Fundamental Theorem of Calculus in Lebesgue Theory€¦ · Math 212a (Fall 2017) Yum-Tong Siu 1 Fundamental Theorem of Calculus in Lebesgue Theory For Riemann integration there are

Math 212a (Fall 2017) Yum-Tong Siu 1

Fundamental Theorem of Calculus inLebesgue Theory

For Riemann integration there are the following two parts in the funda-mental theorem of calculus. The first part is that for a continuous functionf on the open interval (a, b) in R and for a ∈ (a, b),

d

dx

∫ x

a

f(t)dt = f(x)

for x ∈ (a, b). The second part is that for a function f on on the open interval(a, b) in R and for a ∈ (a, b) whose derivative f ′(x) is continuous on (a, b)and for a < b in (a, b), ∫ b

a

f ′(x)dx = f(b)− f(a).

In Lebesgue’s theory of integration the two parts of the fundamental theoremof calculus hold for more general functions. The first part is that for anintegrable function f on the open interval (a, b) in R and for a ∈ (a, b), thefunction

x 7→∫ x

a

f

is differentiable almost everywhere on (a, b) and is equal to f almost every-where on (a, b), where the integral

∫ x

af in the sense of Lebesgue is interpreted

as −∫ a

xf when x < a. The second part is that for an absolutely continuous

function f on the open interval (a, b) in R and for a < b in (a, b), the deriva-tive f ′(x) exists almost everywhere on [a, b] and is an integrable function on[a, b] and ∫ b

a

f ′(x)dx = f(b)− f(a).

Here the meaning of f being absolutely continuous on (a, b) is as follows.Given any ε > 0 there exists δε > 0 such that for any finite collection ofdisjoint open subintervals (a1, b1), · · · , (ak, bk) of (a, b),

k∑j=1

|f(bj)− f(aj)| < ε

whenever∑k

j=1(bj − aj) < δε.


In the first part of the fundamental theorem of calculus, we first have toprove that the indefinite integral of an integrable function is differentiablealmost everywhere. Since an integrable function f can be written as thedifference of two nonnegative integrable functions f+ and f−, to prove thealmost everywhere differentiability of an indefinite integral of an integrablefunction, it suffices to do the proof for the special case where the integrandis a nonnegative integrable function so that the indefinite integral is nonde-creasing function.

The method to prove the almost everywhere differentiability of an in-creasing function to use the technique of cumulative effects of errors appliedto the errors of not all four derivates (or Dini numbers) being equal which arethe lim inf and lim sup of the left-hand and right-hand difference quotients.The tool used to apply this technique of cumulative effect of errors is Vitali’scovering technique which we now discuss and which seeks to cover, up to anyprescribed positive number, a set E of positive exterior measure m∗(E) bya finite number of disjoint intervals which are selected from the given set ofopen intervals, with each point of E being the left end-point of at least oneof them. These open intervals will be the ones occurring in the differencequotients which describe the non-coincidence of all four derivates.

Intuitive Discussion of Vitali’s Covering Lemma and Proof. Before we rigor-ously present and prove Vitali’s covering lemma, we first intuitively explainwhat it is about. When every point x of a set E of positive exterior mea-sure in R is assigned some positive numbers {rx}, a finite disjoint union ofopen intervals (x, rx) can be selected whose union covers E with an error lessthan a prescribed positive number. First, the general case is reduced to thecase of E bounded and infx∈E sup rx > 0. We choose x sufficiently close toinf E and replace E by E − (x, rx) to repeat the argument which must endin an a priori finite number of steps (which depends on the bound of E andinfx∈E sup rx. The sufficient closeness used in each of the finite number ofsteps depends on this a priori finite number (of steps). We now begin thedetailed rigorous treatment of the Vitali covering technique.

Exterior Measure of Limit of Increasing Sets. In the case of Ek ↗ E inR, the conclusion m∗(E) = limk→∞m∗(Ek) holds also without the assump-tion on the measurability of Ek. The idea of the verification is to reduceit back to the case of a limit of increasing measurable sets by approximat-ing, in exterior measure, each Ek by an open superset. Since Ek ⊂ E, the


inequality limk→∞m∗(Ek) ≤ m∗(E) is trivial. To get the other inequalitym∗(E) ≤ limk→∞m∗(Ek), we take an arbitrary ε > 0. Choose an open setOk containing Ek with m∗(Ok) ≤ m∗(Ek) + ε. Let Sk =

∩ℓ≥k Oℓ so that

Sk ⊂ Sk+1. Let S =∪

k∈N Sk. Then each Sk (being a countable intersectionof open sets) is measurable and Sk ↗ S so thatm(S) = limk→∞m(Sk). FromEk ⊂ Sk it follows that E ⊂ S and

m∗(E) ≤ m(S) = limk→∞

m(Sk) ≤ limk→∞

(m∗(Ek) + ε) = ε+ limk→∞

m∗(Ek).

