = 0 and gn1Nc = fn1Nc ↗ f1Nc . Thus, g := f1Nc is

F-measurable.

Finally if f is any F-measurable function, write f = f1 − f2 + i(f3 − f4) for non-

negative F-measurable functions f1, . . . , f4. For any j, let gj be a non-negative F-

measurable function and Nj ∈ F be such that gj1Ncj

= fj1Ncj

and µ(Nj) = 0. Then

for g := g1 − g2 + i(g3 − g4) and N :=⋃j Nj we have that g is F-measurable and

g1Nc = f1Nc with µ(N) = 0. :) X

� Exercise 10.8. [Folland, p.59, ex.24] Let (X,F , µ) be a measure space of finite

measure. Let (X, F , µ) be its completion. Let f : X → R be a bounded non-negative

function.

Show that f is F-measurable if and only if there exist sequences (gn)n, (hn)n of F-

measurable simple functions such that gn ≤ f ≤ hn and∫

(hn − gn)dµ < 1n for all

n.

Show that in this case,∫fdµ = limn→∞

∫gndµ = limn→∞

∫hndµ.

96


If f is F-measurable, then we can find a set N ∈ F such that µ(N) = 0 and f1Nc is

F-measurable. Let M > 0 be such that 0 ≤ f ≤M (as f is bounded). Taking

gn := 1µ(X)nbµ(X)nf1Ncc and hn := 1

µ(X)ndµ(X)nf1Nce+M1N ,

we have that

gn ≤ f1Nc ≤ f = f1Nc + f1N ≤ hn.

Also, since gn(X), hn(X) ⊂{

1µ(X)nk : 0 ≤ k ≤ µ(X)nM, k ∈ N

}we have that gn, hn

are simple functions. Finally, for x 6∈ N , (hn − gn)(x) ≤ 1µ(X)n so

∫(hn − gn)dµ =∫

Nc(hn − gn)dµ ≤ n−1, as required.

For the other direction, assume that gn, hn are as in the statement of the exercise.

Let An ={hn − gn > n−1/2

}.

Note that if a ≤ f ≤ b then for every n, either a − n−1/2 ≤ gn ≤ hn ≤ b + n−1/2 or

a ≤ f ≤ b and hn − gn > n−1/2. Also, if for every n, a− n−1/2 ≤ gn ≤ hn ≤ b+ n−1/2,

then

a = limn

(a− n−1/2) ≤ f ≤ limn

(b+ n−1/2) = b.

Thus,

{a ≤ f ≤ b} =⋂n

({a− n−1/2 ≤ gn ≤ hn ≤ b+ n−1/2

}⋃({a ≤ f ≤ b} ∩An)

).

Since{a− n−1/2 ≤ gn ≤ hn ≤ b+ n−1/2

}∈ F we only need to show that the set

⋂nAn

has µ-measure 0. So it suffices to show that infn µ(An) = 0. To this end,

µ(An) ≤∫An

√n · (hn − gn)dµ ≤ √n · n−1 → 0.

Thus, {a ≤ f ≤ b} is the union of a F-measurable set and a subset of a µ-null set,

and so f is F-measurable.

Now if gn ≤ f ≤ hn as in the statement of the exercise, since f is F-measurable, we

may find a set N ∈ F such that µ(N) = 0 and f1Nc is F-measurable. By replacing

gn with supk≤n gk we may assume without loss of generality that gn1Nc ↗ f1Nc , so by

monotone convergence,∫gndµ→

∫f1Ncdµ and

∫gndµ→

∫fdµ. Also,∫

gndµ ≤∫hndµ ≤

∫gndµ+ n−1.

97

So∫hndµ→

∫f1Ncdµ as well.

Hence we are left with showing that∫f1Ncdµ =

∫fdµ, and for this it suffices to prove

that for all n,∫gndµ =

∫gndµ. Because gn are simple functions and F-measurable for

all n, by linearity it suffices that for any set A ∈ F , we have∫

1Adµ = µ(A) = µ(A) =∫1Adµ. :) X

10.4. Riemann vs. Lebesgue integration

Recall the Riemann integral of a bounded function f : [a, b]→ R:

For every partition P = (xk)nk=0, a = x0 < x1 < · · · < xn = b, define

SP f =n∑k=1

supx∈[xk−1,xk]

f(x)·(xk−xk−1) and sP f =n∑k=1

infx∈[xk−1,xk]

f(x)·(xk−xk−1).

Define Ibaf = inf SP f and ibaf = sup sP f where the infimum and supremum are over all

partitions P of [a, b].

If Ibaf = ibaf we say that f is Riemann integrable and define∫ ba fdx to be the common

value.

Given a partition of [a, b], P = (xk)nk=0, define the functions

ΨP f =

n∑k=1

supx∈[xk−1,xk]

f(x) · 1[xk−1,xk] and ψP f =n∑k=1

infx∈[xk−1,xk]

f(x) · 1[xk−1,xk].

Note that

SP f =

∫ΨP fdλ and sP f =

∫ψP fdλ.

If f is Riemann integrable, we may choose a sequence of partitions (Pk)k such that

limSPkf = lim sPkf =∫ ba fdx. We may also choose the partitions as refinements of the

previous ones in the sequence. In this case the functions Ψk := ΨPkf form a decreasing

sequence and ψk := ψPkf form an increasing sequence. So there exist limiting functions

Ψ = lim Ψk and ψ = limψk. Note that ψ ≤ f ≤ Ψ.

Since f is bounded, and since λ([a, b]) <∞, we get that (ψk)k, (Ψk)k are dominated

sequences, and so by Dominated Convergence,∫ψdλ =

∫Ψdλ =

∫ b

afdx.

So∫|Ψ−ψ|dλ =

∫(Ψ−ψ)dλ = 0 and so Ψ = ψ a.e., which implies that Ψ = f = ψ a.e.

98

Thus, f1A = Ψ1A for some A such that λ(Ac) = 0. Since λ is a complete measure on

(R,L), we have that f is Lebesgue measurable. Also,∫[a,b]

fdλ =

∫[a,b]

Ψdλ =

∫ b

afdx.

Conclusion: If f is Riemann integrable on [a, b] then it is also Lebesgue measurable

on [a, b] and∫

[a,b] fdλ =∫ ba fdx.

� Exercise 10.9. Given a bounded function f : [a, b]→ R define

Θ(x) = limn→∞

sup|y−x|≤n−1

f(y) and θ(x) = limn→∞

inf|y−x|≤n−1

f(y).

Show that f is continuous at x if and only if Θ(x) = θ(x).

Show that a.e. ∫[a,b]

Θdλ = Ibaf and

∫[a,b]

θdλ = ibaf.

Conclude that f is Reimann integrable if and only if

λ({x : f is not continuous at x}) = 0.


Write

Θn(x) = sup|y−x|≤n−1

f(y) and θn(x) = inf|y−x|≤n−1

f(y).

Assume that Θ(x) = θ(x). Let ε > 0. Choose n large enough so that |Θn(x)−Θ(x)| <ε/2 and |θn(x)− θ(x)| < ε/2. So if |y − x| < 1

n we have that

θ(x)− ε/2 < θn(x) ≤ f(y) ≤ Θn(x) < Θ(x) + ε/2.

Since Θ(x)− θ(x), for all |y − x| < 1n we have that |f(x)− f(y)| < ε.

That is, for all ε > 0 there exists n large enough so that for all |y − x| < 1n we have

|f(y)− f(x)| < ε. So f is continuous at x.

99

For the other direction, if f is continuous at x, then for any ε > 0 there exists large

enough n such that we have for all |y − x| < 1n that |f(y) − f(x)| < ε. Thus, for

all ε > 0 there exists n0 such that if n > n0 we have |Θn(x) − θn(x)| < ε. Since

Θn(x)→ Θ(x), θn(x)→ θ(x), we get that Θ(x) = θ(x).

Now, consider a partition P = (xk)mk=0. Let |P | = maxk |xk − xk−1|. If 2|P | < 1

n , for

any x ∈ [xk−1, xk], we have that

supy∈[xk−1,xk]

f(y) ≤ Θn(x) and infy∈[xk−1,xk]

f(y) ≥ θn(x).

Thus,

Ibaf ≤ SP f =m∑k=1

supy∈[xk−1,xk]

f(y)

∫[xk−1,xk]

dλ ≤∫

[a,b]Θndλ,

and similarly,

ibaf ≥ sP f =m∑k=1

infy∈[xk−1,xk]

f(y)

∫[xk−1,xk]

dλ ≥∫

[a,b]θndλ.

On the other hand, if n is very large and x ∈ [xk−1 + 1n , xk − 1

n ] then

Θn(x) ≤ supy∈[xk−1,xk]

f(y) and θn(x) ≥ infy∈[xk−1,xk]

f(y).

So since f is bounded, say by |f | ≤M , we have that∫[xk−1,xk]

Θndλ ≤ supy∈[xk−1,xk]

f(y) · (xk − xk−1) +M · 2n ,

and ∫[a,b]

Θndλ ≤ SP f +m ·M · 2n ,

where m is the number of intervals in the partition. Similarly,∫[a,b]

θndλ ≥ sP f −m ·M · 2n .

By Dominated Convergence, taking n→∞ we have that for any partition P ,∫[a,b]

Θdλ ≤ SP f and

∫[a, b]θdλ ≥ sP f.

Taking infimum and supremum respectively, we have that∫[a,b]

Θdλ = Ibaf and

∫[a,b]

θdλ = ibaf.

100

Finally, note that f is Riemann integrable if and only if∫

[a,b](Θ− θ)dλ = 0 which is

if and only if Θ = θ a.e. on [a, b], which is if and only if f is continuous λ-a.e. on [a, b].

:) X

� Exercise 10.10. Let f ∈ L1(R,B, µ) and F (x) :=∫

(−∞,x] fdµ.

• Show that if f ≥ 0 then F is right continuous and F is continuous at x if and

only if f(x)µ({x}) = 0.

• Show that if µ({x}) = 0 then F is continuous at x.


For any x and ε, note that

F (x+ ε)− F (x) =

∫(x,x+ε]

fdµ.

First assume that f is non-negative. Then, ν(A) :=∫A fdµ is a measure. If xn ↘ x

then the sets An = (x, xn] are decreasing and

F (xn)− F (x) = ν(An)→ ν(⋂n

An) = ν(∅) = 0.

So F is right continuous.

If xn ↗ x then

F (x)− F (xn)→ ν({x}) =

∫{x}

fdµ = f(x)µ({x}).

because⋂n(xn, x] = {x}. So F is left continuous as well if and only if f(x)µ({x}) = 0.

This resolves the case where f is non-negative.

For general f ∈ L1(µ), write f = f1 − f2 + i(f3 − f4) for non-negative fj . Then

F = F1 − F2 + i(F3 − F4) where Fj(x) =∫

(−∞,x] fjdµ.

If µ({x}) = 0 then for every j the function Fj is continuous at x, and thus so is F . :)

X

101



102

Measure Theory

Ariel Yadin

Lecture 11: Product measures

Recall that if (X,F), (Y,G) are measurable spaces then F ⊗ G = σ(A × B : A ∈F , B ∈ G) is a σ-algebra on X × Y .

An example is B1 ⊗ B1 = B2.

When we constructed Lebesgue measure in Rd “hands on” we took boxes as basic

building blocks; these are just products of intervals which where the basic building

blocks in dimension 1. We now give a general construction of a product measure on

(X,F)⊗ (Y,G).

• Definition 11.1. For measurable spaces (X,F), (Y,G), a box is a set of the form

A×B where A ∈ F , B ∈ G.

� Exercise 11.1. Show that for A,C ⊂ X and B,D ⊂ Y we have

(A×B)∩(C×D) = (A∩C)×(B∩D) and (A×B)c = (Ac×Y )](A×Bc) = (X×Bc)](Ac×B).

� Exercise 11.2. Show that the family of all finite disjoint unions of boxes (from

(X,F), (Y,G) forms an algebra.

Show that the σ-algebra generated by this algebra is F ⊗ G.

� Exercise 11.3. Show that 1A×B(x, y) = 1A(x)1B(y).

103

• Proposition 11.2. Let (X,F , µ), (Y,G, ν) be measure spaces.

If A × B =⊎n(An × Bn) where A, (An)n are F-measurable and B, (Bn)n are G-

measurable, then

µ(A)ν(B) =∑n

µ(An)ν(Bn).

Proof. This is actually just poor man’s Fubini.

We have by the assumption A×B =⊎n(An ×Bn) that

1A(x)1B(y) =∑n

1An(x)1Bn(y).

Integrate this function over X for fixed y to get that for all y,

µ(A)1B(y) =∑n

µ(An)1Bn(y),

where we have used poor man’s Fubini once. Integrate this again over Y to get by poor

man’s Fubini again,

µ(A)ν(B) =∑n

µ(An)ν(Bn).

ut

� Exercise 11.4. Let (X,F , µ), (Y,G, ν) be measure spaces.

Show that if ⊎n

(An ×Bn) =⊎n

(Cn ×Dn)

where An, Cn ∈ F , Bn, Dn ∈ G, then∑n

µ(An)ν(Bn) =∑n

µ(Cn)ν(Dn).

104


We have

(Aj ×Bj) ∩ (Ck ×Dk) = (Aj ∩ Ck)× (Bj ×Dk),

so

Aj ×Bj =⊎n

(Aj ∩ Cn)× (Bj ×Dn),

which implies that

µ(Aj)ν(Bj) =∑n

µ(Aj ∩ Cn)ν(Bj ∩Dn).

Similarly,

µ(Ck)ν(Dk) =∑n

µ(An ∩ Ck)ν(Bn ∩Dk).

Thus, ∑n

µ(An)ν(Bn) =∑n,k

µ(An ∩ Ck)ν(Bn ∩Dk) =∑k

µ(Ck)ν(Dk).

:) X

� Exercise 11.5. Let E =⊎nj=1(Aj × Bj) ∈ A. Define ρ(E) =

∑nj=1 µ(Aj)ν(Bj).

Then ρ is well defined, and defines a pre-measure on A.

By Charatheodory’s Extension Theorem we may extend ρ above to a measure on all

of σ(A) = F ⊗ G. Also note that if X and Y are σ-finite with respect to µ and ν, then

X × Y is σ-finite with respect to ρ. So in this case the extension, denoted µ⊗ ν, is the

unique measure on (X,F)⊗ (Y,G) satisfying

µ⊗ ν(A×B) = µ(A)ν(B) ∀ A ∈ F , B ∈ G.

� Exercise 11.6. Show that if X and Y are σ-finite with respect to µ and ν, then

X × Y is σ-finite with respect to ρ.

105


If X =⊎nAn, Y =

⊎nBn where µ(An) <∞, ν(Bn) <∞, then X×Y =

⊎n,k(An×Bk)

and ρ(An ×Bk) = µ(An)ν(Bk) <∞. :) X

� Exercise 11.7. In this exercise the product measure is generalized to further prod-

ucts. Let (Xj ,Fj , µj) be σ-finite measure spaces, j = 1, 2, . . ..

Show that (µ1 ⊗ µ2)⊗ µ3 = µ1 ⊗ (µ2 ⊗ µ3).

Show that for all n there exists a unique measure⊗n

j=1 µj on⊗n

j=1(Xj ,Fj) satisfying

that for all Aj ∈ Fj , j = 1, 2, . . . , n,

n⊗j=1

µj(A1 × · · · ×An) =

n∏j=1

µj(Aj).

11.1. Sections

• Definition 11.3. Let E ⊂ X × Y . Define the sections of E for every x ∈ X, y ∈ Yas

Ex = {y ∈ Y : (x, y) ∈ E} = πY (E ∩ ({x} × Y ))

Ey = {x ∈ X : (x, y) ∈ E} = πX(E ∩ (X × {y})),

where πX : X × Y → X,πY : X × Y → Y are the natural projections.

For a function f : X × Y → Z define the sections of f for every x ∈ X, y ∈ Y as the

functions fx : Y → Z, fy : X → Z by the formula fx(y) = fy(x) = f(x, y).

� Exercise 11.8. Show that (1E)x = 1Ex and (1E)y = 1Ey .

Show that (fx)−1(A) = (f−1(A))x and (fy)−1(A) = (f−1(A))y.

106

Y

X

EEx

x

y

Ey

Figure 4. Depiction of Ex, Ey.


We have for all x, y,

(1E)x(y) = 1{(x,y)∈E} = 1{y∈Ex} = 1Ex(y),

and similarly for (1E)y = 1Ey .

Also,

y ∈ (fx)−1(A) ⇐⇒ fx(y) ∈ A ⇐⇒ f(x, y) ∈ A ⇐⇒ (x, y) ∈ f−1(A) ⇐⇒ y ∈ (f−1(A))x,

and similarly for (fy)−1(A) = (f−1(A))y. :) X

� Exercise 11.9. Show that (⋃nEn)x =

⋃n(En)x and (

⋃nEn)y =

⋃n(Ey) .

Show that (Ec)x = (Ex)c and (Ec)y = (Ey)c.

• Proposition 11.4. Let (X,F), (Y,G) be measurable spaces. If E ∈ F ⊗ G then for

all x ∈ X, y ∈ Y , Ey ∈ F and Ex ∈ G.

107

Also if f is a function on X × Y that is F ⊗ G-measurable, then fy is F-measurable

and fx is G-measurable.

