1 Introduction

Definition 1.1 Let X be a set. An algebra A X is a collection of subsetsof X such that

A E A Ec X\E A A,B A A B = A

A algebra F is an algebra which is closed under the formation of count-able unions: E1, E2, E3, . . .

i=1Ei F

Definition 1.2 A measurable space is a pair (X,F) where X is a set andF is a -algebra of subsets of X.Elements of F are called measurable sets.

Definition 1.3 Let (X,F) be a measurable space. A measure on (X,F) isa function : (X,F) [0,] such that

() = 0 is additive : ifA1, A2, A3, . . . F are pairwise disjoint then(i=1(Ai)) =

i=1 (Ai)

Proposition 1.0.1 (basic properties of measures) Let (X,F, ) be a mea-sure space.

1. Additivity:A1, A2, . . . , An F and are disjoint, (

ni=1(Ai)) =

ni=1 (Ai)

2. Monotonicity:A,B F and A B (A) (B)

3. Inclusion-Exclusion Principle:If A,B F and (AB)

Proposition 1.0.2 Let (X,F, ) be measure space, and assume that E1, E2, . . . F.

1. if Ei E then (E) = limi (Ei)2. if Ei E and there exist i0 such that (Ei0) < , then (E) =

limi (Ei)

Definition 1.5 Let X be a set. A collection S 2X is called a semi-algebra(of subset of X) if

1. ,X S2. A,B S A B S3. For all A S,Ac can be written as a finite pairwise disjoint union of

elements of S.

Proposition 1.0.3 The collection of intervals Int is a semi-algebra of sub-sets of R (it is not a algebra)

Definition 1.6 Let l : Int [0,] be the length function given byl(< a, b >) := b a, where < a, b > represents all possibilities of intervals

Lemma 1.1 1. finite additivity of l:if I =

Nk=1 Ik where I, Ik Int, then l(I) =

Nk=1 l(Ik)

2. subadditivity : if I k=1 IkwhereI, Ik Int, then l(I) k=1 l(Ik)

Proposition 1.0.4 l : Int [0,] is -additive in Int: if I1, I2, . . . Intand I =

k=1 Ik then l(I) =

k=1 l(Ik)

Definition 1.7 (Lebesgue Outer Measure) This is the function m :2R [0,] given by

m(A) = inf

{ j=1

l(Ij) : A j=1

Ij , whereIj Int}

Definition 1.8 Let X be some set. An outer measure on X is a function : 2X [0,] such that:

1. () = 0

2. monotonicity : A B (A) (B)3. subadditivity : (i=1Ai) i=1 (Ai)

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Remark An outer measure is in general not a measure.

Proposition 1.0.5 Lebesgue outer measure is an outer measure

Definition 1.9 A set E R is called Lebesgue measurable if it satisfies thefollowing condition:

For all T R,m(T ) = m(T E) +m(T Ec)This condition is called Caratheodorys Criterion.

Remark

1. T is called a test set

2. Caratheodorys Criterion is symmetric (E Ec)3. The inequalitym(T ) m(TE)+m(TEc) is always true because

of the sub-additivity of m.

Proposition 1.0.6 The Lebesgue Outer Measure of an interval is its length.Furthermore if {Ii}i are pairwise disjoint intervals thenm(

i Ii) =

i l(Ii)

Proposition 1.0.7 The set B0 = {E R : E is Lebesgue measurable} isan algebra.

Remark Sketch Proof:The main part is to show E1, E2 B0 E1 E2 B0 i.e. we needm(T ) = m(T (E1 E2)) +m(T (E1 E2)c)

We already have by sub additivity. To show we note thatm((T Ec1)Ec2) = m(T Ec1)m((T Ec1)E2) by using T

= (T Ec1)

as the test set.

m(T(E1E2)) = m(T(E1unionmulti(Ec1E2))) m(TE1)+m(T(Ec1E2))m(T (E1 E2)c) = m((T Ec1)Ec2) = m(T Ec1)m((T Ec1)E2)