From the arbitrariness of ε > 0, we conclude that m∗(E) ≤ limk→∞m∗(Ek).

Vitali’s Covering by Disjoint Intervals. Let E be an arbitrary set in R withfinite exterior measure m∗(E) <∞. Assume that to every x ∈ E is assigneda nonempty subset Ax of the set R+ of all positive numbers. We are going toprove that for every ε > 0 a finite number of points x1, · · · , xk of E and atthe same time elements rj of Axj

can be selected (for 1 ≤ j ≤ k) such that

(i) the k intervals (xj, xj + rj) for 1 ≤ j ≤ k are disjoint and

(ii) the exterior measure of the intersection E ∩(∪k

j=1(xj, xj + rj))with

E of the union of these k disjoint intervals is ≥ m∗(E)− ε.

Proof. The first step is to reduce the general case to the case of E beingbounded. For n ∈ N let En be the set of points of x of En such thatx ∈ [−n, n] and supAx ≥ 1

n. Then En ↗ E as n → ∞. By the above result

on the exterior measure of the limit of a countable union of increasing sets,we can find nE ∈ N such that m∗ (EnE

) > m∗(E)− ε2. This step replaces E

by the bounded set EnEwhen ε

2is used instead of ε.

Take δ > 0 which we will later specify more precisely. Pick x1 ∈ EnE

with x1 < inf EnE+ δ. Choose r1 ∈ Ax1 with r1 ≥ 1

2nE. If EnE

⊂ (−∞, x1 +r1), then we stop, otherwise we pick x2 ∈ EnE

− (−∞, x1 + r1) with x2 <inf (EnE

− (−∞, x1 + r1)) + δ. Choose r2 ∈ Ax2 with r2 ≥ 12nE

.

Inductively, suppose we have chosen x1, · · · , xν and r1, · · · , rν . If EnE⊂

(−∞, xν + rν), then we stop, otherwise we pick xν+1 ∈ EnE− (−∞, xν + rν)

withxν+1 < inf (EnE

− (−∞, xν + rν)) + δ.


Choose rν+1 ∈ Axν+1 with rν+1 ≥ 12nE

. Since xν+1 ≥ xν + rν and each

rν ≥ 12nE

and since EnE⊂ [−nE, nE], it follows

−nE + (ν − 1)1

2nE

≤ xν ≤ nE

and ν must be ≤ 4(nE)2 + 1 for xν to be chosen. Let k be the largest ν for

which xν can be chosen. Let E ′ be the intersection of EnEand the union of

the k disjoint open intervals∪

1≤ν≤k(xν , xν + rν). Then

m∗(E′) ≥ m∗(EnE

)− kδ

≥ m∗(EnE)− (4(nE)

2 + 1)δ

≥ m∗(E)−ε

2− (4(nE)

2 + 1)δ

and it suffices to choose

δ <ε

2 ((4(nE)2 + 1) + 1)

at the beginning to get m∗(E′) ≥ m∗(E)− ε. Q.E.D.

This procedure in Vitali’s covering technique we describe as cover bydisjoint open intervals to the right up to exterior measure < ε.

Intuitive Description of Using Vitali Covering Technique to Prove AlmostEverywhere Differentiability of Monotone Function. The Vitali coveringtechnique is introduced to prove the almost everywhere differentiability ofa monotone function. Before we give the detailed rigorous presentation, wewould like to first intuitively explain how the proof works for a nondecreasingfunction f on [a, b]. As described earlier, Vitali’s covering technique enablesus to use the technique of cumulative effects of errors applied to the errorsof not all four derivates (or Dini numbers) being equal which are the lim infand lim sup of the left-hand and right-hand difference quotients.