Proof. Let

H = {E ⊂ X × Y : ∀ x ∈ X, y ∈ Y , Ey ∈ F , Ex ∈ G} .

One may verify that H is a σ-algebra (exercise!).

If A × B ∈ F ⊗ G is a box then (A × B)x = B for x ∈ A, and (A × B)y = A for

y ∈ B, and (A × B)x = ∅ = (A × B)y if x 6∈ A or y 6∈ B. So all boxes are in H. Thus,

F ⊗ G ⊂ H, which proves the first assertion.

If f is F ⊗ G-measurable then for any set A in the target σ-algebra, (fx)−1(A) =

(f−1(A))x which is in G because f−1(A) ∈ F⊗G. Similarly for (fy)−1(A) = (f−1(A))y ∈F . ut

11.2. Product integrals

11.2.1. Monotone classes and algebras. We require some technical notions.

X A monotone class on X is a family C ⊂ 2X that is closed under countable

increasing unions and countable decreasing intersections; i.e. for (En)n ⊂ C, if En ⊂En+1 for all n then

⋃n ∈ C, and if En ⊃ En+1 for all n then

⋂n ∈ C.

� Exercise 11.10. Show that an intersection of monotone classes is a monotone

class.

Show that for any collection of set K ⊂ 2X there is a monotone class C(K) that is the

minimal monotone class containing K; i.e. K ⊂ C(K) and for any monotone class C such

that K ⊂ C we have C(K) ⊂ C.

� Exercise 11.11. Show that any σ-algebra is a monotone class.

Show that if C is a monotone class and an algebra, then C is a σ-algebra.

108


A σ-algebra is closed under countable unions and intersections, so this is much more

than is needed for it to be a monotone class. This proves the first assertion.

Now suppose that C is a monotone class and an algebra. Let (An)n be a sequence in

C. We need to prove that⋃nAn ∈ C.

Indeed, set Bn =⋃nj=1Aj . Because C is an algebra, (Bn)n is an increasing sequence

in C. Because C is a monotone class,⋃nAn =

⋃nBn ∈ C. :) X

� Exercise 11.12. Let C be a monotone class. Fix A ∈ C and define

CA = {B ∈ C : A \B,B \A,A ∩B are all in C} .

Show that CA ⊂ C is a monotone class.


If (Bn)n is an increasing sequence in CA then (Bn\A)n, (Bn∩A)n are increasing sequences

in C. So,⋃n

Bn \A =⋃n

(Bn \A) ∈ C and⋃n

Bn ∩A =⋃n

(Bn ∩A) ∈ C.

Also, (A \Bn)n is a decreasing sequence in C so

A \⋃n

Bn = A ∩⋂n

Bcn =

⋂n

(A \Bn) ∈ C.

Thus,⋃nBn ∈ CA as well. Similarly, if (Bn)n is a decreasing sequence in CA, then

(Bn \ A)n, (Bn ∩ A)n are decreasing sequences in C and (A \ Bn)n is an increasing

sequence in C, which leads to⋂n

Bn \A =⋂n

(Bn \A) ∈ C and⋂n

Bn ∩A =⋂n

(Bn ∩A) ∈ C,

109

and

A \⋂n

Bn = A ∩⋃n

Bcn =

⋃n

(A \Bn) ∈ C.

So⋂nBn ∈ CA in this case. :) X

• Lemma 11.5 (Monotone Class Lemma). For an algebra A the monotone class C(A) =

σ(A).

Proof. Since σ(A) is a monotone class continuing A, we have that C(A) ⊂ σ(A). So it

suffices to prove the other inclusion. Since A ⊂ C(A) it suffices to show that C(A) is a

σ-algebra.

For A ∈ C(A) define

CA = {B ∈ C(A) : A \B,B \A,A ∩B are all in C(A)} .

We have that CA is a monotone class.

If A ∈ A then since A is an algebra we have that A ⊂ CA. By the minimality of C(A)

we get that CA = C(A) for all A ∈ A.

Note that by definition, B ∈ CA if and only if A ∈ CB. This implies that for all A ∈ Aand any B ∈ C(A) = CA we have that A ∈ CB. Thus, A ⊂ CB for any B ∈ C(A). By

minimality of C(A) again, CB = C(A) for any B ∈ C(A).

We conclude that if A,B ∈ C(A) then A \ B,B \ A,A ∩ B are all in C(A). Since

X = ∅c ∈ A ⊂ C(A) we have that C(A) is closed under complements as well. So C(A) is

an algebra and a monotone class. Thus C(A) is a σ-algebra, and we are done. ut

� Exercise 11.13. Let (X,F) be a measurable space such that F = σ(A) where Ais an algebra. Let L∞ be the vector space of all bounded measurable functions from X

to R. Let V be a subspace of L∞ such that:

• 1 ∈ V and 1A ∈ V for all A ∈ A.

• If (fn)n is a monotone increasing sequence in V such that fn ↗ f and such that

f is bounded, then f ∈ V as well.

110

Show that V = L∞.


Define G = {A : 1A ∈ V }. By the assumptions, A ⊂ G. Also, if (An)n is an increasing

sequence in G, then (1An)n is an increasing sequence in V such that 1An ↗ 1A where

A =⋃nAn. Since 1A is bounded, we have that 1A ∈ V , which implies that A ∈ G. Note

that G is closed under complements: If A ∈ G then 1A ∈ V and so 1Ac = 1 − 1A ∈ V(because V is a vector space). Thus, if (An)n is a decreasing sequence in G, then (Acn)n

is an increasing sequence in G, and⋂nAn =

(⋃nA

cn

)c ∈ G.

We conclude that G is a monotone class. Thus, by the Monotone Class Lemma,

F = C(A) ⊂ G.

So 1A ∈ V for all A ∈ F . Since V is a vector space, all simple functions are also in

V .

Now, if f ≥ 0 is a bounded measurable function, then there exists a monotone se-

quence of simple functions ϕn ↗ f . Since f is bounded, the assumptions on V tell us

that f ∈ V . So V contains all non-negative bounded measurable functions.

For a general real-valued bounded measurable function, note that f+, f− ∈ V so also

f = f+ − f−. :) X

11.2.2. Fubini-Tonelli for indicators. The next theorem tells us how to compute the

product measure of a set. First measure each section in one axis, and then integrate all

measures of sections over the other axis.

••• Theorem 11.6. Let (X,F , µ), (Y,G, ν) be σ-finite measure spaces. For any E ∈F ⊗ G we have that the functions

x 7→ ν(Ex) and y 7→ µ(Ey)

are measurable. Moreover,

µ⊗ ν(E) =

∫ν(Ex)dµ =

∫µ(Ey)dν.

111

Proof. Step I. First assume that µ, ν are finite measures.

Define C to be the set of all set E ∈ F ⊗ G such that the theorem holds for E. We

need to show that C = F ⊗ G.

If E = A × B for A ∈ F , B ∈ G then Ex = B and Ey = A for (x, y) ∈ E and

Ex = ∅ = Ey for (x, y) 6∈ E. So x 7→ ν(Ex) is the function ν(B)1A and y 7→ µ(Ey) is

the function µ(A)1B. These are of course measurable. Also, µ⊗ ν(A×B) = µ(A)ν(B)

by definition, which is

µ⊗ ν(A×B) =

∫ν(B)1Adµ =

∫µ(A)1Bdν.

Thus, all boxes A×B are in C.We have already seen that if E =

⊎nj=1(Aj ×Bj) then

µ⊗ ν(E) =n∑j=1

µ(Aj)ν(Bj).

Since in this case

Ex =

n⊎j=1

(Aj ×Bj)x and Ey =

n⊎j=1

(Aj ×Bj)y,

the functions x 7→ ν(Ex) and y 7→ µ(Ey) are just∑n

j=1 ν(Bj)1Aj and∑n

j=1 µ(Aj)1Bj ,

which are measurable and satisfy the conclusion of the theorem by additivity of the

integral.

Thus all finite disjoint unions of boxes are in C, which is to say that C contains the

algebra that generates F ⊗ G. If C was a monotone class then we would have by the

Monotone Class Lemma that F ⊗ G ⊂ C.So we are left with showing that C is a monotone class.

To this end, let (En)n be an increasing sequence in C. Let E =⋃nEn. Define

fn(x) = ν((En)x). This is a measurable function satisfying∫fndµ = µ ⊗ ν(En). Also,

(fn)n form an increasing sequence in L+(X,F , µ), because ((En)x)n is an increasing

sequence. Note that Ex =⋃n(En)x which implies that fn ↗ f for f(x) = ν(Ex) by

continuity of measures. So f ∈ L+(X,F , µ). By Monotone Convergence and continuity

112

of the measure µ⊗ ν,∫fdµ = lim

n→∞

∫fndµ = lim

n→∞µ⊗ ν(En) = µ⊗ ν(E).

Similarly, the functions gn(y) = µ((En)y) converge monotonically gn ↗ g for g(y) =

µ(Ey), and using Monotone Convergence as before,∫gdν = µ⊗ ν(E).

Thus, E ∈ C.Now for the case of a decreasing sequence (En)n in C. Let E =

⋂nEn. Again we define

fn(x) = ν((En)x) and gn(y) = µ((En)y). These are decreasing sequences that converge

to f(x) = ν(Ex) and g(y) = µ(Ey) respectively. So f, g are measurable. Also, for all

n, |fn| ≤ ν(Y ) < ∞ and |gn| ≤ µ(X) < ∞. Since the measures are finite, constant

functions are in L1 for both µ and ν. Thus, we may use Dominated Convergence of

conclude that ∫fdµ = lim

n→∞

∫fndµ = lim

n→∞µ⊗ ν(En) = µ⊗ ν(E),

and similarly, ∫gdν = µ⊗ ν(E).

So E ∈ C in the decreasing case as well.

This shows that C is a monotone class, completing the proof in the case µ, ν are finite

measures.

Step II. For general µ, ν which are σ-finite, let X × Y =⋃n(Xn × Yn) where

(Xn)n, (Yn)n are increasing sequences of measurable sets, and µ(Xn) <∞, ν(Yn) <∞.

Let E ∈ F ⊗ G. For every n, consider E ∩ (Xn × Yn). The functions µn(A) =

µ(A ∩ Xn), νn(B) = ν(B ∩ Yn) are finite measures. For any box A × B we have that

µn⊗νn(A×B) = µn(A)νn(B) = µ⊗ν(A×B∩Xn×Yn). Thus, ρn(C) := µ⊗ν(C∩Xn×Yn)

is the unique σ-finite product measure ρn = µn⊗νn on F ⊗G. By the previous step, the

functions x 7→ νn(Ex) = ν(Ex∩Yn) and y 7→ µn(Ey) = µ(Ey∩Xn) are measurable. Since

(Xn)n, (Yn)n are increasing sequences, continuity of measures gives νn(Ex) → ν(Ex)

113

and µn(Ey) → µ(Ey), so that x 7→ ν(Ex) and y 7→ µ(Ey) are measurable as limits of

measurable functions.

� Exercise 11.14. Let (Z,H, α) be a measure space. Let W ∈ H and define

β(A) = α(A ∩W ) for all A ∈ H.

Show that (Z,H, β) is a measure space.

Show that for any f ∈ L+(Z,H) we have∫fdβ =

∫Wfdα.


β is a measure since β(∅) = α(∅) = 0 and

β(⊎n

An) = α(⊎n

(An ∩W )) =∑n

α(An ∩W ) =∑n

β(An).

If f = 1A then ∫fdβ = β(A) = α(A ∩W ) =

∫W

1Adα.

If f is a simple function, then∫fdβ =

∫W fdα by linearity.

If f ∈ L+, then let ϕn ↗ f be a sequence of simple functions approximating f . Then,

by monotone convergence,∫fdβ ←

∫ϕndβ =

∫Wϕndα→

∫Wfdα.

:) X

Using the above exercise, we conclude that

µ⊗ ν(E ∩Xn × Yn) = µn ⊗ νn(E) =

∫ν(Ex ∩ Yn)dµn =

∫µ(Ey ∩Xn)dνn

=

∫Xn

ν(Ex ∩ Yn)dµ =

∫Yn

µ(Ey ∩Xn)dν.

114

Since ν(Ex ∩ Yn)1Xn(x) ↗ ν(Ex) and µ(Ey ∩ Yn)1Yn(y) ↗ µ(Ey), we use Monotone

Convergence to get that

µ⊗ ν(E) limn→∞

µ⊗ ν(E ∩Xn × Yn) =

∫ν(Ex)dµ =

∫µ(Ey)dν.

ut

11.3. The Fubini-Tonelli Theorem

??? THEOREM 11.7 (Fubini-Tonelli). Let (X,F , µ), (Y,G, ν) be σ-finite measure

spaces.

• Let f ∈ L+((X,F , µ) ⊗ (Y,G, ν)). Then the functions x 7→∫fxdν and y 7→∫

fydµ are in L+.

• Let f ∈ L1((X,F , µ) ⊗ (Y,G, ν)). Then the functions x 7→∫fxdν and y 7→∫

fydµ are defined a.e. and in L1.

In both cases we have∫fd(µ⊗ ν) =

∫ (∫fxdν

)dµ(x) =

∫ (∫fydµ

)dν(y).

Proof. Start with f ∈ L+. If f = 1E this is just the previous theorem. Since (f + g)x =

fx + gx and (f + g)y = fy + gy, we have the theorem for all simple functions. If f ∈ L+

is general, let ϕn ↗ f be an increasing sequence of approximating simple functions. It

may be easily verified that (ϕn)x ↗ fx and (ϕn)y ↗ fy. So fx, fy are indeed in L+.

Also, (ϕn)x, (ϕn)y are simple functions. Note that∫

(ϕn)xdν and∫

(ϕn)ydµ are also

increasing sequences of functions in L+. So by Monotone Convergence,∫fd(µ⊗ ν) = lim

n→∞

∫ϕnd(µ⊗ ν) = lim

n→∞

∫ ∫(ϕn)xdνdµ =

∫ ∫fxdνdµ.

Similarly for ∫fd(µ⊗ ν) =

∫ ∫fydµdν.

Now for f ∈ L1. Write f = f1 − f2 + i(f3 − f4) for fj ∈ L+. Note that (fx)j = (fj)x

and (fy)j = (fj)y. Since f is integrable, so are fj for all j. Thus, the first part of the

theorem tells us that∫ (∫(fj)xdν

)dµ =

∫ (∫(fj)

ydµ

)dν =

∫fjd(µ⊗ ν) <∞,(11.1)

115

which gives that∫

(fj)xdν,∫

(fj)ydµ are a.e. finite, and integrable as functions of x and

y respectively. Since |∫fxdν| ≤

∑4j=1

∫(fj)xdν, we have that x 7→

∫fxdν is an L1

function. Similarly y 7→∫fydµ is an L1 function. We also have by summing (11.1) that∫ (∫

fxdν

)dµ =

∫ (∫fydµ

)dν =

∫fd(µ⊗ ν).

ut

� Exercise 11.15. [Folland, p.69, ex.51] Let (X,F , µ), (Y,G, ν) be measure spaces

that are not necessarily σ-finite. Show that:

• If f : X → C is F-measurable and g : Y → C is G-measurable then h(x, y) :=

f(x)g(y) is F ⊗ G-measurable.

• If f ∈ L1(X,F , µ) and g ∈ L1(Y,G, ν) then h ∈ L1(X × Y,F ⊗ G, µ⊗ ν) and∫hd(µ⊗ ν) =

∫fdµ ·

∫gdν.


Note that he function φ : C2 → C defined by φ(z, w) = zw is continuous and thus

measurable. Also, the function H(x, y) := (f(x), g(y)) from X × Y to C2 is F ⊗ G-

measurable since the Borel σ-algebra on C2 is generated by sets of the form B1 × B2

where B1, B2 are Borel in C, and for any such set B1 ×B2,

H−1(B1×B2) = {(x, y) : f(x) ∈ B1, g(y) ∈ B2} = f−1(B1)×g−1(B2) ∈ F×G ⊂ F⊗G.

Hence h(x, y) = φ(f(x), g(y)) = φ ◦H(x, y) is F ⊗ G-measurable as a composition of

measurable functions.

For the second assertion, suppose first that f = 1A, g = 1B. Then clearly h = 1A×B

and ∫hd(µ× ν) = (µ× ν)(A×B) = µ(A)ν(B) =

∫fdµ ·

∫gdν.

116

If f =∑n

k=1 ak1Ak and g =∑m

j=1 bj1Bj are simple functions where (Ak)k and (Bj)j

are both pairwise disjoint, then

h =n∑k=1

m∑j=1

akbj1Ak×Bj

is a simple function and ∫hd(µ× ν) =

∫fdµ ·

∫gdν

follows by linearity.

If f ≥ 0, g ≥ 0 then we may choose fn ↗ f, gn ↗ g where fn, gn are simple for all n.

Since multiplication is continuous, fngn ↗ h. Thus, by monotone convergence,∫hd(µ× ν) =

∫fdµ ·

∫gdν.

If f, g are real-valued integrable functions, then

h(x, y) = (f+(x)− f−(x)) · (g+(y)− g−(y))

= f+(x)g+(y) + f−(x)g−(y)− f+(x)g−(y)− f−(x)g+(y).