Adding them together givesm(T (E1E2)+m(T (E1E2)c) m(T E1)+m(T Ec1) = m(T )Lemma 1.2 Suppose E1 . . . En are Lebesgue measurable and pairwise dis-joint. Then for every T R,m(T ni=1Ei) =ni=1m(T Ei) ()Proof We use induction on n. For n = 1 there is nothing to prove. Sup-pose (**) holds for n. Let E1 . . . En+1 B0 be pairwise disjoint. Thenm(T n+1i=1 Ei) = m(T n+1i=1 Ei En+1) +m(T n+1i=1 Ei Ecn+1)= m(T En+1) +m(T

ni=1Ei) =

n+1i=1 m

(T Ei)

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Remark From this lemma we can see that m is finitely additive on B0by taking T = R.

Proposition 1.0.8 B0 is a algebra

Idea of proof We already know that B0 is an algebra so we need it to beclosed under countable set unions i.e. {Ei}i=1 B0

i=1Ei B0

The trick is to construct a set of pairwise disjoint sets {Ej}j=1 by definingE

j := Ej\

j1i=1 Ei B0

So then we have:

1. Ei E

j = for i 6= j

2.i=1Ei =

i=1E

i

So we may assume that {Ei}i=1 B0are pairwise disjoint. Estimatei=1Ei

withni=1Ei.

Then use the previous lemma when you consider in m(T ) = m(T ni=1Ei) +m

(T\ni=1Ei). Then take n to at the end. subadditivity of m gives .

Proposition 1.0.9 m is additive on B0

Proof Using the same notation as the previous proposition: We need toshow m(

i=1Ei) =

i=1m

(Ei)We show as follows: m(i=1Ei) m(ni=1Ei) = ni=1m(Ei), thentake n to

Proposition 1.0.10 Every interval is Lebesgue measurable.

Definition 1.10 m|B0 is called Lebesgues measure on R

Theorem 1.1 There exists a algebra B0 of subsets of R, and a set func-tion m : B0 [0,) such that

1. B0 contains the intervals

2. m is additive on B0

3. m(interval)= length of the interval

This theorem puts the previous propositions together

Proposition 1.1.1 If E B0 and x R then m(x+ E) = m(E)

Definition 1.11 A measure space (X,F, ) is called finite if there existsFi F such that X =

i=1 Fi and (Fi)

Remark (X,F, ) is called a probability space if (Fi) = 1.

Proposition 1.1.2 (R,B0,m) is finite

Definition 1.12 Let (X,F, ) be a measure space

A X is called negligible or null or measure zero set ifinf{(E) : E F, E A} = 0

(X,F, ) is called complete if if all negligible sets are in F

Proposition 1.1.3 (R,B0,m) is complete

Idea of proof Firstly show A R is null m(A) = 0. Then showm(T ) = m(T A) +m(T Ac).

Lemma 1.3 (Lindelof) Every open subset of R is equal to a countableunion of open intervals

Idea of proof Take your open set U and for every x U find open intevals(x , x + ) that covers each x. Then use completeness of R to find evensmaller intervals (x x, x + x) where x, x Q. Then use countabilityof Q to show countable union.

Definition 1.13 The smallest algebra of subsets of R which contains allopen sets is called the Borel algebra and is denoted by B(R). Elements ofB(R) are called Borel sets.

Lemma 1.4 All intervals, open sets, closed sets are Borel sets.

Proposition 1.1.4 Open sets and closed sets are Lebesgue measureable.

Remark If we can show B(R) B0 then we are done.

Proof We show B(R) B0.Define F := {E B(R) : E is Lebesgue measurable}. We claim the follow-ing:

1. open sets F2. F is a algebra

Remark This implies that B(R) F because by definition, B(R) is thesmallest algebra which contains the open sets.

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Proof of claim

1. Take U R open. By Lindelof there are open intervals {Ik}k=1 suchthat U =

k=1 Ik. Intervals are Lebesgue measurable, and B0 is a

algebra.Therefore U B0. Open sets B(R) so U B(R) and thusU F.