Let D+f(x0) be the lim inf of the right difference quotient at x0 and letD+f(x0) be the lim sup of the right difference quotient of f at x0. Sim-ilarly denote by Let D−f(x0) (respectively D−f(x0)) be the lim inf (re-spectively lim sup) of the right difference quotient of f at x0. We needto show that for almost all x0 the four derivates (lower right derivate, upperright derivate, lower left derivate, upper left derivate) D+f(x0), D

+f(x0),


D−f(x0), D−f(x0) agree. For given 0 < α < β rational, we derive a con-

tradiction from the positive exterior measure of the set E of all x withD+f(x0) < α < β < D+f(x0). To prove that four numbers agree, weneed to verify three statements of equality. The other three similar state-ments needed follow by similar arguments. Choose rx > 0 (respectivelysx > 0) with difference quotient of f on [x, rx] < α (respectively on [x, x+sx]> β). Apply Vitali’s covering lemma to get (xj, xj + rxj

) and then to eachE∩ (xj, xj + rxj

) to get (xj,k, xj,k+sxj,k) ⊂ (xj, xj + rxj

) to get the contradic-

tion that the f -increment on ∪j

(xj, xj + rxj

)is < α while the f -increment

on its nonempty subset ∪j,k

(xj,k, xj,k − sxj,k

)is > β.

We now present the detailed rigorous proof of the almost everywheredifferentiability of a monotone function. We introduce the following notationin Vitali’s covering procedure of covering a set E of exterior measure zeroby disjoint open intervals to the right up to exterior measure less than aprescribed positive number ε. We denote by A the map x 7→ Ax ⊂ Rand denote by Vright,ε(E,A) the disjoint union

∪1≤ν≤k(xν , xν + rν) of open

intervals.

Likewise, we can perform this procedure to get a cover by disjoint openintervals to the left up to exterior measure < ε, namely, under the sameassumption we can select for any given ε > 0 a finite number of pointsx′1, · · · , x′k′ of E and at the same time elements r′j of Axj

(for 1 ≤ j ≤ k′)such that

(i) the k′ intervals (x′j − r′j, x′j) for 1 ≤ j ≤ k′ are disjoint and

(ii) the exterior measure of the intersection E ∩(∪k′

j=1(x′j − r′j, x

′j))with

E of the union of these k′ disjoint intervals is ≥ m∗(E)− ε.

To verify the statement (with “right” replaced by “left”), we can just straight-forwardly modify the above argument or use the transformation negR : x 7→−x for x ∈ R to apply the above argument to the image of E under negRand then get the images of the disjoint open intervals under negR.

We denote by Vleft,ε(E,A) the disjoint union∪k′

j=1(x′j − r′j, x

′j) of open

intervals.

In our application of Vright,ε(E,A) we will use an open subset O of Ewith m(O) < m∗(E)+ ε such that (x, x+ r) is contained in O for x ∈ E and


r ∈ Ax. Then Vright,ε(E,A) is contained in O and the measure m∗(E′) of the

intersection E ′ of Vright,ε(E,A) with E is no more than m(O) < m∗(E) + ε.In such a case

m∗(E)− ε ≤ m∗(E′) ≤ m (Vright,ε(E,A)) < m∗(E) + ε.

Likewise, in our application of Vleft,ε(E,A) we will use an open subset O of Ewith m(O) < m∗(E)+ ε such that (x− r, x) is contained in O for x ∈ E andr ∈ Ax. Then Vleft,ε(E,A) is contained in O and the measure m∗(E

′′) of theintersection E ′′ of Vleft,ε(E,A) with E is no more than m(O) < m∗(E) + ε.In such a case

m∗(E)− ε ≤ m∗(E′′) ≤ m (Vleft,ε(E,A)) < m∗(E) + ε.

Application of Vitali’s Covering Argument. For a function f(x) defined onsome open interval in R containing x0 we define

D+f(x0) = lim infx→x0+

f(x)− f(x0)

x− x0, D+f(x0) = lim sup

x→x0+

f(x)− f(x0)

x− x0,

D−f(x0) = lim infx→x0−

f(x)− f(x0)

x− x0, D−f(x0) = lim sup

x→x0−

f(x)− f(x0)

x− x0

so that f ′(x0) exists if and only if the four numbers D+f(x0), D+f(x0),

D−f(x0), D−f(x0) are equal.