This is a linear combination of 4 products of non-negative functions. Since∫fα(x)gβ(y)d(µ× ν) =

∫fαdµ ·

∫gβdν

for any choice of α, β ∈ {+,−}, we have by linearity∫|h|d(µ× ν) ≤

∑α,β∈{+,−}

∫fαdµ ·

∫gβdν <∞,

so h is integrable. Also by linearity, and by the above for non-negative functions,∫hd(µ× ν) =

∑α,β∈{+,−}

∫αβfα(x)gβ(y)d(µ× ν)

=∑

α,β∈{+,−}αβ

∫fαdµ ·

∫gβdν =

∫fdµ ·

∫gdν.

Finally for complex valued functions in L1 write f = f1 + if2 and g = g1 + ig2, and

note that if h(x, y) = f(x)g(y) then

h(x, y) = f1(x)g1(y)− f2(x)g2(y) + i ·[f1(x)g2(y) + f2(x)g1(y)

].

117

So the real part of h and imaginary part of h fall into the category of the previous

assertion, and so satisfy that they are in L1 (because f1, f2, g1, g2 are all in L1) and by

linearity∫hd(µ× ν) =

∫f1dµ ·

∫g1dν −

∫f2dµ ·

∫g2dν + i ·

∫f1dµ ·

∫g2dν + i ·

∫f2dµ ·

∫g1dν

=

∫fdµ ·

∫gdν.

:) X

� Exercise 11.16. Let µ be the counting measure on ([0, 1],B([0, 1])) (for any Borel

set A, µ(A) = |A|). Let λ be Lebesgue measure. Let D ={

(x, x) ∈ [0, 1]2}

. Compute∫ (∫1D(x, y)dµ(x)

)dλ(y),

∫ (∫1D(x, y)dλ(y)

)dµ(x), and show that they are not equal.

Use the fact that µ⊗ λ can be defined as an extension (although non-unique) of the

pre-measure on boxes to compute∫

1Dd(µ ⊗ λ) and show this is also not equal to the

other two integrals.


For any y ∈ [0, 1], ∫1D(x, y)dλ(y) = λ({x}) = 0,

and ∫1D(x, y)dµ(x) = µ({y}) = 1.

Thus,∫ (∫1D(x, y)dµ(x)

)dλ(y) = 1 and

∫ (∫1D(x, y)dλ(y)

)dµ(x) = 0.

Also,∫

1Dd(µ ⊗ λ) = (µ ⊗ λ)∗(D), where (µ ⊗ λ)∗ is the outer measure induced by

the pre-measure on the algebra of finite disjoint unions of boxes. This pre-measure is

defined by (µ⊗ λ)(A×B) = µ(A)λ(B). Thus,

(µ⊗ λ)∗(D) = inf

{∑n

µ(An)λ(Bn) : D ⊂⋃n

An ×Bn}.

118

If D ⊂ ⋃nAn × Bn then for any x ∈ [0, 1] we have that there exists some n for which

(x, x) ∈ An × Bn, so x ∈ Bn ∩ An. Thus, [0, 1] ⊂ ⋃n(Bn ∩ An), so∑

n λ(Bn ∩ An) ≥λ([0, 1]) = 1. So there must exist k such that |Bk ∩ Ak| has λ(Bk ∩ Ak) > 0, and

specifically Bk ∩Ak is infinite. Now, for this k we have that µ(Ak) ≥ µ(Ak ∩Bk) =∞.

Hence, for any D ⊂ ⋃nAn ×Bn there exists k such that∑n

µ(An)λ(Bn) ≥ µ(Ak)λ(Bk) =∞.

Thus, (µ⊗ λ)∗(D) =∞. :) X

� Exercise 11.17. [Folland p.69] Let X be a linearly ordered set that is uncountable,

but for any x ∈ X the set {y ∈ X : y < x} is countable. Let F be the countable-co-

countable σ-algebra on X; i.e. A ∈ F if A is countable or if Ac is countable. Define

E = {(x, y) ∈ X ×X : x < y}.Show that Ex, E

y are measurable for all x, y ∈ X.

For A ∈ F define µ(A) = 0 if A is countable and µ(A) = 1 if Ac is countable. Show

that µ is a measure on (X,F).

Show that∫ (∫

1E(x, y)dµ(x))dµ(y),

∫ (∫1E(x, y)dµ(y)

)dµ(x) exist and are not equal.


For any x ∈ X let Cx = {y : y < x}. By assumption Cx is countable for all x and

(Cx)c is uncountable for all x.

For any x, y ∈ X we have Ex = {y : y > x} = (Cx ] {x})c, so (Ex)c is countable,

and thus Ex is measurable and µ(Ex) = 1.

Also, Ey = {x : x < y} = Cy which is countable, so Ey is measurable and µ(Ey) = 0.

µ is easily verified to be a measure.

Finally, ∫ (∫1E(x, y)dµ(x)

)dµ(y) =

∫µ(Ey)dµ(y) = 0,

119∫ (∫1E(x, y)dµ(y)

)dµ(x) =

∫µ(Ex)dµ(x) = µ(X) = 1.

:) X

� Exercise 11.18. [Folland p.69] Consider (N, 2N, µ) where µ is the counting measure.

Define

f(n, k) =

1 n = k

−1 n = k + 1

0 otherwise.

Show that∫|f |d(µ⊗µ) =∞ and that

∫ (∫f(x, y)dµ(x)

)dµ(y),

∫ (∫f(x, y)dµ(y)

)dµ(x)

exist but are not equal.


f(n, k) = 1{n=k} − 1{n=k+1}.

|f(n, k)| = 1{n=k} + 1{n=k+1} so∫|f |d(µ⊗ µ) =

∑(n,k)∈N2

(1{n=k} + 1{n=k+1}) = # {(n, n), (n+ 1, n) : n ∈ N} =∞.

For any k ∈ N, ∫f(n, k)dµ(n) =

∑n

1{n=k} − 1{n=k+1} = 0.

However, for any n ∈ N,

∫f(n, k)dµ(k) =

∑k

1{n=k} − 1{n=k+1} =

0 n > 0

1 n = 0.

So∫ ∫f(n, k)dµ(n)dµ(k) = 0 and

∫ ∫f(n, k)dµ(k)dµ(n) =

∫1{0}(n)dµ(n) = 1.

:) X

120

� Exercise 11.19. [Folland p.69] Let (X,F , µ) be a σ-finite measure space. Let

f ∈ L+(X,F , µ). Set Γf := {(x, y) ∈ X × [0,∞] : y ≤ f(x)}.Show that Γf ∈ F ⊗ B([0,∞]). Show that (µ⊗ λ)(Γf ) =

∫fdµ.


The functions ϕ(x, y) = f(x) − y is F ⊗ B([0,∞])-measurable, as a composition of the

continuous function subtraction on the measurable function (x, y) 7→ (f(x), y). Thus,

Γf = {(x, y) : f(x)− y ≥ 0} = ϕ−1([0,∞]) ∈ F ⊗ B([0,∞]).

Note that

(Γf )x = {y : y ≤ f(x)} = 1[0,f(x)].

Now using Fubini-Tonelli,

(µ⊗ λ)(Γf ) =

∫ ∫(Γf )x(y)dλ(y)dµ(x) =

∫λ([0, f(x)])dµ(x) =

∫f(x)dµ(x).

:) X



121

Measure Theory

Ariel Yadin

Lecture 12: Change of variables

12.1. Linear transformations of Lebesgue measure

Recall that we showed that if L is an invertible linear transformation on Rd, for any

A ⊂ Rd that is Lebesgue measurable, we have that L(A) is also Lebesgue measurable and

λ(L(A)) = |detL|λ(A). The proof of this fact was not so simple and kind of technical.

We now give a more general statement, and also a shorter proof using the tools we

have developed.

••• Theorem 12.1. Let L be an invertible linear transformation of Rd and let f : Rd →C. If f is Lebesgue measurable then so is f ◦ L. Also, if f ≥ 0 or if f ∈ L1, then so is

f ◦ L and ∫fdλ = | detL|

∫f ◦ Ldλ.

Specifically, if A ⊂ Rd is Lebesgue measurable then L(A) is Lebesgue measurable and

λ(L(A)) = |detL|λ(A).

Proof. The main idea is as in the first proof we gave for Lebesgue measure of sets.

First suppose that f is Borel. Then f ◦ L is also because L is continuous.

If the theorem holds for linear maps L,M then it also holds for L ◦M as:∫f = |detL|

∫f ◦ L = |detL| · | detM |

∫(f ◦ L) ◦M = |detL ◦M |

∫f ◦ L ◦M.

Thus, we want to decompose linear transformations into elementary types that we know

how to deal with.

Every invertible linear transformation can be written as composition of the following

three types: M which is multiplication of coordinate j by a non-zero scalar α, A which

is adding coordinate k to coordinate j, and S which is swapping coordinates k and j.

So we only need to prove the theorem for M,A, S.

122

Now detM = α,detA = 1, detS = −1. Thus, using the Fubini-Tonneli Theorem, we

can integrate the coordinates in any order, so∫f ◦M(x1, . . . , xd)dλ

d =

∫f(x1, . . . , αxj , . . . , xd)dλ

d =

∫ ∫gx(αxj)dλ(xj)dλ

d−1(x),

where x = (x1, . . . , xj−1, xj+1, . . . , xd) and gx(xj) = f(x1, . . . , xd). The statement∫g(x)dλ(x) = |α| ·

∫g(αx)dλ(x),

is just the d = 1 case, which we will prove shortly. Given the d = 1 case we have

|α|∫f ◦Mdλd =

∫|α|∫gx(αxj)dλ(xj)dλ

d−1(x)

=

∫ ∫gx(xj)dλ(xj)dλ

d−1(x) =

∫fdλd.

Similarly for A and S: Given the d = 1 case we have∫f ◦Adλd =

∫ ∫gx(xj + xk)dλ(xj)dλ

d−1(x)

=

∫ ∫gx(xj)dλ(xj)dλ

d−1(x) =

∫fdλd.

For S we get by changing the order of integration (Fubini-Tonelli),∫f ◦ S(x1, . . . , xj , . . . , xk, . . . , xd)dλ

d

=

∫f(x1, . . . , xk, . . . , xj , . . . , xd)dλ(x1) · · · dλ(xj) · · · dλ(xk) · · · dλ(xd) =

∫fdλd.

So we are left with verifying the d = 1 cases of M and A:∫gdλ = |α| ·

∫g(αx)dλ(x) and

∫gdλ =

∫g(x+ a)dλ(x).

If g = 1A then this is just |α|λ(α−1A) = λ(A) and λ(A + a) = λ(A). By additivity

this extends to simple functions. Monotone Convergence gives this for non-negative

functions. Decomposing g into real and imaginary parts and those into positive and

negative parts gives the general case.

This completes the theorem for Borel functions.

If f is Lebesgue then there exists a Borel function g and a Borel set N such that

λ(N) = 0 and g1Nc = f1Nc . We have essentially already proved this in an exercise, but

123

the next exercise also reproves this. Since

(f◦L)(x)·(1Nc◦L−1)(x) = f(L(x))·1Nc(L(x)) = g(L(x))·1Nc(L(x)) = (g◦L)(x)·(1Nc◦L−1)(x),

we have that (f ◦ L) · (1Nc ◦ L−1) is Borel. So f ◦ L is Lebesgue. Moreover, because

λ(N) = λ(L−1(N)) = 0,∫fdλ =

∫gdλ = |detL|

∫(g ◦ L)dλ = |detL|

∫(f ◦ L)dλ.

ut

� Exercise 12.1. Show that a function f is Lebesgue if and only if there exists a

Borel function g and a Borel set N such that λ(N) = 0 and g1Nc = f1Nc .


If there exists a Borel function g and a Borel set N such that λ(N) = 0 and g1Nc = f1Nc ,

then for any Borel set B in the image, f(x) ∈ B if and only if g(x) ∈ B, x 6∈ N or

f(x) ∈ B, x ∈ N . Thus f−1(B) = g−1(B) ∩N c ] f−1(B) ∩N . Since f−1(B) ∩N ⊂ N

which has measure 0, we get that f−1(B)∩N is a Lebesgue null set, and thus Lebesgue

measurable. Since g−1(B) ∩ N c is Borel, we have that f−1(B) is Lebesgue. So in this

case f is a Lebesgue function.

For the other direction, if f is Lebesgue, consider the case f = 1A for a Lebesgue set

A. Since A = B ∪ F for a Borel set B and a null set F (not necessarily Borel). Since

λ∗(F ) = 0, there exists a Borel set N ⊃ F such that λ(N) = 0. Thus, 1A1Nc = 1B1Nc ,

and we are done taking g = 1B.

By additivity this extends to simple functions: If f =∑n

j=1 aj1Aj for Lebesgue

(Aj)j , then for every j there exist a Borel set Bj and a Borel null set Nj such that

1Aj1(Nj)c = 1Bj1(Nj)c . Taking N =⋃nj=1Nj and g =

∑nj=1 aj1Bj , we have that g is

Borel and f1Nc = g1Nc .

Now if f ≥ 0, let ϕk ↗ f be a monotone sequence of Lebesgue simple functions

approximating f . For every k there exists a Borel null set Nk and a Borel (simple)

124

function gk such that ϕk1(Nk)c = gk1(Nk)c . Taking N =⋃kNk which is a Borel null set,

we have that (gk1Nc = ϕk1Nc)k form an increasing sequence of Borel functions such

that gk1Nc ↗ f1Nc . So g = f1Nc is a Borel function.

Finally, if f is a general Lebesgue function, write f = f1 − f2 + i(f3 − f4) where fj

are all non-negative Lebesgue functions. Then, there exist Borel functions gj and Borel

null sets Nj , j = 1, 2, 3, 4, such that gj1(Nj)c = fj1(Nj)c . Taking N = N1 ∪N2 ∪N3 ∪N4

which is a Borel null set, and g = g1 − g2 + i(g3 − g4) we have that g1Nc = f1Nc . :) X

� Exercise 12.2. Complete the details of the previous theorem; that is, show that

∫gdλ = |α| ·

∫g(αx)dλ(x) and

∫gdλ =

∫g(x+ a)dλ(x).

� Exercise 12.3. Show that if λ is Lebesgue measure on Rd and U is a unitary

operator on Rd then λ(U(A)) = λ(A).

Specifically, Lebesgue measure is invariant under rotations.

12.2. Change of variables formula

Let U ⊂ Rd be an open set. Suppose that Ψ : Rd → Rd is a map such that for

Ψ = (ψ1, . . . , ψd) all the maps ψi have continuous partial derivatives ∂ψi∂xj

on U , in all

coordinates xj . (i.e. ψi are all C1 in U). For any x ∈ Rd define a matrix DΨ(x) by

(DΨ(x))i,j = ∂ψi∂xj

(x).

125

Ψ is called a C1 diffeomorphism (on U) if Ψ is 1-1 and DΨ(x) is invertible for all

x ∈ U . In this case by the inverse function theorem, Ψ−1 : Ψ(U) → U is also a C1

diffeomorphism, and DΨ−1(y) = (DΨ(Ψ−1(y)))−1.

� Exercise 12.4. Show that if L is an invertible linear map then it is a diffeomor-

phism with DL(x) = L(x) for all x ∈ Rd.

Also, show that for any diffeomorphism Ψ, D(L ◦Ψ)(x) = L(DΨ(x)).

••• Theorem 12.2. Let Ψ : U → Rd be a C1 diffeomorphism for an open set U ⊂ Rd.

If f : Ψ(U)→ C is Lebesgue, then f ◦Ψ is Lebesgue on U . If in addition f ≥ 0 or f is

integrable, then ∫Ψ(U)

fdλ =

∫U

(f ◦Ψ)(x) · | detDΨ(x)|dλ(x).

Proof. Start with Borel f . Because Ψ is continuous, f ◦Ψ is Borel.

We consider the sup-norm ||x|| = maxj |xj | and for a matrix M ∈ GLd(R), the

operator norm ||M || = maxi∑

j |Mi,j |. So ||Mx|| ≤ ||M ||||x||.For a ∈ Rd and ε > 0 write Q(a, ε) = {x : ||x− a|| ≤ ε}. Note that Q(a, ε) is a box.

Step I. We show that for Q = Q(a, ε),

λ(Ψ(Q)) ≤∫Q|detDΨ(x)|dλ(x).

Since ψj are all C1 functions the mean value theorem tells us that

ψi(x)− ψi(a) =

d∑j=1

(xj − aj)∂ψi∂xj(y),

for some y on the line segment between x and a. Thus, for any x ∈ Q = Q(a, ε),

||Ψ(x)−Ψ(a)|| ≤ ε · supy∈Q

maxi

∑j

(DΨ(y))i,j = ε · supy∈Q||DΨ(y)||.

Thus,

Ψ(Q) ⊂ Q(a, ε · supy∈Q||DΨ(y)||).

126

By the scaling of Lebesgue measure

λ(Ψ(Q)) ≤ λ(Q(a, ε · supy∈Q||DΨ(y)||)) = λ(sup

y∈Q||DΨ(y)|| ·Q) = (sup

y∈Q||DΨ(y)||)d · λ(Q).

For any invertible linear map L ∈ GLd(R),

λ(Ψ(Q)) = | detL|λ(L−1 ◦Ψ(Q)) ≤ |detL| · (supy∈Q||L−1DΨ(y)||)d · λ(Q).