2. F trivial E F Ec F: because E F E B(R) B0 (but sinceB(R) and B0 are algebras.) Ec B(R) B0 Ec F.

{Ei}i=1 F i=1Ei F : because {Ei}i=1 F For all i

Ei B(R) B0 i=1Ei B(R) B0

i=1Ei F

Theorem 1.2 (Regularity) If E B0 then for all > 0 there are setF E U such that F is closed, U is open and m(U\F ) < . Moreover,if m(E) 0 then RHS =, andif m(A1) = 0, then RHS = 0. In either case RHS 6= 1

Step 3 in the proof given in lectures provide us such Ak. These Ak aretranslates of a fixed set A. But Step 2 gives translation invariance. Thusall Ak would have the same measure if they are Lebesgue measurable. Thenwe get the contradiction above.

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Notation: If f : X R is a function, then[f < t] := {x X: f(x) < t},[a f < b] := {x X: a f(x) < b},[f B] := {x X: f(x) B} etc.

Definition 1.14 Let (X,F) be a measurable space.A function f : X [,] is called F measurable if for every t R,[f < t] F.

Examle In the case when X = RIf F = B0, f is called Lebesgue measurableIf F = B(R), f is called Borel measurable

Proposition 1.3.1 Let (X,F) be a measurable space. The following areequivalent:

1. f : X R is F measurable2. For all t ( [f > t] F)3. For all a, b ( [a f < b] F)4. For all B B(R), f1(B) F and f1(), f1() F

Proposition 1.3.2 Suppose f1, f2, . . . , fn : X R are F-measurable. If{(f1(x), f2(x), . . . , fn(x)) R : x X} U where U Rn is open, and if : U R, where : (t1, t2, . . . , tn) 7 (t1, t2, . . . , tn) is continious, thenx 7 (f1(x), f2(x), . . . , fn(x)) is F-measurable

Notation If {an} is a sequence a limit point is a limit of a subsequence {ank} limnan = sup{limitpoints} = max{limitpoints} limnan = inf{limitpoints} = min{limitpoints}Note that lim = lim = lim when limit of {an} exists.

Proposition 1.3.3 If {fn}n=1 are F-measurable, then the following func-tions are again F-measurable:

1. f(x) := supn{fn(x) : n 1},f(x) := infn{fn(x) : n 1},

2. f(x) := limnfn(x) = limnsup

3. f(x) := limnfn(x) = limninf

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4. fc(x) :={ limnfn(x) when limit exists

c if limit does not exist

5. f(x) :=

i=1 fn(x) (if sum converges)f(x) :=

i=1 fn(x) (if product converges)

Examples of measurable functions

1. Definition 1.15 Let A be a set. The indicator function of A is definedas:

1A(x) :={ 1 x A

0 x 6 AThe indicator function is B(R) measurable if and only if A is Borel.

This is true because [1A t] ={ t 0Ac 0 < t < 1R t 1

2. Every continuous function f : [a, b] R is Borel measurable. This isbecause [f < t] is open for all t thus [f < t] B(R)

Remark If a function is Borel measurable then it also is Lebesgue mea-surable

Definition 1.16 Let (X,F) be a measurable space.A simple function is f : X R of the form f(x) := ni=1 i1Ei(x) whereEi F and 1Ei is the indicator of Ei

Caution This is not the same as a step function.

Theorem 1.4 Basic Approximation: Let (X,F) be a measurable space.Then f : X R is bounded measurable if and only if > 0 a simplefunction : X R such that |f(x) (x)| < x X

Terminology Let (X,F, ) be a measure space, and P some property ofx X. We say that property holds almost everywhere if {x X|property P fails}=0

Examples

1. fn(x) f(x) almost everywhere {x|fn(x) 6 f(x)}=02. f(x) = g(x) almost everywhere [f 6= g] = 0

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Proposition 1.4.1 If f : [a, b] [,] is Lebesgue measurable and fi-nite a.e, then , there is a cts function : [a, b] R for which m{x [a, b] : |f(x) (x)| > } <