Let O be an open subset of R containing x0. For γ ∈ R we define the sets

A+(x0, γ,O) =

{r > 0

∣∣∣ f(x0 + r)− f(x0)

r< γ and (x0, x0 + r) ⊂ O

},

A+(x0, γ,O) =

{r > 0

∣∣∣ f(x0 + r)− f(x0)

r> γ and (x0, x0 + r) ⊂ O

},

A−(x0, γ,O) =

{r > 0

∣∣∣ f(x0 − r)− f(x0)

−r< γ and (x0 − r, x0) ⊂ O

},

A−(x0, γ,O) =

{r > 0

∣∣∣ f(x0 − r)− f(x0)

−r> γ and (x0 − r, x0) ⊂ O

}.

In other words, the four sets A+(x0, γ,O), A+(x0, γ,O), A−(x0, γ,O), andA−(x0, γ,O) can be described as follows.


(i) the set A+(x0, γ,O) is the set of all r > 0 such that (x0, x0 + r) ⊂ Oand the difference quotient of the function f at the point x0 and thepoint x0 + r (to the right of x0) is < γ.

(ii) The set A+(x0, γ,O) is the set of all r > 0 such that (x0, x0 + r) ⊂ Oand the difference quotient of the function f at the point x0 and thepoint x0 + r (to the right of x0) is > γ.

(iii) The set A−(x0, γ,O) is the set of all r > 0 such that (x0 − r, x0) ⊂ Oand the difference quotient of the function f at the point x0 and thepoint x0 − r (to the left of x0) is < γ.

(iv) The set A−(x0, γ,O) is the set of all r > 0 such that (x0 − r, x0) ⊂ Oand the difference quotient of the function f at the point x0 and thepoint x0 − r (to the left of x0) is > γ.

The key setup is that we have the following four statements.

(1) When D+f(x0) < γ, by definition for every ε > 0 the set A+(x0, γ,O)contains some positive number < ε.

(2) When D+f(x0) > γ, by definition for every ε > 0 the set A+(x0, γ,O)contains some positive number < ε.

(3) When D−f(x0) < r, by definition for every ε > 0 the set A−(x0, γ,O)contains some positive number < ε.

(4) When D−f(x0) > r, by definition for every ε > 0 the set A−(x0, γ,O)contains some positive number < ε.

Suppose we have a subset E of R of finite exterior measure m∗(E) <∞.

(a) We denote by A+(γ,O) the collection x 7→ A+(x, γ,O) ⊂ R+ for x ∈ Eto get Vright,ε(E,A+(γ,O)) of disjoint intervals.

(b) We denote by A+(γ,O) the collection x 7→ A+(x, γ,O) ⊂ R+ for x ∈ Eto get Vright,ε(E,A+(γ,O)) of disjoint intervals.

(c) We denote by A−(γ,O) the collection x 7→ A−(x, γ,O) ⊂ R+ for x ∈ Eto get Vleft,ε(E,A−(γ,O)) of disjoint intervals.

(d) We denote by A−(γ,O) the collection x 7→ A−(x, γ,O) ⊂ R+ for x ∈ Eto get Vleft,ε(E,A−(γ,O)) of disjoint intervals.


Almost Differentiability of Non Decreasing Function. Let f be a nondecreas-ing function on (a, b). We are going to prove that f is almost everywheredifferentiable on (a, b) in the following steps.

(i) For 0 ≤ α < β rational, the set of points where D+f < α < β < D+fis of measure zero. Thus, D+f ≥ D+f almost everywhere on (a, b).

(ii) For 0 ≤ α < β rational, the set of points where D+f < α < β < D−fis of measure zero. Thus, D+f ≥ D−f almost everywhere on (a, b).

(iii) For 0 ≤ α < β rational, the set of points where D−f < α < β < D+fis of measure zero. Thus, D−f ≥ D+f almost everywhere on (a, b).

Suppose we have already verified these three steps (i), (ii), and (iii). Thenthe following commutative diagram

D+f = D+f≥ ≤

D−f ≥ D−f

holds almost everywhere on (a, b), because

(1) the top horizontal line comes from Step (i) and the automatic inequalityD+f ≤ D+f ,

(2) the left vertical line comes from Step (ii),

(3) the right vertical line comes from Step (iii), and

(4) the bottom horizonal line is an automatic inequality.

Starting from the upper left corner of the commutative diagram and followingit in the counter-clockwise sense, we get

D+f ≥ D−f ≥ D−f ≥ D+f = D+f

almost everywhere on (a, b), which means that

D+f = D−f = D−f = D+f

almost everywhere on (a, b) and f ′ exists almost everywhere on (a, b). Wenow verify the three steps (i), (ii), and (iii).