Now, y 7→ DΨ(y) is a continuous function, and so uniformly continuous on the com-

pact set Q. Thus, for any η > 0 there exists k > 0 such that for all ||z−y|| ≤ 1k , z, y ∈ Q

we have ||(DΨ(z))−1DΨ(y)||d ≤ 1 + η. Write Q =⋃nkj=1Qk,j where Qk,j = Q(xk,j ,

1k )

and all have disjoint interiors. Then,

λ(Ψ(Q)) =

nk∑j=1

λ(Ψ(Qk,j))

≤nk∑j=1

|detDΨ(xk,j)| · ( supy∈Qj

||(DΨ(xk,j))−1DΨ(y)||)d · λ(Qk,j)

≤ (1 + η)

∫ nk∑j=1

|detDΨ(xk,j)|1Qk,j (x)dλ(x).

This holds for any Q =⋃nkj=1Qk,j where Qk,jQ(xk,j ,

1k ), and all have disjoint interiors.

The integrand∑nk

j=1 | detDΨ(xk,j)|1Qk,j (x) is continuous, and on Q tends uniformly

to | detDΨ(x)|1Q(x) as k →∞. By Dominated Convergence that

λ(Ψ(Q)) ≤ (1 + η) limk→∞

nk∑j=1

∫Qk,j

| detDΨ(xk,j)|dλ(x) = (1 + η)

∫Q| detDΨ(x)|dλ(x).

Thus, sending η → 0,

λ(Ψ(Q)) ≤∫Q|detDΨ(x)|dλ(x).

Step II. We show that for any open set U ,

λ(Ψ(U)) ≤∫U| detDΨ(x)|dλ(x).

Indeed, if U is an open set, we may write U =⋃nQn where Qn are almost disjoint

boxes (i.e. have disjoint interiors). Thus,

λ(Ψ(U)) ≤∑n

λ(Ψ(Qn)) ≤∑n

∫Qn

| detDΨ(x)|dλ(x) =

∫U|detDΨ(x)|dλ(x).

127

Step III. We show that for any Borel set A,

λ(Ψ(A)) ≤∫A|detDΨ(x)|dλ(x).

Indeed, for a Borel set A with λ(A) <∞, there exist open sets (Un)n such that A ⊂ Unfor all n and λ(Un \A)→ 0. By replacing Un with

⋂nj=1 Uj we may assume that (Un)n

is a decreasing sequence. Thus, continuity of measure gives that for U =⋂n Un we have

A ⊂ U and λ(U \A) = 0. So 1Un → 1A a.e., and the Dominated Convergence Theorem

now gives that

λ(Ψ(A)) ≤ λ(Ψ(U)) = limn→∞

λ(Ψ(Un))

≤ limn→∞

∫Un

| detDΨ(x)|dλ(x) =

∫A|detDΨ(x)|dλ(x).

Now if A is a general Borel set then A =⊎nAn where (An)n all have finite measure

(this is just σ-finiteness). So

λ(Ψ(A)) ≤∑n

λ(Ψ(An)) ≤∑n

∫An

|detDΨ(x)|dλ(x) =

∫A| detDΨ(x)|dλ(x).

Step IV. If f =∑n

j=1 aj1Aj is a Borel simple function defined on Ψ(U),∫Ψ(U)

fdλ =

n∑j=1

ajλ(Aj) ≤n∑j=1

aj

∫1Ψ−1(Aj)(x) · | detDΨ(x)|dλ(x)

=

∫U|detDΨ(x)| ·

n∑j=1

aj1Aj (Ψ(x))dλ(x) =

∫U


Taking limits with the monotone convergence theorem gives that for any Borel f ≥ 0

defined on Ψ(U), ∫Ψ(U)

fdλ ≤∫U


Replacing f with the function (f ◦Ψ)(x) · | detDΨ(x)| which is defined on U we have∫U

(f ◦Ψ)(x) · | detDΨ(x)|dλ(x) ≤∫

Ψ(U)(f ◦Ψ ◦Ψ−1)(x) · | detDΨ(Ψ−1(x))| · | detDΨ−1(x)|dλ(x).

Since DΨ(Ψ−1(x)) = (DΨ−1(x))−1, we have that | detDΨ(Ψ−1(x))|·| detDΨ−1(x)| = 1,

giving∫U

(f ◦Ψ)(x) · | detDΨ(x)|dλ(x) ≤∫

Ψ(U)fdλ ≤

∫U


128

This establishes the theorem for all non-negative Borel f .

To get general integrable Borel f just decompose into real, imaginary, positive and

negative parts. To get Lebesgue f find a Borel function that is a.e. identical to f . We

omit the details. ut

Example 12.3. Let us compute I :=∫∞−∞ e

−x2dx. Consider

I2 =

∫ ∫e−x

2e−y

2dxdy =

∫R2

f(x, y)dλ2(x, y),

where f(x, y) = e−(x2+y2), by Fubini.

Let U = (0,∞)× (−π, π). Let Ψ : U → R2 be Ψ(r, θ) = (r cos θ, r sin θ). So Ψ(U) =

R2 \ ((−∞, 0]× {0}). Since λ2((−∞, 0]× {0}) = 0 we have that I2 =∫

Ψ(U) fdλ2.

Note that

DΨ(r, θ) =

cos θ −r sin θ

sin θ r cos θ

.So |detDΨ(r, θ)| = r.

We get using Fubini again, since f(Ψ(r, θ)) = e−r2,

I2 =

∫Ψ(U)

fdλ2 =

∫Uf(Ψ(r, θ)) · rdλ(r, θ)

=

∫ π

−π

∫ ∞0

re−r2drdθ = 2π ·

∫ ∞0

(−12) ddre

−r2dr = π.

So I =√π. 454



129

Measure Theory

Ariel Yadin

Lecture 13: Lebesgue-Radon-Nykodim

13.1. Signed measures

• Definition 13.1. Let (X,F) be a measurable space. A signed measure on (X,F)

is a function ν : F → [−∞,∞] such that

• ν(∅) = 0;

• ν(F) ⊂ (−∞,∞] or ν(F) ⊂ [−∞,∞); that is, ν takes on at most one of the

values −∞,∞ (but not both);

• If (An)n is a sequence of pairwise disjoint sets in F then ν(⊎nAn) =

∑n ν(An),

and if |ν(⊎nAn)| <∞ then

∑n |ν(An)| <∞.

When we say positive measure we stress that the measure in question is a measure

that is not signed; i.e. in the usual sense.

We really only require signed measures in order to prove differentiation theorems.

� Exercise 13.1. Show that if µ1, µ2 are measures on (X,F) and at least one of

them is a finite measure, then µ1 − µ2 is a signed measure.

� Exercise 13.2. Let f be a measurable function f : (X,F) → [−∞,∞]. Assume

that one of the integrals∫f+dµ,

∫f−dµ is finite, in which case the integral

∫fdµ exists.

Show that ν(A) =∫A fdµ is a signed measure.

130

� Exercise 13.3. Let ν be a signed measure on (X,F). Let (An)n be a sequence in

F .

Show that if (An)n is increasing, then

ν(⋃n

An) = limn→∞

ν(An).

Show that if (An)n is decreasing and |ν(A1)| <∞ then

ν(⋂n

An) = limn→∞

ν(An).

• Definition 13.2. Let ν be a signed measure on (X,F). We say that A ∈ F is

positive (respectively negative) if for every B ⊂ A such that B ∈ F we have ν(B) ≥ 0

(respectively ν(B) ≤ 0).

If A is both positive and negative we say A is null.

� Exercise 13.4. Let µ be a positive measure on (X,F). Let f be a measurable

function f : (X,F)→ [−∞,∞] such that the integral∫fdµ exists. Let ν(A) =

∫A fdµ,

which we know is a signed measure.

Show that A ∈ F is ν-positive (respectively negative, null) if and only if f1A ≥ 0 a.e.

(respectively f1A ≤ 0, f1A = 0 a.e.).


Let A ∈ F .

If A is positive then fix ε > 0 and let B = {f1A ≤ −ε} which is measurable because

f1A is measurable. Note that B = A ∩ {f ≤ −ε} ⊂ A, so ν(B) ≥ 0. Also, f1B ≤

131

(−ε)1B, and

0 ≤ ν(B) =

∫f1Bdµ ≤ −ε · µ(B).

So µ(B) = 0 (which also implies that ν(B) = 0.)

On the other hand, if f1A ≥ 0 a.e. then for any F 3 B ⊂ A we have that f1B =

f1A1B ≥ 0 a.e., so

ν(B) =

∫f1Bdµ ≥ 0.

For the negative case, note that A ∈ F is ν-negative if and only if A is (−ν)-positive,

where −ν is the signed measure defined by −ν(B) = −∫B fdµ =

∫B(−f)dµ, we have

that A is negative if and only if (−f)1A ≥ 0 a.e., which is if and only if f1A ≤ 0 a.e.

For the null case, A ∈ F is null if and only if it is both positive and negative, which

is if and only if 0 ≤ f1A ≤ 0 a.e. :) X

� Exercise 13.5. Show that if A is positive then any B ∈ F such that B ⊂ A is

also positive.

Show that a countable union of positive sets is positive.


The first assertion is easy.

For the second, let (An)n be a sequence of positive sets and let A =⋃nAn. Then,

Bn = An \⋃n−1j=1 Aj ⊂ An is also positive. Since A =

⊎nBn, for any measurable C ⊂ A

we have that

ν(C) =∑n

ν(C ∩Bn) ≥ 0.

:) X

••• Theorem 13.3 (Hahn Decomposition Theorem). If ν is a signed measure on (X,F)

then there exist disjoint measurable sets P∩N = ∅ such that X = P]N and P is positive

and N is negative.

This is called a Hahn decomposition of X.

132

If P ′, N ′ are another Hahn decomposition of X we have that P4P ′ = N4N ′ is a

null set.

Proof. By possible considering −ν instead of ν, we may assume that ν(A) < ∞ for all

A ∈ F .

Let m = sup {ν(P ) : P is positive }. Let (Pn)n be a sequence of positive sets such

that ν(Pn)→ m. Let P =⋃n Pn. P is positive and

m ≥ ν(P ) = limn→∞

ν(

n⋃j=1

Pj) ≥ limn→∞

ν(Pn) = m.

So m = ν(P ) <∞.

Let N = X \ P .

For any F 3 A ⊂ N , if A is positive then for any F 3 B ⊂ A, also B ] P is positive,

so ν(P ) = m ≥ ν(B ]P ) = ν(B) + ν(P ), and so ν(B) = 0, which implies that A is null.

That is, any positive subset of N is null.

Now let F 3 A ⊂ N . If ν(A) > 0 then A is not null, and since A is not positive

there exists F 3 B ⊂ A such that ν(B) < 0. So for C = A \ B ⊂ A we have that

ν(C) = ν(A)− ν(B) > ν(A). So for any A ⊂ N such that ν(A) > 0 define

nA := inf{n ≥ 1 : ∃ B ⊂ A , ν(B) > ν(A) + n−1

},

and let BA ⊂ A be a set such that ν(BA) > ν(A) + 1nA

.

So assume for a contradiction that N is not negative. So there exists F 3 A0 ⊂ N

such that ν(A0) > 0. Given Ak define inductively nk+1 = nAk and Ak+1 = BAk . So for

all k ≥ 1 we have that ν(Ak) > ν(Ak−1) + 1nk

and Ak ⊂ Ak−1. Note that

ν(Ak) > ν(Ak−1) + 1nk> · · · >

k∑j=1

1nj.

Since ν(A0) <∞, for A =⋂k Ak,

∞ > ν(A) = limk→∞

ν(Ak) ≥∑k

1nk> 0.

133

Thus,∑

k1nk<∞ so nk →∞. Also, ν(A) > 0. Take k large enough so that nk > 2nA.

Then, B ⊂ A ⊂ Ak−1. By the definition of nk = nAk−1we have that

ν(A) + 1nA

< ν(B) ≤ ν(Ak−1) + 1nk−1 ≤ ν(Ak−1) + 1

2nA.

So ν(Ak−1) > ν(A) + 12nA

for all k such that nk > 2nA. Thus,

ν(A) = limk→∞

ν(Ak) ≥ ν(A) + 12nA

,

a contradiction!

Thus, N is negative.

Finally if P ′, N ′ is a Hahn decomposition, then P \ P ′ ⊂ N ′ ∩ P , so must be both

positive and negative. Similarly, P ′ \ P ⊂ N ∩ P ′. ut

••• Theorem 13.4 (Jordan Decomposition Theorem). If ν is a signed measure on

(X,F) then there exist unique positive measures ν+, ν− such that ν = ν+−ν− and such

that X = P ]N where ν+(N) = ν−(P ) = 0.

Proof. Let X = P ]N be a Hahn decomposition. Define

ν+(A) = ν(A ∩ P ) and ν−(A) = −ν(A ∩N).

These are positive measures because P is positive and N is negative. It is immediate

that ν = ν+ − ν− and that ν+(N) = ν−(P ) = 0.

Now for uniqueness: Suppose that ν = µ+−µ− for positive measure µ+, µ− and X =

P ′ ]N ′ where µ+(N ′) = µ−(P ′) = 0. For any A ⊂ P ′ we have that ν(A) = µ+(A) ≥ 0

and for any B ⊂ N ′ we have ν(B) = −µ−(B) ≤ 0. So P ′, N ′ for a Hahn decomposition,

and so P4P ′ is ν-null. For any A ∈ F , since (A ∩ P )4(A ∩ P ′) ⊂ P4P ′,

µ+(A) = µ+(A ∩ P ′) = ν(A ∩ P ′) = ν(A ∩ P ) = ν+(A ∩ P ) = ν+(A),

and similarly,

µ−(A) = µ−(A ∩N ′) = −ν(A ∩N ′) = −ν(A ∩N) = ν−(A ∩N) = ν−(A).

ut

134

• Definition 13.5. ν = ν+ − ν− is called the Jordan decomposition of ν. ν+ is the

positive part and ν− is the negative part.

We also define the total variation or absolute value of ν as |ν| := ν+ + ν−.

ν is called finite (respectively, σ-finite) if |ν is a finite (respectively, σ-finite) measure.

� Exercise 13.6. Show that A is ν-null if and only if ν+(A) = ν−(A) = 0, which is

if and only if |ν|(A) = 0.

� Exercise 13.7. Show that if P,N is a Hahn decomposition then ν+(N) = ν−(P ) =

0.

� Exercise 13.8. Show that for all A ∈ F ,

ν(A) =

∫A

(1P − 1N )d|ν|,

where P,N are any Hahn decomposition.


If P,N are a Hahn decomposition, then ν+(N) = ν−(P ) = 0. So

ν(A) = ν+(A ∩ P )− ν−(A ∩N) = |ν|(A ∩ P )− |ν|(A ∩N) =

∫A

1Pd|ν| −∫A

1Nd|ν|.

We may combine the integrals into one since at least one of them is finite, because either

ν+ or ν− is a finite measure. :) X

� Exercise 13.9. Show that L1(ν+) ∩ L1(ν−) = L1(|ν|).

135

• Definition 13.6. For a signed measure and f ∈ L1(|ν|) we define∫fdν =

∫fdν+ −

∫fdν−.

� Exercise 13.10. Show that for f ∈ L1(|ν|),∣∣ ∫ fdν∣∣ ≤ ∫ |f |d|ν|.

� Exercise 13.11. Show that for measurable A,

|ν|(A) = sup

{∣∣ ∫Afdν

∣∣ : |f | ≤ 1

}.

� Exercise 13.12. Show that if ν = µ−ρ where µ, ρ are positive measures (of which

one is finite), then µ ≥ ν+, ρ ≥ ν−.

� Exercise 13.13. Let ν1, ν2 be signed measures such that ν1 <∞, ν2 <∞.

Show that |ν1 + ν2| ≤ |ν1|+ |ν2|.Show that | − ν1| = |ν1|.

136

13.2. The Lebesgue-Radon-Nikodym Theorem

• Definition 13.7. Let ν, µ be signed measures on (X,F). We say that ν, µ are mu-

tually singular (or just singular), denoted ν ⊥ µ, if X = A]B where A is ν-null and

B is µ-null.

� Exercise 13.14. Show that ν+ ⊥ ν−.

Show that ν ⊥ µ if and only if |ν| ⊥ µ if and only if ν+ ⊥ µ, ν− ⊥ µ.

• Definition 13.8. Let ν, µ be signed measures on (X,F). We say that ν is absolutely

continuous with respect to µ, denoted ν � µ, if every µ-null set is also a ν-null set.

� Exercise 13.15. Show that ν � µ if and only if |ν| � |µ| if and only if

ν+ � |µ|, ν− � |µ|.

� Exercise 13.16. Show that if ν � µ and ν ⊥ µ then ν ≡ 0.

� Exercise 13.17. Let ν(A) =∫A fdµ for some measurable f such that f <∞ (but

perhaps f can take the value −∞). Show that ν � µ. Show that ν is finite if and only

if f is integrable.


If µ(A) = 0 then f1A = 0 µ-a.e. So ν(A) =∫f1Adµ = 0. :) X

137

• Proposition 13.9. Let ν be a finite signed measure and µ a positive measure on

(X,F). ν � µ if and only if for every ε > 0 there exists δ > 0 such that |ν(A)| < ε for

all A ∈ F such that µ(A) < δ.