(i.e the set where f(x) is not close to (x), is small)

Theorem 1.5 (Egoroff) Suppose fn : E R are Lebesgue measurableand that E B0 has finite measure. Assume that fn(x) f(x) a.e in E.Then > 0 K E compact s.t m(E\K) < and fn f uniformly onK.

i.e N s.t (n > N x K we have |fn(x) f(x)| < )

Theorem 1.6 (Lusin) If f : E R is Lebesgue measurable, and E B0has finite measure, then > 0 K E compact s.t m(E\K) < andf : K R is uniformly cts

i.e s.t (x, x K, |x x | < |f(x) f(x)| < )

Reminder:Regulated integral

1. A step function is of the form f =n

i=1 i.1Ii where {Ii} are intervals.2. For a step function:

f :=

i.l(Ii)

3. A regulated function is a uniform limit of step function

4. If f is regulated, thenf := limn

fn where fn f uniformly

and fn are step functions.

Lebesgue Integration

1. A simple function is f =n

i=1 i.1Ei where {Ei} are measurable2. If f is simple then

f :=

ni=1 im(Ei)

It is natural to defineE f :=

ni=1 i.m(E Ei)

3. A bounded measurable is a uniform limit of simple functions.

4. So if f is bounded measurable, take fn simple such that fn funiformly and define

f := limn

fn

Definition 1.17 A canonical representation of a simple fuction f isf(x) :=

yf(R) y.1[f=y](x)

Definition 1.18 If f is a (Lebesgue measurable) simple function and E B0 of finite measure, then the integral of f over E (w.r.t m) isE fdm :=

yf(R) y.m(E [f = y])

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Proposition 1.6.1 If f , g are simple and E B0 has finite measure, then:1. Linearity:

E(f + g) =

E f +

E g

2. Monotonicity: f g E f E g3. Absolute value: | E f | E |f |4. f = g a.e on E E f = E g5. If E = E1 unionmulti E2 with E1, E2 B0, then

E f =

E1f +

E2f

Integrating Bounded Measurable Functions

Proposition 1.6.2 Let f : E R be bounded measurable and E B0 havefinite measure.There exists simple functions n : E R such that n funiformly on E and

{E n}n=1 always converges. its limit is independent of the choice of n

Definition 1.19 The integral of f over E is defined to beE f := limn

E n

Proposition 1.6.3 (Properties of the integral) Suppose f,g are boundedmeasurable, and that E B0 has finite measure.

1. Linearity:E(f + g) =

E f +

E g

2. Monotonicity: f g E f E g3. Absolute value: | E f | E |f |4.E f.1G =

EG f for all G B0

5. f = g a.e on E E f = E g6. If E = E1 unionmulti E2 with E1, E2 measurable, then

E f =

E1f +

E2f

Theorem 1.7 Suppose {fn}n=1 are measurable, E B0 has finite measure,and that there exists M such that |fn(x)| M for all n N, x E. Iffn f(x) for all x E, then

E fn

E f

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Integrating unbounded functions

Definition 1.20 Suppose f : E [0,] measurable, E B0(not neces-sarily of finite measure).

ThenE fdm := sup

{[h 6=0] h

0 h f.1Em[h 6= 0] h bounded Lebesgue measurable

}Proposition 1.7.1 Suppose f, g : R [0,] are measurable and E B0(not necessarily of finite measure).Then we have:

1. linearity: , 0 E(f + g) = E f + E g2. monotonicity: f g E f E g3.E f.1G =

EG f for all G B0

4. f = g a.e E f = E gTheorem 1.8 (Fatous Lemma) Suppose {fn}n=1 are non-negative mea-surable functions.Then for every E B0

E limnfn limn

E fn

Lemma 1.5 If f : R [0,] is measurable, and E f

Proposition 1.8.1 Suppose f, g are absolutely integrable on E B01. linearity: for all , R f+g is also integrable on E and E(f+

g) = E +

E g

2. monotonicity: f g E f E g3.E f.1G =

EG f for all G B0

4. f = g almost everywhere on E E f = E g5. if E = E1 unionmulti E2 where E1, E2 B0, then