Proof. The ε, δ condition implies that for all A ∈ F such that µ(A) = 0, we have that

for any ε > 0, |ν(A)| < ε. Thus, µ(A) = 0 implies ν(A) = 0. So if A is a µ-null set, then

for any B ⊂ A we have µ(B) = 0 so also ν(B) = 0. This implies that A is a ν-null set.

For the other direction, assume that ν is a positive measure. Suppose the ε, δ condition

fails. So there exists ε > 0 such that for any n there exists a set An ∈ F with µ(An) <

2−n but ν(An) ≥ ε. Define

F = lim supAn =⋂n

⋃k≥n

Ak.

Since⋃k≥nAk is a decreasing sequence, and since ν is a finite measure,

ν(F ) = limn→∞

ν(⋃k≥n

Ak) ≥ limn→∞

ν(An) ≥ ε.

Also, for any n,

µ(⋃k≥n

Ak) ≤∑k≥n

µ(Ak) ≤∑k≥n

2−k = 2−n+1.

Thus, µ(F ) ≤ infn 2−n+1 ≤ 0. So F is a µ-null set and ν(F ) > 0, which implies that it

is not the case that ν � µ.

Now, if ν is a signed measure then ν � µ if and only if ν+, ν− � µ. So for any

ε > 0 there exists δ > 0 such that if µ(A) < δ then ν+(A), ν−(A) < ε. This implies that

|ν(A)| = ν+(A) + ν−(A) < ε. ut

� Exercise 13.18. Show that for any integrable f , for every ε > 0 there exists δ > 0

such that if µ(A) < δ then∣∣ ∫A fdµ| < ε.

We are almost ready to prove the basic differentiation theorem. A technical lemma

first:

138

• Lemma 13.10. Let ν, µ be finite positive measures on (X,F). Either ν ⊥ µ or there

exist some ε > 0 and A ∈ F such that µ(A) > 0 and A is positive for the signed measure

ν − εµ.

Proof. Let X = Pn ]Nn be a Hahn decomposition for the signed measure ν − 1nµ. Let

P =⋃n Pn and N =

⋂nNn = P c. For any n we have that N ⊂ Nn is negative for

ν − 1nµ. That is, 0 ≤ ν(N) ≤ 1

nµ(N) for all n, and so ν(N) = 0.

If µ(P ) = 0 then ν ⊥ µ.

Otherwise, there exists some n for which µ(Pn) > 0. Since Pn is positive for ν − 1nµ,

we can take ε = 1n and A = Pn. ut

We have seen that measures of the form A 7→∫A fdµ are absolutely continuous with

respect to µ. The next theorem tells us that these are the only examples.

??? THEOREM 13.11 (Lebesgue-Radon-Nikodym Theorem). Let ν be a σ-finite

signed measure and let µ be a σ-finite positive measure on (X,F). There exist unique

signed measures σ, ρ such that σ ⊥ µ, ρ � µ, and ν = σ + ρ. Moreover, there exists

a measurable function f : X → [−∞,∞] such that for all A ∈ F , ρ(A) =∫A fdµ

(specifically the integral exists).

Moreover, ρ is positive if and only if f is positive. ρ is finite if and only if f is

integrable. σ, ρ are positive if and only if ν is positive. σ, ρ are finite if and only if ν is

finite.

X The decomposition ν = σ + ρ is called the Lebesgue decomposition. We use

the notation dρdµ to denote the function f given by the theorem. This function is µ-a.e.

unique, and called the Radon-Nikodym derivative of ρ with respect to µ.

Proof. Case I. ν, µ are finite and positive. Define

M =

{f : X → [0,∞] :

∫Afdµ ≤ ν(A) ∀ A ∈ F

}.

0 ∈M so M 6= ∅. If f, g ∈M then for B = {f > g} and for any A ∈ F ,∫A

(f ∨ g)dµ =

∫A∩B

fdµ+

∫A∩Bc

gdµ ≤ ν(A ∩B) + ν(A ∩Bc) = ν(A).

139

So f ∨ g ∈M as well.

Set m = supf∈M∫fdµ ≤ ν(X) < ∞. Let (fn)n ⊂ M such that

∫fndµ → m. Let

gn = max {f1, . . . , fn}. So (gn)n is an increasing sequence such that gn ↗ f := supn fn.

Note that gn ∈M for all n. So

m = limn→∞

∫fndµ ≤ lim

n→∞

∫gndµ ≤ m.

By Monotone Convergence, m = limn→∞∫gndµ =

∫fdµ. So by perhaps modifying f

on a set of measure 0, we may assume that f <∞. Also, for any A ∈ F , by Monotone

Convegence, ∫Afdµ = lim

n→∞

∫Agndµ ≤ ν(A),

so f ∈M .

Consider the signed measure σ(A) := ν(A)−∫A fdµ. Note that σ is positive because

f ∈M . Since σ, µ are finite positive measures, if σ is not singular with respect to µ then

there exist ε > 0 and A ∈ F such that µ(A) > 0 and σ(B) ≥ εµ(B) for all measurable

B ⊂ A. This implies that for any C ∈ F ,∫C

(f + ε1A)dµ =

∫Cfdµ+ εµ(A ∩ C) ≤

∫Cfdµ+ ν(A ∩ C)−

∫A∩C

fdµ

=

∫C∩Ac

fdµ+ ν(C ∩A) ≤ ν(C ∩Ac) + ν(C ∩A) = ν(C).

So f + ε1A ∈M which implies that

m ≥∫

(f + ε1A)dµ =

∫fdµ+ εµ(A) > m,

a contradiction. So σ ⊥ µ. Setting ρ(A) =∫A fdµ, since ρ� µ, we have the decompo-

sition for positive finite ν, µ. Uniqueness of the decomposition is proved in an exercise

following.

Case II. ν, µ are σ-finite positive measures. In this case we can write X =⊎nXn

where ν(Xn) <∞, µ(Xn) <∞. For every n set νn(A) = ν(A∩Xn), µn(A) = µ(A∩Xn).

These are finite measures, so we have th? Lebesgue decomposition νn = σn + ρn where

σn ⊥ µn and ρn(A) =∫A fndµn for some integrable fn. We also know that µn(Xc

n) =

νn(Xcn) = 0, so ∫

Afndµn =

∫Afn1Xndµn,

140

and by replacing fn with fn1Xn we may assume that fn = 0 off Xn. It is an exercise to

show that∫A fndµ =

∫A fndµn. Also, σn(Xc

n) = νn(Xcn)−

∫Xcnfndµn = 0.

Set σ =∑

n σn It is another exercise to show that σ ⊥ µ. Set f =∑

n fn and

ρ(A) =∫A fdµ =

∑n

∫A fndµn =

∑n ρn(A). So

ν(A) =∑n

νn(A) =∑n

(σn(A) + ρn(A)) = σ(A) + ρ(A).

This completes the proof for σ-finite positive measures.

Case III. For σ-finite signed measure ν, write the Jordan decomposition ν = ν+−ν−.

We have the Lebesgue decomposition ν± = σ±+ρ± and then ν = (σ+−σ−)+(ρ+−ρ−)

which satisfy the conditions of the theorem. ut

� Exercise 13.19. Show that the Lebesgue decomposition is unique.


Suppose that ν = σ+ ρ = σ′+ ρ′ where σ, σ′ ⊥ µ and ρ, ρ′ � µ. If ν is a finite measure,

then σ − σ′ = ρ′ − ρ are well defined signed measures. Since ρ′ − ρ� µ and σ − σ′ ⊥ µwe have that σ − σ′ = ρ′ − ρ ≡ 0.

Now, if ν is σ-finite, write X =⊎Xn where ν(Xn) < ∞. Then, define νn(A) =

ν(A ∩Xn). In this case, if ν = σ + ρ is a Lebesgue decomposition of ν with respect to

µ, then for any measurable A,

νn(A) = ν(A ∩Xn) = σ(A ∩Xn) + ρ(A ∩Xn).

Since the signed measure ρn(A) := ρ(A ∩Xn) is absolutely continuous with respect to

µ, and since the signed measure σn(A) := σ(A ∩Xn) is singular with respect to µ, this

forms a Lebesgue decomposition of νn with respect to µ, which is unique because νn is

a finite measure.

141

Thus, if ν = σ+ρ = σ′+ρ′ where σ, σ′ ⊥ µ and ρ, ρ′ � µ, then σ(A∩Xn) = σ′(A∩Xn)

and ρ(A ∩Xn) = ρ(A ∩Xn), which implies that

σ(A) =∑n

σ(A ∩Xn) =∑n

σ′(A ∩Xn) = σ′(A),

ρ(A) =∑n

ρ(A ∩Xn) =∑n

ρ′(A ∩Xn) = ρ′(A).

:) X

� Exercise 13.20. Let µ be a positive measure and let Y be a measurable set of

finite measure. Set µY (A) = µ(A ∩ Y ) for all measurable A. Let f be a measurable

function such that f = 0 off Y .

Show that∫A fdµ =

∫A fdµY .

� Exercise 13.21. Let (νn)n be a sequence of positive measures. Let µ be a positive

measure.

Show that if νn ⊥ µ for all n then∑

n νn ⊥ µ.

Show that if νn � µ for all n then∑

n νn � µ.

• Proposition 13.12 (Chain rule). Suppose that ν � µ � ρ for σ-finite positive

measures ν, µ, ρ. If f ∈ L1(ν) then f dνdµ ∈ L1(µ) and∫fdν =

∫fdν

dµdµ.

Also, ν � ρ and dνdρ = dν

dµ ·dµdρ , ρ-a.e.

Proof. If f = 1A then∫fdν =

∫f dνdµdµ holds by definition of the Radon-Nikodym

derivative. This extends by additivity to simple functions and by Monotone convergence

to non-negative functions. Taking real, imaginary, positive and negative parts proves

this for all f ∈ L1(ν).

142

For the second assertion note that for any measurable A,∫A

dν

dρdρ = ν(A) =

∫A

dν

dµdµ =

∫A

dν

dµ· dµdρdρ.

So the functions inside the integrals on both sides must be equal ρ-a.e. ut

� Exercise 13.22. Show that if ν � µ and µ � ν then dνdµ ·

dµdν = 1 for ν-a.e. and

µ-a.e.

� Exercise 13.23. Let νj � µ for a positive σ-finite µ and signed σ-finite νj , j = 1, 2.

Show that d(ν1+ν2)dµ = dν1

dµ + dν2dµ .

� Exercise 13.24. Let νj � µj be σ-finite measures on (Xj ,Fj) for j = 1, 2. Then

ν1 ⊗ ν2 � µ1 ⊗ µ2 and for a.e. (x, y) ∈ X1 ×X2,

d(ν1 ⊗ ν2)

d(µ1 ⊗ µ2)(x, y) =

dν1

dµ1(x) · dν2

dµ2(y).

� Exercise 13.25. Let µ be the counting measure on ([0, 1],B([0, 1])) and λ Lebesgue

measure.

Show that λ� µ but dλdµ does not exist. (What fails in the Radon-Nikodym Theorem?)

Show µ has no Lebesgue decomposition with respect to λ.

143


Note that µ(A) = 0 if and only if A = ∅. So µ(A) = 0 implies A = ∅ which implies that

λ(A) = 0. That is, λ� µ.

Assume for a contradiction dλdµ exists. Then, for any x ∈ [0, 1], since {x} is Borel,

0 = λ({x}) =

∫dλdµ1{x}dµ = dλ

dµ(x)µ({x}) = dλdµ(x).

So dλdµ ≡ 0 which is impossible.

The Radon-Nikodym theorem requires σ-finiteness, and µ is not σ-finite.

Now, assume that µ = σ + ρ where σ ⊥ λ and ρ � λ. Note that for any Borel set

A, if A 6= ∅ and x ∈ A then σ(A) ≥ σ({x}) = µ({x})− ρ({x}) = 1 because λ({x}) = 0.

However this implies that σ(A) > 0 for any non-empty Borel set A. Since σ ⊥ λ there

exist Borel sets [0, 1] = A ] B such that σ(A) = 0 and λ(B) = 0. But then A = ∅ and

B = [0, 1] which contradicts λ(B) = 0. :) X

� Exercise 13.26. Let µ, ν be σ-finite measures on (X,F) such that ν � µ. Let

λ = ν + µ.

(a) Show that ν � λ.

(b) Show that λ� µ.

(c) Show that 0 ≤ dνdλ(x) < 1 for µ-a.e. x ∈ X.

(d) Show that µ-a.e.

dν

dµ=

dνdλ

1− dνdλ

.


(a) If A ∈ F is such that ν(A) > 0 then µ(A) > 0, because ν � µ. So λ(A) =

µ(A) + ν(A) > 0. Thus, if λ(A) = 0 then ν(A) = 0. Since this holds for all

A ∈ F we get that ν � λ.

144

(b) If A ∈ F is such that µ(A) = 0 then ν(A) = 0 because ν � µ. Thus, λ(A) =

µ(A) + ν(A) = 0. Since this holds for all A ∈ F we have that λ� µ.

(c) Because λ, ν are positive measure we have that dνdλ ≥ 0. Let A =

{dνdλ ≥ 1

}.

Then

ν(A) =

∫A

dν

dλdλ ≥ λ(A) = ν(A) + µ(A).

So µ(A) = 0.

(d) Since dλdµ = dν

dµ + dµdµ = dν

dµ + 1, we get that

dν

dµ=dν

dλ· dλdµ

=dν

dλ· dνdµ

+dν

dλ.

So,

dν

dµ·(

1− dν

dλ

)=dν

dλ.

By (c) we get that µ-a.e. 0 < 1 − dνdλ ≤ 1, so we can divide by this function to

complete the proof.

:) X

� Exercise 13.27. Let (X,F , µ) be a σ-finite measure space. Let G ⊂ F be a sub-

σ-algebra of F . Let ν = µ∣∣G .

(a) Suppose that f ∈ L1(X,F , µ). Show that there exists g ∈ L1(X,G, ν) such that

for every A ∈ G, ∫Afdµ =

∫Agdν.

(b) Suppose that for f ∈ L1(X,F , µ) there are two such functions g, g′ ∈ L1(X,G, ν)

such that for all A ∈ G,∫Afdµ =

∫Agdν =

∫Ag′dν.

Show that g = g′ ν-a.e.

145


(a) Because f ∈ L1, we know that |f | <∞ µ-a.e., so we may assume that |f | <∞.

First assume that f is positive and µ is finite. In this case, consider the

function

ρ(A) :=

∫Afdµ

defined for all A ∈ G. First of all, we showed in class that this defines a finite

positive measure on (X,G). Moreover, if ν(A) = 0 for some A ∈ G, since

ν = µ∣∣G we have that µ(A) = 0, and so ρ(A) =

∫A fdµ = 0. Since this holds

for all A ∈ G, the signed measure ρ is absolutely continuous with respect to the

measure ν. Since µ is finite, so is ν. Thus, by the Radon-Nykodim Theorem

there exists a positive integrable g = dρdν ∈ L1(X,G, ν) such that dρ = gdν; that

is, for all A ∈ G, ∫Afdµ = ρ(A) =

∫Adρ =

∫Agdν.

Now, if µ is only σ-finite, then write X =⊎nXn with µ(Xn) <∞. Consider

νn(A) := µ(A ∩Xn) for all A ∈ G. So ν =∑

n νn. Define ρn(A) :=∫A∩Xn fdµ.

Sincen∑j=1

f1A∩Xj ↗ f1A,

by monotone convergence we get that

ρ(A) :=

∫Afdµ =

∑n

∫A∩Xn

fdµ =∑n

ρn(A).

Also, as above, if νn(A) = 0 then µ(A ∩ Xn) = 0 and so ρn(A) = 0. So

ρn � νn. Since νn is finite, gn := dρnνn

exists and is in L1(X,G, νn). Specifically,

gn is G-measurable. Also, since ρn(A) = 0 for A ∩Xn = ∅, we have that gn can

be chosen such that it is supported on Xn. Define g =∑

n gn. Since (Xn)n are

disjoint and so gn have disjoint support, we get that g is always finite and well

defined. Also, since gn = gn1Xn , by monotone convergence again∫Agdν =

∫ ∑n

gn1A∩Xndν =∑n

∫A∩Xn

gndνn =∑n

ρn(A) = ρ(A).

146

Specifically, ∫Xgdν = ρ(X) =

∫Xfdµ <∞,

so g ∈ L1(X,G, ν).

Now, for the case that f is a general (not necessarily positive) function in L1.

Write f = (f1 − f2) + i(f3 − f4) for fj ∈ L1 positive. By the previous case,

there exist real-valued functions gj ∈ L1(X,G, ν) such that for any A ∈ G and

j = 1, 2, 3, 4 we have ∫Afjdµ =

∫Agjdν.

By linearity of the integral we get that for all A ∈ G,∫Afdµ =

∫Af1dµ−

∫Af2dµ+ i ·

∫Af3dµ− i ·

∫Af4dµ =

∫A

(g1 − g2) + i(g3 − g4)dν.

So we may choose g = g1 − g2 + i(g3 − g4) which is a function in L1(X,G, ν).

(b) Suppose that g, g′ are as in the question. Then for all A ∈ G,∫Agdν =

∫Ag′dν.

We have shown in class that this implies that g = g′ ν-a.e.

:) X



147

Measure Theory

Ariel Yadin

Lecture 14: Convergence

Until now we have only considered the convergence of a sequence of functions (fn)n

to a limit f is a pointwise sense: f→f means that fn(x)→ f for all x; we also extended

this a bit to include a.e. convergence. We had a glimpse at some points of uniform

convergence. The introduction of measure gives us many other topologies to consider.

14.1. Modes of convergence

• Definition 14.1. Let (fn)n, f be a sequence of measurable functions on a measure

space (X,F , µ).

• We say that (fn)n converges to f a.e., denoted fna.e.−→ f , if µ {fn 6→ f} = 0.

• We say that (fn)n converges to f in L1, denoted fnL1

−→ f , if∫|fn − f |dµ → 0

as n→∞.

• We say that (fn)n is Cauchy in measure if for every ε > 0 µ {|fn − fm| > ε} →0 as m,n→∞.

• We say that (fn)n converges to f in measure, denoted fnµ−→ f , if for every

ε > 0 µ {|fn − f | > ε} → 0 as n→∞.

Example 14.2. Let us consider a few examples on (R,B, λ). fn = 1n1[0,n], gn = 1[n,n+1],

`n = n1[0,1/n] and

hn = 1[j2−k,(j+1)2−k] for n = 2k + j , 0 ≤ j < 2k.

(That is, k = blog2 nc and j = n− 2k.) So that

h1 = 1[0,1] h2 = 1[0,

12 ]

h3 = 1[12 ,1]

etc.

We have

fna.e.−→ 0 gn

a.e.−→ 0 `na.e.−→ 0.

148

However ∫|fn|dλ =

∫|gn|dλ =

∫|`n|dλ = 1.

So fn 6 L1

−→ 0, gn 6 L1

−→ 0, `n 6 L1

−→ 0.

As for hn, note that∫|hn|dλ = 2−blog2 nc. So hn

L1

−→ 0. However, (hn)n does not

converge pointwise, because for any x ∈ (0, 1) there are infinitely many n such that

hn(x) = 0 and infinitely many n for which hn(x) = 1.

Note that (gn)n is not Cauchy in measure. Indeed, |gn(x)− gm(x)| > ε if and only if

x ∈ [n, n+ 1]4[m,m+ 1]. So

λ {|gn − gm| > ε} = λ([n, n+ 1]4[m,m+ 1]) = 2.

It is also a sequence that does not converge in measure, as we will see later.

On the other hand, fnλ−→ 0, `n

λ−→ 0 and hnλ−→ 0: For all n > 1

ε we have that

|fn| ≤ 1n < ε, so λ {|fn| > ε} = λ(∅) = 0. `n(x) > ε if and only if x ∈ [0, 1/n] and n > ε.

So

limn→∞

λ {|`n| > ε} = limn→∞

λ([0, 1/n]) = 0.

|hn(x)| > ε if and only if x ∈ [j2−k, (j + 1)2−k] for k = blog2 nc and j = n− 2k. So

λ {|hn| > ε} = λ([j2−k, (j + 1)2−k]) = 2−blog2 nc → 0.

To sum up:

a.e. L1 measure

fn X X X

gn X X X

`n X X X

hn X X X

454

By these examples, it is not always true that a.e. convergence implies L1 convergence,

or vice-versa. Recall however the Dominated Convergence Theorem which actually

relates a.e. convergence to L1 convergence.

� Exercise 14.1. Show that if fna.e.−→ f and |fn| ≤ g ∈ L1 for all n then f ∈ L1 and

fnL1

−→ f .

149

Show that if fna.e.−→ f are bounded functions and the measure space is finite, then

fnL1

−→ f .

• Proposition 14.3 (L1 implies measure). If fnL1

−→ f then fn → µµf .

Proof. Set An,ε = {|fn − f | > ε}. Then,

ε · µ(An,ε) ≤∫An,ε

|fn − f |dµ ≤∫|fn − f |dµ→ 0.

ut

The converse is false as is seen by fn, `n above.

• Proposition 14.4 (a.e. implies measure). If fna.e.−→ f then fn → µµf .

Proof. For any ε > 0 let An,ε = {|fn − f | > ε}. Note that if x ∈ An,ε then fn(x) 6→ f(x).

So An,ε ⊂ N where N = {x : fn(x) 6→ f(x)}. Thus, µ(An,ε) ≤ µ(N) = 0. ut

The converse is false as shown by hn above.

However, we can still relate convergence in measure to a.e. convergence of a subse-

quence.

• Proposition 14.5. (fn)n is Cauchy in measure if and only if there is a measurable

function f such that fnµ−→ f .

Moreover, if fnµ−→ f then there exists a subsequence (fnk)k such that fnk

a.e.−→ f a.e.

as k →∞.

Proof. Suppose that (fn)n is Cauchy in measure. For any k there exists nk such that

for all n,m ≥ nk, µ{|fn − fm| > 2−k

}< 2−k. Let gk = fnk . Set

F = lim supk

{|fnk − fnk+1| > 2−k

}=⋂n

⋃k≥n

{|fnk − fnk+1| > 2−k

}.

Since for any n,

µ(F ) ≤ µ( ⋃k≥n

{|fnk − fnk+1| > 2−k

})≤∑k≥n

2−k = 2−n+1,

150

we have that µ(F ) = 0. Note that for any x 6∈ F we have that (gk(x))k is a Cauchy

sequence in C. So we may define f(x) = limk→∞ gk(x) for x 6∈ F and f(x) = 0 for

x ∈ F . We had an exercise proving that such a function is measurable. Since µ(F ) = 0

we have that fnk = gka.e.−→ f , and so gk

µ−→ f . But then,

µ {|fn − f | > ε} ≤ µ({|fn − gk| > ε/2}

⋃{|gk − f | > ε/2}

)≤ µ {|fn − gk| > ε/2}+ µ {|gk − f | > ε/2} → 0 as n, k →∞.

So fnµ−→ f .

For the other direction, if fnµ−→ f then for any ε > 0,

µ {|fn − fm| > ε} ≤ µ {|fn − f | > ε/2}+ µ {|fm − f | > ε/2} → 0 as m,n→∞.

So (fn)n is Cauchy in measure. ut

� Exercise 14.2. Show that if fnµ−→ f and fn

µ−→ g then f = g a.e.

Conclude that if (fn)n converges to f and to g in any one of the three modes (a.e.,

L1, measure), possibly different modes for f and for g, then f = g a.e.


If (fn)n converges to f and g then fnµ−→ f and fn

µ−→ g. So it suffices to work with

this assumption.

For any m > 0,

µ{|f − g| > 1

m

}≤ µ

{|f − fn| > 1

m

}+ µ

{|g − fn| > 1

m

}→ 0 as n→∞.

So

µ {f 6= g} ≤∑m

µ{|f − g| > 1

m

}= 0.

:) X

� Exercise 14.3. Let (X,F , µ) be a measure space. Let (fn)n be a sequence of

151

non-negative measurable functions, and let f be a measurable function such that (fn)n

converges to f in measure.

Show that ∫fdµ ≤ lim inf

n→∞

∫fndµ.


Let (fnk)k be a subsequence such that

limk→∞

∫fnkdµ = lim inf

n

∫fndµ.

So we want to show that∫fdµ ≤ limk→∞

∫fnkdµ. Since (fnk)k is a subsequence, we

have that for all ε > 0,

limk→∞

µ {|fnk − f | > ε} ≤ lim supn

µ {|fn − f | > ε} = 0.

So (fnk)k converges in measure to f .

Let gk := fnk for all k, which convergen in measure to f . By a theorem in class we

now have that there is a further subsequence (gkj )j such that limj→∞ gkj = f a.e. Since

these are all non-negative functions, Fatou’s Lemma tells us that∫fdµ ≤ lim inf

j

∫gkjdµ.

However, the sequence (∫gkjdµ)j is a subsequence of the converging sequence (

∫fnkdµ)k

which converges to lim infn∫fndµ. So the limit is∫

fdµ ≤ lim infj

∫gkjdµ ≤ lim

j→∞

∫gkjdµ = lim

k→∞

∫fnkdµ = lim inf

n

∫fndµ.

:) X

� Exercise 14.4. Let (X,F , µ) be a measure space. Let (fn)n be a sequence of

measurable functions, and let f be a measurable function such that (fn)n converges to

f in measure.

Suppose that g ∈ L1(X,F , µ) such that for all n, |fn| ≤ g.

152

Show that (fn)n converges to f in L1.


Since (fn)n converges to f in measure, there is a subsequence (nk)k such that (fnk)k

converges to f µ-a.e. Since |fn| ≤ g for all n, we get that |fnk−f | → 0 µ-a.e. as k →∞.

Also |f | = limk |fnk | ≤ g. Thus, |fn − f | ≤ 2g for all n.

Consider the sequence In :=∫|fn − f |dµ. We want to show that lim supn→∞ In = 0.

Let (fnk)k be a subsequence such that Ink → lim supn In as k →∞. Set hk := |fnk−f |.Since this is a subsequence, we have that for all ε > 0,

limk→∞

µ {hk > ε} ≤ lim supn→∞

µ {|fn − f | > ε} = 0.

So (hk)k converges to 0 in measure.

We saw in class that there exists a subsequence (hkj )j such that limj→∞ hkj = 0 µ-a.e.

Because (hkj )j is a subsequence of (hk)k = (|fnk − f |)k which is a subsequence of

(|fn − f |)n, we have that hkj ≤ 2g ∈ L1 for all j. Since (hkj )j converges to 0 a.e. and is

a dominated sequence, we have by the dominated convergence theorem that

limj→∞

∫hkjdµ = 0.

But by definition, (hkj )j is a subsequence of (|fnk − f |)k; so the sequence (∫hkjdµ)j is

a subsequence of the converging sequence (Ink)k. Thus these sequences have the same

limit which is

lim supn

In = limk→∞

Ink = limj→∞

∫hkjdµ = 0.

:) X

� Exercise 14.5. [Folland p.63] Let (X,F , µ) be a measure space. Let (An)n be a

sequence of measurable sets with finite measure. Suppose that 1AnL1

−→ f .

Show that f = 1A a.e. for some A ∈ F .

153

14.2. Uniform and almost uniform convergence

• Definition 14.6. Let (fn)n, f be a sequence of measurable functions on a measure

space (X,F , µ).

We say that (fn)n converges uniformly to f if supx |fn(x)−f(x)| → 0 as n→∞; that

is, for every ε > 0 there exists n0 such that for all n > n0 and any x, |fn(x)− f(x)| < ε.

We say that (fn)n converges uniformly to f on A if (fn1A)n converges uniformly to

f1A; that is supx∈A |fn(x)− f(x)| → 0 as n→∞.

We say that (fn)n converges almost uniformly to f if for every ε > 0 there exists

a measurable A such that µ(Ac) < ε and (fn)n converges uniformly to f on A.

••• Theorem 14.7 (Egoroff’s Theorem). Let (X,F , µ) be a finite measure space. Sup-

pose that fna.e.−→ f . Then (fn)n converges almost uniformly to f .

Proof. By augmenting fn, f on a set of measure 0 we may assume without loss of gen-

erality that fn(x)→ f(x) for every x.

Let

Bn,k =⋃m≥n

{|fm − f | > 1

k

}.

For any x ∈ Bk :=⋂nBn,k we have that for all n there exists m ≥ n such that

|fm(x) − f(x)| > 1k . That is, lim supn→∞ |fn(x) − f(x)| ≥ 1

k > 0. So fn(x) 6→ f(x).

That is, Bk ⊂ {fn 6→ f}.For fixed k, the sequence (Bn,k)n is decreasing. Since µ is a finite measure,

limn→∞

µ(Bn,k) = µ(Bk) ≤ µ {fn 6→ f} = 0.

For any ε > 0 and any k ≥ 1 let nk,ε be large enough so that µ(Bnk,ε,k) < ε2−k. Let

B =⋃k Bnk,ε,k. Then µ(B) ≤ ∑k µ(Bnk,ε,k) ≤ ε. For A = Bc we have that for any

η > 0 taking k = dη−1e, there exists n0 = nk,ε such that for all n ≥ n0 we have that for

all x ∈ A, |fn(x)− f(x)| ≤ 1k ≤ η. Thus, (fn)n converges uniformly to f on A.

To conclude: for any ε > 0 we can find A such that (fn)n converges uniformly to f

on A and µ(Ac) < ε. This is almost uniform convergence. ut

� Exercise 14.6. Show that if (fn)n converges almost uniformly to f then fna.e.−→ f .

154

� Exercise 14.7. A version of Egoroff’s theorem in non-finite settings:

Let (X,F , µ) be a measure space. Suppose that fna.e.−→ f . Suppose that |fn| ≤ g ∈ L1.

Show that (fn)n converges almost uniformly to f .

� Exercise 14.8. Suppose that (X,F , µ) is a σ-finite measure space. Suppose that

fna.e.−→ f .

Show that there exists a sequence of measurable sets (Ak)k, such that µ(⋂k A

ck) = 0

and such that for any k, (fn)n converges uniformly to f on Ak.

� Exercise 14.9. Show that if (fn)n are continuous functions into C on some Borel

measure space, and if (fn)n converges to f uniformly, then f is continuous as well.


Let ε > 0. Take n0 = n0(ε) so that for all n > n0 we have supx |fn(x)− f(x)| < ε/2.

Now, if (xk)k is any sequence xk → x, then fn(xk)→ fn(x) for all n. For all n > n0(ε),

|f(xk)− f(x)| ≤ |f(xk)− fn(xk)|+ |f(x)− fn(x)|+ |fn(xk)− fn(x)|

≤ 2 supx|f(x)− fn(x)|+ |fn(xk)− fn(x)| < ε+ |fn(xk)− fn(x)|.

Taking k → ∞ we have that for any ε > 0 lim supk |f(xk) − f(x)| ≤ ε. Thus, f is

continuous at x, for all x. :) X

155

• Lemma 14.8. Let A ⊂ R be a Lebesgue set of finite Lebesgue measure. For any ε > 0

there exists a continuous function ψ such that 0 ≤ ψ ≤ 1, ψ vanishes outside an interval

and ∫|ψ − 1A|dλ < ε.

Proof. For any interval I = [a, b], let

ψε,I(x) =

0 x 6∈ (a, b),

1 x ∈ [a+ ε, b− ε],x−aε x ∈ [a, a+ ε],

b−xε x ∈ [b− ε, b].

So ψε,I is continuous and vanishes outside I, and is 1 in [a+ ε, b− ε] and 0 ≤ ψ(x) ≤ 1

for x ∈ [a, b] \ [a+ ε, b− ε]. Thus,∫|1I − ψε,I |dλ ≤ 2ε.

If A ⊂ [a, b] then λ(A) = supA⊃K compact λ(K). Fix ε > 0. Choose a compact K ⊂ Aso that λ(K) ≥ λ(A)−ε. Now we may find a sequence of almost disjoint closed intervals

(i.e. with disjoint interiors) (In)n such that λ(In) ≤ λ(K) + ε and K ⊂ ⋃n In. We may

assume without loss of generality that In ∩K 6= ∅ for all n.

For every n let I ′n be the open ε · 2−n enlargement of In; that is, if In = [xn, yn] then

I ′n = (xn − ε · 2−n, yn + ε · 2−n). So K ⊂ ⋃n I′n is an open cover. So there exists n such

that K ⊂ ⋃nj=1 I

′j . Note that

λ(

n⋃j=1

I ′j) ≤n∑j=1

λ(I ′j) ≤ ε+

∞∑j=1

λ(In) ≤ λ(K) + 2ε.

For every j = 1, . . . , n let ψj = ψε·2−j ,Ij . So∫|ψj − 1Ij |dλ ≤ 2 · 2−j . Thus for

ϕ =∑n

j=1 1Ij and ψ =∑n

j=1 ψj we have that

∫|ψ − ϕ|dλ ≤

n∑j=1

∫|ψj − 1Ij |dλ ≤

n∑j=1

ε2−j ≤ ε.

156

Also, ∫|1A − ϕ|dλ ≤

n∑j=1

∫|1K∩In − 1In |dλ+

∫|1A − 1K |dλ

≤n∑j=1

λ(In \K) + λ(A \K) ≤ λ(n⋃j=1

Ij \K) + λ(A \K)

≤ λ(n⋃j=1

I ′j)− λ(K) + λ(A)− λ(K) ≤ 3ε.

Note that ψ is a function that vanishes outside⋃nj=1 Ij , which is contained in some

interval I = [a, b]. ut

� Exercise 14.10. Show that if f : R → C is Lebesgue and integrable then for

every ε > 0 there exists a continuous function ψ such that ψ vanishes outside a bounded

interval and ∫|f − ψ|dλ < ε.


If f =∑n

j=1 aj1Aj is a simple function, where (Aj)nj=1 are pairwise disjoint, the for

each j find a continuous function ψj vanishing outside some bounded interval such that∫|ψj − 1Aj |dλ < ε

naj. Then, for ψ =

∑nj=1 ajψj we have that

∫|f − ψ|dλ ≤

n∑j=1

aj

∫|1Aj − ψj |dλ < ε.

Note that ψ vanishes outside some interval (which just is the smallest interval containing

the union of the bounded intervals supporting the ψj ’s).