E f :=

E1f +

E2f

Theorem 1.9 (Lebesgue Dominated Convergence Theorem (DCT))Let{fn}n=1 be a sequence of measurable functions such that fn(x) f(x)for all x E B0. If |fn(x)| g(x) where g is absolutely integrable on E,then

E limnfn = limn

E fn

Product Measures

Definition 1.23 Let (X,C, ), (Y,D, ) be measure spaces. A measure rect-angle is a set of the form C D where C C, D D. Denote by R thecollection of measurable rectangles.

Proposition 1.9.1 R is a semi-algebra

Proposition 1.9.2 The algebra generated by R isA(R) := {ni=1Ci Di| Ci C, Di D, n N}Definition 1.24 : A(R) [0,] is defined as (ni=1Ci Di) :=n

i=1 (Ci)(Di) (here 0. = 0)

Proposition 1.9.3 This is

1. properly defined:ni=1Ci Di =

ni=1C

i D

i

ni=1 (Ci)(Di) =n

i=1 (Ci)(D

i)

2. additive on A(R)

3. is finite (if are finite)

Definition 1.25 The product algebra C D is the minimal algebrawhich contains {C D : C C, D D}

Theorem 1.10 Let (X,C, ), (Y,D, ) be two finite measure spaces. Thereexists a unique measure called the product measure on (X Y,CD)such that ( )(C D) = (C)(D) for all C C, D D. This measureis finite.

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Definition 1.26 The completion of (,F, ) is denoted (,F0, 0) where

1. F0 = {E0 | E F such that (E4E0) = 0}2. 0 := |F0

Note (R,B0,m) is the completion of (R,B(R),m)

Definition 1.27 (Lebesgue Measure on Rn) This is the measure space(Rn, (B0 . . .B0

n

)0, (m . . .m n

)0)i.e the completion of the product spaceni=1(R,B0,m)

Definition 1.28 Let E X Y and x X, y Y1. the x-section of E is Ex := {y Y |(x, y) E}2. the y-section of E is Ey := {x X|(x, y) E}

Remark .x-sections are subsets of Y , y-sections are subsets of X,(A B)x = Ax Bx, (A B)x = Ax Bx, (Ac)x = (Ax)c,(A\B)x = Ax\Bx, (A)x = (A)x etc..

Proposition 1.10.1 If E C D, then for all x X (Ex D), for ally Y (Ey C)

Proposition 1.10.2 Suppose (X,C, ), (Y,D, ) are finite. If E CD,then

1. x 7 (Ex) is C-measurable andX (Ex)d(x) = ( )(E)

2. y 7 (Ey) is D-measurable and Y (Ey)d(y) = ( )(E)Theorem 1.11 Let (X,C, ), (Y,D, ) be finite measure spaces, if E CD, then

1. x 7 (Ex), y 7 (Ey) are measurable.2.X (Ex)d(x) = ( )(E) =

Y (E

y)d(y)

Theorem 1.12 (Fubinis Theorem) Supose (X,C, ), (Y,D, ) are fi-nite and that f : XY [,] is CD measurable and |f |d

3. x 7 Y f(x, y)d(y) is C measurable and -absolutely integrable.y 7 X f(x, y)d(x) is D measurable and -absolutely integrable.

4.X

( Y f(x, y)d(y)

)d(x) =

XY fd =

Y

( X f(x, y)d(x)

)d(y)

Theorem 1.13 (Tornellis Theorem) Suppose (X,C, ), (Y,D, ) are finite and that f : XY [,] is CD measurable. If X ( Y |f |d)d < then XY fd

Lp spaces

Definition 1.29 A function : [a, b] R is called convex if x, y [a, b],t [0, 1)

(tx+ (1 t)y) t(x) + (1 t)(y)Proposition 1.13.1 is convex in [a, b] if and only if a x < y z