Now if f is any non-negative measurable function, let ϕn ↗ f be an approximating

sequence of simple functions. By Dominated Convergence,∫|ϕn − f |dλ → 0. So there

exist n large enough so that∫|ϕn−f |dλ < ε

2 . for this n let ψn be a continuous function

157

supported in a bounded interval such that∫|ϕn − ψn|dλ < ε

2 . Note that∫|ψn − f |dλ ≤

∫|ϕn − f |dλ+

∫|ψn − ϕn|dλ < ε.

Now if f is a general Lebesgue measurable function (into C), write f = f1 − f2 +

i(f3 − f4), for non-negative Lebesgue fj . Then choose continuous functions ψj that

are supported in a bounded interval and∫|fj − ψj |dλ < ε/4. Summing we have that

ψ = ψ1 − ψ2 + i(ψ3 − ψ4) is supported in a bounded interval and continuous, and∫|f − ψ|dλ ≤

∫ √|f1 − f2 − (ψ1 − ψ2)|2 + |f3 − f4 − (ψ3 − ψ4)|2dλ

≤∫|f1 − f2 − (ψ1 − ψ2)|dλ+

∫|f3 − f4 − (ψ3 − ψ4)|dλ

≤4∑j=1

∫|fj − ψj |dλ < ε.

:) X

••• Theorem 14.9 (Lusin’s Theorem). Let f : [a, b]→ C be Lebesgue. Let ε > 0. Then

there exists a compact set K ⊂ [a, b] such that λ([a, b] \K) < ε and f∣∣K

is continuous.

Proof. Let (ψn)n be a sequence of continuous functions each supported in a bounded

closed interval In ⊂ [a, b] such that∫|f −ψn|dλ < 1

n . Thus, ψnL1

−→ f , and so ψnλ−→ f .

Let (gk = ψnk)k be a subsequence such that gka.e.−→ f . Since λ([a, b]) <∞,

Egoroff’s Theorem tells us that (gk)k converges almost uniformly to f . That is, for

any ε > 0 there exists a Lebesgue set Aε such that λ([a, b]\Aε) < ε/2 and (gk)k converges

uniformly to f on Aε. Let Kε ⊂ Aε be a compact subset such that λ(Aε \Kε) < ε/2.

So λ([a, b] \Kε) ≤ λ([a, b] \Aε) + λ(Aε \Kε) < ε. Also, (gk)k converges uniformly to f

on Kε, so f is continuous on Kε. ut

Lusin’s Theorem is informally the statement that Lebesgue functions are almost con-

tinuous.



158

Measure Theory

Ariel Yadin

Lecture 15: Differentiation

We now turn to understanding differentiation in Rd. We will work throughout in the

measure space (Rd,Bd, λ = λd). B(x, r) denotes the open ball of radius r around x (in

L2-norm).

15.1. Hardy-Littlewood Maximal Theorem

A technical lemma:

• Lemma 15.1. Let U =⋃αBα be a union of a family of open balls Bα = B(xα, rα)

for every α. Then, for any c < λ(U) there exist finitely many pairwise disjoint balls

Bj = Bαj , j = 1, . . . , n, such that∑n

j=1 λ(Bj) > 3−dc.

Proof. There exists a compact K ⊂ U such that λ(K) > c, by inner regularity. Since

K ⊂ ⋃αBα is an open cover, there exists finitely many Aj = Bαj , j = 1, . . . ,m, that

cover the compact K, K ⊂ ⋃mj=1Aj .

Reorder A1, . . . , Am so that the radii are decreasing. Set B1 = A1 and for all j ≥ 2

let Bj be Ak for the smallest k such that Ak is disjoint from B1 ∪ · ∪Bj−1.

Suppose that Bj = B(xj , rj) for all j. These are disjoint by definition.

Now, if x ∈ Ak, the for some j we have that Ak ∩ Bj 6= ∅. If j is the smallest such

possible index, then Ak is disjoint from B1 ∪ · ∪ Bj−1, so the radius of Ak is at most

that of Bj ; otherwise we would have wanted to choose Ak instead of Bj . Thus, since

there is some x ∈ Ak ∩ Bj , we have that all elements in Aj are at distance at most

2rad(Ak) ≤ 2rj from x, and so Aj ⊂ B(x, 2rj) ⊂ B(xj , 3rj).

Since K ⊂ ⋃mj=1Aj ⊂

⋃nj=1B(xj , 3rj) we have that

c < λ(K) ≤n∑j=1

λ(B(xj , 3rj)) = 3dn∑j=1

λ(Bj).

ut

159

• Definition 15.2. A measurable function f : Rd → C is called locally integrable if

for every bounded measurable set B ⊂ Rd,∫B |f |dλ <∞.

L1loc = L1

loc(Rd,Bd, λ) denotes the space of locally integrable functions.

For any x ∈ Rd and r > 0 we define the average of f ∈ L1loc on B(x, r) by

(Arf)(x) =1

λ(B(x, r))

∫B(x,r)

fdλ.

• Proposition 15.3. For any f ∈ L1loc, Arf is jointly continuous in x, r.

Proof. Let xn → x, rn → r.

Then 1B(xn,rn)(x)→ 1B(x,r) for all x 6∈ B[x, r]\B(x, r). Since λ(B[x, r]\B(x, r)) = 0

we get that f1B(xn,rn)a.e.−→ f1B(x,r). Also, since f ∈ L1

loc, for all n such that |rn − r| <1,dist(xn, x) < 1 we have that |f1B(xn,rn)| ≤ |f1B(x,r+2)| ∈ L1. So by Dominated

Convergence,∫B(xn,rn)

fdλ→∫B(x,r)

fdλ and λ(B(xn, rn))→ λ(B(x, r)).

Thus, Arnf(xn)→ Arf(x). ut

This implies that Arf is a measurable function.

� Exercise 15.1. Let xn → x, rn → r. Show that 1B(xn,rn)(x) → 1B(x,r) for all

x 6∈ B[x, r] \B(x, r).

� Exercise 15.2. Show that λ(B[x, r]) = λ(B(x, r)).

• Definition 15.4. Given f ∈ L1loc define the Hardy-Littlewood maximal function

Mf(x) = supr>0

Ar|f |(x) = supr>0

1

λ(B(x, r))

∫B(x,r)

|f |dλ.

160

� Exercise 15.3. Show that Mf is measurable.

••• Theorem 15.5 (Hardy-Littlewood Maximal Theorem). For all α > 0 and all

f ∈ L1,

λ(Hf > α) = λ({x : Hf(x) > α}) ≤ 3d

α

∫|f |dλ.

Proof. Let U = {Hf > α}. For every x ∈ U there exists rx > 0 such that∫B(x,rx) |f |dλ >

α ·λ(B(x, rx)). Since U ⊂x∈U B(x, rx), for any c < λ(U) there exist x1, . . . , xn such that

for rj = rxj the balls (Bj := B(xj , rj))nj=1 are pairwise disjoint and

∑nj=1 λ(Bj) > 3−dc.

Thus,

c < 3dn∑j=1

λ(Bj) ≤ 3d1

α

n∑j=1

∫B(xj ,rj)

|f |dλ ≤ 3d

α·∫|f |dλ.

Taking supremum of c < λ(U) we obtain the result. ut

� Exercise 15.4. Recall for a function ψ : R→ R, the definitions

lim supx→y

ψ(x) := limε→0

sup0<|x−y|<ε

ψ(x) = infε>0

sup0<|x−y|<ε

ψ(x).

Show that lim supx→y |ψ(x) − c| if and only if for every sequence xn → y we have

ψ(xn)→ c. (In this case we say that limx→y ψ(x) = c.)

••• Theorem 15.6 (Basic Differentiation Theorem). For any f ∈ L1loc,

limr→0

Arf(x) = f(x) for λ-a.e. x.

Proof. First assume that f = f1B(0,R) for some R > 0. So f ∈ L1.

For every ε > 0 we can find a continuous function g such that∫|f − g|dλ < ε. So for

any x and δ > 0 there exists r > 0 such that if |y − x| < r then |g(x) − g(y)| < δ. For

161

this r > 0,

|Arg(x)− g(x)| ≤ 1

λ(B(x, r))

∫B(x,r)

|g(y)− g(x)|dλ(y) ≤ δ.

Thus, limr→0Arg(x) = g(x) for all x. We conclude that

lim supr→0

|Arf(x)− f(x)| ≤ lim supr→0

(|Ar(f − g)(x)|+ |Arg(x)− g(x)|+ |g(x)− f(x)|

≤ H(f − g)(x) + |f − g|(x).

So setting

Eα =

{x : lim sup

r→0|Arf(x)− f(x)| > α

},

we have that Eα ⊂ {Hf > α/2} ∪ {|f − g| > α/2}. We have

αλ({|f − g| > α}) ≤∫{|f−g|>α}

|f − g|dλ ≤∫|f − g|dλ < ε,

so by the Hardy-Littlewood Maximal Theorem,

λ(Eα) ≤ 3dε

α+

2ε

α→ 0 as ε→ 0.

Set E =⋃nE1/n. Then λ(E) = 0 and for any x 6∈ E, limr→0Arf(x) = f(x).

This proves the theorem for f that vanishes outside B(0, R).

For general f ∈ L1loc, then for all x ∈ B(x,R) and r < R we have that Arf(x) =

Arg(x) where g = f1B(0,2R). So for a.e. x ∈ B(0, R),

limr→0

Arf(x) = limR>r→0

Arf(x) = limr→0

Arg(x) = g(x) = f(x).

So for every R there is a set NR ⊂ B(0, R) such that λ(NR) = 0 and for x ∈ B(0, R)\NR

we have limr→0Arf(x) = f(x). Set N =⋃RNR. Then λ(N) = 0 and if x 6∈ N then

limr→∞Arf(x) = f(x). ut

15.2. The Lebesgue Differentiation Theorem

• Definition 15.7. For a function f ∈ L1loc define the Lebesgue set of f by

Lf =

{x : 1

λ(B(x,r))

∫B(x,r)

|f(y)− f(x)|dλ(y) = 0

}.

• Proposition 15.8. For any f ∈ L1loc, λ(Lcf ) = 0.

162

Proof. For any z ∈ Q+ iQ ⊂ C, we have a null set Nz such that for x 6∈ Nz, by applying

the Basic Differentiation Theorem to the function y 7→ |f(y)− z|,

limr→0

1

λ(B(x, r))

∫B(x,r)

|f(y)− z|fλ(y) = 0.

Let N =⋃z∈Q+iQNz. So λ(N) = 0.

Let x 6∈ N . Fix ε > 0. Let z ∈ Q+ iQ be such that |f(x)− z| < ε. So |f(y)− f(x)| ≤|f(y)− z|+ ε, and

1

λ(B(x, r))

∫B(x,r)

|f(y)− f(x)|dλ(y) ≤ 1

λ(B(x, r))

∫B(x,r)

|f(y)− z|dλ(y) + ε→ ε.

Taking ε→ 0 we have that x ∈ Lf .

So Lcf ⊂ N . ut

X A family (Ar)r of sets is said to shrink nicely to x if for all r > 0, Ar ⊂ B(x, r)

and λ(Ar) ≥ αλ(B(x, r)) for some constant α.

For example, if A ⊂ B(0, 1) is a Borel set of positive measure α = λ(A) > 0, then

rA+ x ⊂ B(x, r) and λ(rA+ x) = αrd ≥ cαλ(B(x, r)), so (rA+ x)r shrink nicely to x.

••• Theorem 15.9 (Lebesgue Differentiation Theorem). Let f ∈ L1loc and any x ∈ Lf ,

if (Ar)r shrink nicely to x then

limr→0

1

λ(Ar)

∫Ar

|f(y)− f(x)|dλ(y) = 0.

Proof. For some α > 0,

1

λ(Ar)

∫Ar

|f(y)− f(x)|dλ(y) ≤ 1

αλ(B(x, r)

∫B(x,r)

|f(y)− f(x)|dλ(y)→ 0.

ut

15.3. Differentiation and Radon-Nykodim derivative

We now make a connection between differentiation and the Radon-Nykodim deriva-

tive.

163

A measure µ on (Rd,Bd) is called regular if µ(K) <∞ for every compact K and for

every Borel A,

µ(A) = infA⊂U open

µ(U).

� Exercise 15.5. Let f : Rd → [0,∞] be a measurable function, and define dµ = fdλ;

i.e. µ(A) =∫A fdµ.

Show that µ is regular if and only if f ∈ L1loc.


If f ∈ L1loc then since any compact set K is bounded, we have that µ(K) =

∫K fdλ <∞.

Also, for any ball B, f1B ∈ L1, and so the measures µ, λ restricted to B are finite. Thus,

for any ε > 0 there exists δ > 0 so that if λ(A ∩ B) < δ then µ(A ∩ B) < ε. For any

Borel A ⊂ B, and any δ > 0 we can find an open set A ⊂ U ⊂ B such that λ(U \A) < δ.

So µ(U)− µ(A) = µ(U \A) < ε. This proves outer regularity for bounded A.

If A is unbounded, write A =⋃nAn where An is bounded for all n. Then, for any

ε > 0 take an open set Un ⊃ An such that µ(Un) ≤ µ(An) + ε2−n. So for U =⋃n Un

we have µ(U) ≤ µ(A) + ε.

For the other direction, if dµ = fdλ, then for any bounded Borel set A we may find

a compact set K containing A. So∫A fdλ ≤

∫K fdλ = µ(K) <∞. :) X

••• Theorem 15.10. Let µ be a regular Borel measure on (Rd,Bd). Let µ = σ + ρ be

the Lebesgue decomposition with respect to Lebesgue measure λ, so σ ⊥ λ and ρ� λ.

Then, for λ-a.e. x ∈ Rd,

limr→0

µ(Ar)

λ(Ar)=dρ

dλ(x),

for any family (Ar)r that shrinks nicely to x.

Proof. First note that if x ∈ Lf ,

limr→0

ρ(Ar)

λ(Ar)= lim

r→0

1

λ(Ar)

∫Ar

dρ

dλdλ(y) =

dρ

dλ(x),

by the Lebesgue Differentiation Theorem.

164

So we only need to show that for a.e. x,

limr→0

σ(Ar)

λ(Ar)= 0.

Note that since Ar ⊂ B(x, r),

σ(Ar)

λ(Ar)≤ σ(B(x, r))

αλ(B(x, r)),

so we may assume that Ar = B(x, r).

Let E be a Borel set such that σ(E) = 0 and λ(Ec) = 0. For a positive integer k set

Fk =

{x ∈ E : lim sup

r→0

σ(B(x,r))λ(B(x,r)) >

1k

}.

It suffices to prove that λ(Fk) = 0 for all k.

µ is outer regular, so for any ε > 0 there exists an open set Uε ⊃ E such that

σ(Uε) + λ(Uε) = µ(Uε) ≤ µ(E) + ε = λ(E) + ε,

which implies that σ(Uε) ≤ ε. For every x ∈ Fk there exists an open ball Bx ⊂ Uε

such that σ(Bx) > 1kλ(Bx). Set Vε =

⋃x∈Fk Bx ⊂ Uε. So for any c < λ(Vε) there exist

finitely many pairwise disjoint balls Bx1 , . . . , Bxn such that

c < 3dn∑j=1

λ(Bxj ) ≤ 3d 1k

n∑j=1

σ(Bxj ) ≤3d

kσ(Vε) ≤

3d

kε.

So σ(Fk) ≤ 3d

k ε. Since ε > 0 was arbitrary, we get that σ(Fk) = 0. ut

� Exercise 15.6. For f ∈ L1loc, define

H∗f(x) = sup

{1

λ(B)

∫B|f |dλ : B = B(y, r) , x ∈ B

}.

Show that Hf ≤ H∗f ≤ 2dHf .



165

Measure Theory

Ariel Yadin

Lecture 16: The Riesz Representation Theorem

16.1. Compactly supported functions

Let X be some topological space. Let Cc(X) denote the space of all functions f :

X → C that have supp(f) = cl({x : f(x) 6= 0}) is a compact set.

We will use the notation f ≺ U to denote f : X → [0, 1], supp(f) ⊂ U , f ∈ Cc(X).

• Definition 16.1. We say that a space X is Urysohn if for any x ∈ U where U is

open, there exists f ≺ U and an open set V such that x ∈ V ⊂ supp(f) with 1V ≤ f .

Example 16.2. Any locally compact Hausdorff space is Urysohn. (This is a conse-

quence of what is known as Urysohn’s Lemma.)

The notion of locally compact Hausdorff spaces is an important one in topology, but

it is beyond the scope of this course. 454

� Exercise 16.1. Show that Rd is Urysohn.


Let x ∈ U for U open. So there exists an open ball B = B(x, 3ε) such that x ∈B(x, 3ε) ⊂ U .

Define

f(y) =

1 if ||y − x|| ≤ ε,

2− ε−1 · ||y − x|| if ε ≤ ||y − x|| ≤ 2ε,

0 if ||y − x|| ≥ 2ε.

Since the function y 7→ ||y − x|| is continuous, so is f .

166

Since x ∈ B(x, ε) ⊂ clB(x, 2ε) = supp(f) ⊂ U and f∣∣B(x,ε)

≡ 1, we are done. :) X

� Exercise 16.2. Let X be a countable set. Consider X as a topological space

with the discrete topology (i.e. all subsets are open; alternatively, consider the metric

dist(x, y) = 0 for x = y and dist(x, y) = 1 for x 6= y.

Show that a subset K ⊂ X is compact if and only if K is finite.

Show that X is Urysohn.


If K is compact then K ⊂ ⋃x∈K {x} is an open cover, so there exists a finite sub-cover,

that is K ⊂ ⋃nj=1 {xj} for some x1, . . . , xn ∈ K. This implies that K = {x1, . . . , xn} is

finite.

For the other direction, if K is finite and K ⊂ ⋃α Uα is an open cover, then for every

x ∈ K there exists α = α(x) such that x ∈ Ux := Uα. So K ⊂ ⋃x∈K Ux is a finite

sub-cover. Every open cover has a finite sub-cover, so K is compact.

We now show that X is Urysohn. If x ∈ U for open U , then define f = 1{x}. Since

{x} is open and compact we have that x ∈ {x} ⊂ supp(f) ⊂ U . Finally, f is continuous

since for the discrete topology any function is continuous. :) X

� Exercise 16.3. Show that if f ∈ Cc(X) is a non-negative function then g := 1∧ fis compactly supported and continuous; g ∈ Cc(X), g : X → [0, 1].


It is immediate that supp(g) ⊂ supp(f), so we only need to show that g is continuous.

Two proofs:

167

Proof I. If U ⊂ R is an open set, then so is V = U\{1}. If 1 6∈ U then g−1(U) = f−1(U)

which is open. If 1 ∈ U then g(x) ∈ U if and only if either f(x) ∈ U or f(x) > 1. So

g−1(U) = f−1(U)⋃ {f > 1} which is also open.

Proof II. Let xn → x. If f(x) < 1 then because f is continuous, there exists n0 such

that for all n > n0 we have f(xn) < 1. Thus, limn→∞ g(xn) = limn→∞ f(xn) = f(x) =

g(x). If f(x) ≥ 1 then for any ε > 0 there exists n0 such that for all n > n0 we have

|f(xn) − f(x)| < ε, which implies that f(xn) ≥ 1 − ε. This implies that g(xn) ≥ 1 − εfor all n > n0. So lim infn g(xn) ≥ 1 − ε. Since this holds for all ε > 0, we have that

lim supn g(xn) ≤ 1 ≤ lim infn g(xn) so limn→∞ g(xn) = 1 = g(x). :) X

• Proposition 16.3. Let X be Urysohn.

Let K ⊂ U for compact K and open U . Then there exists f ∈ Cc(X) such that

1K ≤ f ≺ U .

Proof. For any x ∈ K ⊂ U we can find an open set Vx and a function fx ≺ U such that

x ∈ Vx ⊂ supp(fx) and 1Vx ≤ f .

Since K is compact and K ⊂ ⋃x∈K Vx is an open cover, we have a finite sub-cover;

that is, there exist x1, . . . , xm such that K ⊂ ⋃mj=1 Vxj .

Set f :=∑m

j=1 fxj . Then 1K ≤∑m

j=1 1Vxj ≤ f and supp(f) ⊂ ⋃nj=1 supp(fxj ) ⊂ U .

So f ∈ Cc(X) and non-negative. Thus, 1 ∧ f ≺ U and 1K ≤ 1 ∧ f . ut

• Proposition 16.4. Let X be an Urysohn space. Let K ⊂ ⋃nj=1 Uj for compact K

and open (Uj)nj=1. Then, there exist functions fj ≺ Uj such that

∑nj=1 fj(x) = 1 for all

x ∈ K.

Proof. For every x ∈ K there exists j such that x ∈ Uj . So we can find a function

gx ≺ Uj and an open set Vx such that x ∈ Vx ⊂ supp(gx) ⊂ Uj and 1Vx ≤ gx.

Since K is compact and K ⊂ ⋃x∈K Vx is an open cover, we have that there exists a fi-

nite sub-cover; i.e. there exist x1, . . . , xm ∈ K such that K ⊂ ⋃mi=1 Vxi ⊂

⋃mi=1 supp(gxi).

For every j set

Kj =⋃{supp(gxi) : supp(gxi) ⊂ Uj} .

Then Kj is a finite union of compact sets, so it is compact. Also, by definition, Kj ⊂ Uj .

168

Thus, we may find a function gj ∈ Cc(X) such that 1Kj ≤ gj ≺ Uj .Let g =

∑nj=1 gj . Since for every xi there exists j such that supp(gxi) ⊂ Uj , then K ⊂⋃m

i=1 supp(gxi) ⊂⋃nj=1Kj . So for all x ∈ K we have that g(x) ≥ 1. Let V = {g > 0}

which is an open set because g is continuous. Since K ⊂ V , we can find h ∈ Cc(X) such

that 1K ≤ h ≺ V .

Define f = g+1−h. Since supp(h) ⊂ {g > 0}, we have that f(x) > 0 for all x. Thus,

we can define fj :=gjf .

For x ∈ K, since h(x) = 1, then∑n

j=1 fj(x) =∑n

j=1gj(x)g(x) = 1. Also, supp(fj) ⊂

supp(gj) ⊂ U and so fj ≺ U . ut

16.2. Linear functionals

• Definition 16.5. Let F be some space of functions f : X → C. A linear functional

I on F is a function I : F → C such that I(αf+g) = αI(f)+I(g) for all α ∈ C, f, g ∈ F .

A linear functional is called positive if for any f ∈ F that is non-negative, we have

I(f) ≥ 0.

X Recall that f ≥ 0 means that f is non-negative. Also, f ≥ g means that f − g ≥ 0.

� Exercise 16.4. Let (X,F , µ) be a measure space. Show that the function

I(f) =∫fdµ is a positive linear functional on L1(X,F , µ).

• Proposition 16.6. Let X be an Urysohn topological space. Let I be a positive linear

functional on Cc(X). Then, for any compact K ⊂ X there exists a constant CK > 0

such that for any f ∈ Cc(X) with supp(f) ⊂ K we have I(f) ≤ CK · ||f ||∞ (recall

||f ||∞ = supx∈X |f(x)|).

Proof. For general f ∈ Cc(X), since f = Ref + iImf we have |I(f)| ≤ |I(Ref)| +|I(Imf)|, so it suffices to consider real-valued f .

169

Given a compact K, Let ϕK ∈ Cc(X) be a function ϕK : X → [0, 1] such that ϕ∣∣K≡ 1

(X is always open, so this is possible by Urysohn).

If supp(f) ⊂ K then |f(x)| ≤ ϕK(x) · ||f ||∞ for any x ∈ X. So ||f ||∞ · ϕK − f ≥ 0

and ||f ||∞ ·ϕK− (−f) ≥ 0 ||f ||∞ · I(ϕK)− I(f) ≥ 0 and ||f ||∞ · I(ϕK)+ I(f) ≥ 0, which

is |I(f)| ≤ I(ϕK) · ||f ||∞. ut

16.3. The Riesz Representation Theorem

• Definition 16.7. Let X be a topological space. A Borel measure on X is a measure

on (X,B(X)).

A Borel measure µ is called Radon if

• For any compact K ⊂ X we have µ(K) <∞.

• µ is outer regular; i.e. for all Borel sets A,

µ(A) = infA⊂U open

µ(U).

• µ is inner regular on open sets; i.e. for any open set U ,

µ(U) = supU⊃K compact

µ(K).

X The notation f ≺ U is short for f ∈ Cc(X), f : X → [0, 1] and supp(f) ⊂ U .

??? THEOREM 16.8 (Riesz Representation Theorem). Let X be an Urysohn space.

Let I be a positive linear functional on Cc(X). Then, there exists a unique Radon

measure µ on X such that I(f) =∫fdµ for all f ∈ Cc(X).

Moreover, the measure µ satisfies

µ(U) = sup {I(f) : f ≺ U} ∀ open U,

µ(K) = inf {I(f) : f ∈ Cc(X) , f ≥ 1K} ∀ compact K.

Proof. The statement of the theorem itself suggest how to define µ: Define

µ(U) := sup {I(f) : f ≺ U}

for any open U ⊂ X.

170

Step I. µ is countably sub-additive on open sets; i.e. if U =⋃n Un for open sets

(Un)n, we claim that µ(U) ≤∑n µ(Un).

Indeed, for any f ∈ Cc(X) such that f ≺ U , we have that K = supp(f) is compact

and K ⊂ ⋃n Un is an open cover. So there exists a finite sub-cover K ⊂ ⋃nj=1 Uj . We

may now find g1, . . . , gn such that gj ≺ Uj and∑n

j=1 gj(x) = 1 for all x ∈ K. This

implies that f =∑n

j=1 fgj and fgj ≺ Uj . So by definition of µ,

I(f) =n∑j=1

I(fgj) ≤n∑j=1

µ(Uj) ≤∑n

µ(Un).

Since this holds for any f ≺ U , we get that µ(U) ≤∑n µ(Un).

For an arbitrary subset A ⊂ X define

µ∗(A) = infA⊂U open

µ(U).

(*) Exercise: Show that µ∗(U) = µ(U) for all open U .

Step II. µ∗ is an outer measure.

Recall the axioms of an outer measure: µ∗(∅) = 0, µ∗(A) ≤ µ∗(B) for all A ⊂ B and

µ∗(⋃nAn) ≤∑n µ

∗(An).

The first two are easy. Now for the third axiom.

Since a countable union of open sets is open, we have that

µ∗(A) = inf

{∑n

µ(Un) : A ⊂⋃n

Un , Un are all open

}.

(*) Exercise: prove this.

Let A =⋃nAn. If µ∗(An) = ∞ for some n there is nothing to prove. So assume

that µ∗(An) < ∞ for all n. Fix ε > 0 and for every n let Un be an open set such that

An ⊂ Un and µ(Un) ≤ µ∗(An) + ε · 2−n. Then, A ⊂ ⋃n Un and

µ∗(A) ≤∑n

µ(Un) ≤∑n

µ∗(An) + ε.

Taking ε→ 0 completes the proof that µ∗ is an outer measure (step II).

Step III. Every open set U is µ∗-measurable.

Recall that U is µ∗-measurable if and only if for any set A such that µ∗(A) <∞, we

have µ∗(A) ≥ µ∗(A ∩ U) + µ∗(A ∩ U c). If A is such a set, fix some ε > 0 and let V be

171

an open set such that A ⊂ V and µ(V ) ≤ µ∗(A) + ε. Since V ∩U is open, by definition

of µ there exists f ∈ Cc(X), f ≺ V ∩ U , such that I(f) ≥ µ(V ∩ U) − ε. Also, the set

K = supp(f) is closed so V ∩Kc is open. So we may find g ∈ Cc(X), g ≺ V ∩Kc, such

that I(g) ≥ µ(V ∩Kc)−ε. Now note that supp(f)∩ supp(g) = ∅ because supp(g) ⊂ Kc.

So 0 ≤ f + g ≤ 1 and supp(f + g) ⊂ K ∪ (V ∩Kc) ⊂ V . Thus, by definition of µ,

µ∗(A) ≥ µ(V )− ε ≥ I(f + g)− ε = I(f) + I(g)− ε

≥ µ(V ∩ U) + µ(V ∩Kc)− 3ε ≥ µ(A ∩ U) + µ(A ∩ U c)− 3ε.

Taking ε→ 0 completes the proof that U is µ∗-measurable (step III).

Step IV. By Charatheodory’s Theorem µ∗ restricted to the Borel sets is a measure.

Denote this measure by µ.

For any Borel set A we have a sequence of open sets (Un)n such that µ(Un)→ µ(A)

and A ⊂ Un for all n. So µ is outer-regular.

Step V. We show that µ(K) = inf {I(f) : f ≥ 1K} for all compact K.

If K is compact and f ≥ 1K then for any ε > 0 let Uε = {f > 1− ε}. Then Uε

is open (because f is continuous). For any g ≺ Uε we have that f > (1 − ε)g, so

I(f) ≥ (1 − ε)I(g). Taking supremum over all g ≺ Uε we have that µ(K) ≤ µ(Uε) ≤(1 − ε)−1I(f). Since this holds for all ε, taking ε → 0 we get that µ(K) ≤ I(f). This

was for any f ≥ 1K , so µ(K) ≤ inf {I(f) : f ≥ 1K}For the other inequality, let U ⊃ K be any open set. Then we can find fU ∈ Cc(X),

such that f∣∣K≡ 1 and fU ≺ U . So fU ≥ 1K and I(fU ) ≤ µ(U). Taking infimum, we

have that

inf {I(f) : f ≥ 1K} ≤ inf {I(fU ) : K ⊂ U open } ≤ inf {µ(U) : K ⊂ U open } = µ(K).

This proves µ(K) = inf {I(f) : f ≥ 1K} for all compact K (step V).

Step VI. µ(K) < ∞ for all compact K. Indeed, there exists f ∈ Cc(X) such that

f : X → [0, 1] and f∣∣K≡ 1. So f ≥ 1K , and I(f) ≤ CK · ||f ||∞ = CK for some constant

CK > 0. Thus, µ(K) ≤ I(f) <∞.

172

Step VII. µ is inner regular on open sets. Indeed, if U is an open set, for any

α < µ(U) let f ∈ Cc(X), f ≺ U be such that I(f) ≥ α. Set K := supp(f) which is

compact and K ⊂ U .

For any g ∈ Cc(X) such that g ≥ 1K we have that g − f ≥ 0, so I(g) ≥ I(f) ≥ α.

Taking infimum over all such g we have that µ(K) ≥ α.

Thus, for any α < µ(U) we have a compact K ⊂ U such that µ(K) ≥ α. Taking

supremum over α gives inner regularity (step VII).

Step VIII. We now show that I(f) =∫fdµ for all f ∈ Cc(X).

If f ∈ Cc(X) write f = f1−f2 +i(f3−f4) for non-negative fj , so it suffices to consider

non-negative f . Since f ∈ Cc(X) is continuous on a compact set, namely supp(f), it

attains a maximum there. So ||f ||∞ < ∞. So replacing f by 1||f ||∞ f , we may assume

without loss of generality that f : X → [0, 1].

Fix n > 0. For any 1 ≤ j ≤ n set Kj = {f ≥ j/n} and K0 = supp(f). For 1 ≤ j ≤ ndefine

fj =1

n∧(f − j−1

n

)+.

That is,

fj(x) =

0 f(x) ≤ j−1

n ,

f(x)− j−1n f(x) ∈ [ j−1

n , jn ].

1n f(x) ≥ j

n .

With this definition 1n1Kj ≤ fj ≤ 1

n1Kj−1 , so 1nµ(Kj) ≤

∫fjdµ ≤ 1

nµ(Kj−1).

Note that fj is continuous, and supp(fj) ⊂ Kj−1, so fj ∈ Cc(X) (this includes the

case j = 1 where K0 = supp(f)).

For any open set U ⊃ Kj−1, we have that nfj ≺ U . So nI(fj) ≤ µ(U). Taking

infimum over all such open sets U ⊃ Kj−1, by outer regularity we have that I(fj) ≤1nµ(Kj−1).

Also, nfj ≥ 1Kj , so 1nµ(Kj) ≤ I(fj).

173

fj + j−1n

jn

j−1n

Figure 5. The function fj raised by j−1n .

Now, note that f =∑n

j=1 fj . (Exercise!) So by additivity of the integral and the

linear functional,

1

n

n∑j=1

µ(Kj) ≤ I(f) ≤ 1

n

n∑j=1

µ(Kj−1) and

1

n

n∑j=1

µ(Kj) ≤∫fdµ ≤ 1

n

n∑j=1

µ(Kj−1).

Taking the differences between these inequalities,

∣∣I(f)−∫fdµ

∣∣ ≤ 1

n

n∑j=1

(µ(Kj−1)− µ(Kj)) =µ(K0)− µ(Kn)

n≤ 1

nµ(supp(f)).

µ is finite on compact sets (as a Radon measure), so µ(supp(f)) < ∞. Thus, taking

n→∞ we have that I(f) =∫fdµ as required (step VIII).

Step IX. Uniqueness. Let ν be another Radon measure satisfying I(f) =∫fdν for

all f ∈ Cc(X). For any open set U , if f ≺ U then I(f) =∫U fdν ≤ ν(U). This holds

for any f ≺ U so

ν(U) ≥ sup {I(f) : f ≺ U} .

174

Since ν is inner regular on open sets, we can find a sequence of compact sets (Kn)n such

that Kn ⊂ U and ν(Kn) → ν(U). For every n, Urysohn guaranties that there exists a

function fn ∈ Cc(X) such that fn∣∣Kn≡ 1 and fn ≺ U . So

ν(Kn) ≤∫Kn

fndν ≤∫fndν = I(fn) ≤ sup {I(f) : f ≺ U} .

Taking n→∞ we get that ν(U) = sup {I(f) : f ≺ U} = µ(U).

So ν(U) = µ(U) for all open sets U . Since ν, µ are outer regular, this implies that

they are equal on all Borel sets. ut

� Exercise 16.5. Let X be an Urysohn space. Let F ⊂ X be a closed subset. Let

µ be a Radon measure on F . Define I(f) =∫f∣∣Fdµ for all f ∈ Cc(X).

Show that I is a positive linear functional on Cc(X).

Show that if I(f) =∫fdν for the Radon measure ν guarantied by the Riesz Repre-

sentation Theorem, then ν(A) = µ(A ∩ F ) for all Borel A.



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