The verification of the three steps (i), (ii), (iii) is simply reproducing thesimple intuitive proof, but in rigorous details with extensive use of notations.

Verification of Step (i). Let Eα,β be the set of points x ∈ (a, b) withD+f(x) <α < β < D+f(x). Assume that m∗ (Eα,β) > 0 and we are going to derivea contradiction. Choose an arbitrary ε > 0 on which we will add someconditions later. Let O be an open subset of R which contains Eα,β suchthat m(O) < m∗(Eα,β) + ε. Consider Vright,ε(Eα,β,A+(α,O)). Let E ′

α,β beEα,β ∩ Vright,ε(Eα,β,A+(α,O)). Then

(∗) m∗(Eα,β)−ε < m∗(E ′

α,β

)≤ m (Vright,ε(Eα,β,A+(α,O))) ≤ m∗(Eα,β)+ε.

Let O′ = Vright,ε(Eα,β,A+(α,O)). Now consider Vright,ε(E′α,β,A+(β,O′)).

Then

(∗∗) m∗(E′α,β)− ε < m

(Vright,ε(E

′α,β,A+(β,O′))

).

For x ∈ Eα,β and r ∈ A+(x, α,O) the image of (x, x + r) under the non-decreasing function f is an interval (which may be open, half-closed, orclosed) whose length is < αr. Hence there is the following inequality for themeasure of the image f (Vright,ε(Eα,β,A+(α,O))) of Vright,ε(Eα,β,A+(α,O))under the map f .

m (f (Vright,ε(Eα,β,A+(α,O)))) < αm (Vright,ε(Eα,β,A+(α,O))) .

Likewise, for x ∈ Eα,β and r ∈ A+(x, β,O′) the image of (x, x+ r) under thenon-decreasing function f is an interval (which may be open, half-closed, orclosed) whose length is > βr. Hence there is the following inequality for themeasure of the image f (Vright,ε(Eα,β,A+(β,O′))) of Vright,ε(Eα,β,A+(β,O′))under the map f .

m(f(Vright,ε(Eα,β,A+(β,O′))

))> βm

(Vright,ε(Eα,β,A+(β,O′))

).

SinceVright,ε(Eα,β,A+(β,O′)) ⊂ O′ = Vright,ε(Eα,β,A+(α,O)),

it follows that after the application of the map f the inclusion

f(Vright,ε(Eα,β,A+(β,O′))

)⊂ f (Vright,ε(Eα,β,A+(α,O)))

holds and


))≤ m (f (Vright,ε(Eα,β,A+(α,O)))) ,


which implies that

βm(Vright,ε(Eα,β,A+(β,O′))

)< αm (Vright,ε(Eα,β,A+(α,O))) .

From (∗) and (∗∗) we have

βm (m∗(Eα,β)− 2ε) < αm (m∗(Eα,β) + ε) ,

which leads to a contradiction when ε > 0 is chosen so small that

β (m∗(Eα,β)− 2ε) > α (m∗(Eα,β) + ε))

holds from m∗(Eα,β) > 0 and α < β.

The verification of Step (ii) and Step (iii) are similar and we give thecompletely analogous details below.

Verification of Step (ii). Let Eα,β be the set of points x ∈ (a, b) withD+f(x) < α < β < D−f(x). Assume that m∗ (Eα,β) > 0 and we aregoing to derive a contradiction. Choose an arbitrary ε > 0 on which we willadd some conditions later. Let O be an open subset of R which containsEα,β such that m(O) < m∗(Eα,β) + ε. Consider Vright,ε(Eα,β,A+(α,O)). LetE ′

α,β be E ∩ Vright,ε(Eα,β,A+(α,O)). Then

(∗)′ m∗(Eα,β)−ε < m∗(E′α,β) ≤ m (Vright,ε(Eα,β,A+(α,O))) ≤ m∗(Eα,β)+ε.

LetO′ = Vright,ε(Eα,β,A+(α,O)). Now consider Vleft,ε(E′α,β,A−(β,O′)). Then

(∗∗)′ m∗(E′α,β)− ε < m

(Vleft,ε(E

′α,β,A−(β,O′))

).

For x ∈ Eα,β and r ∈ A+(x, α,O) the image of (x, x + r) under the non-decreasing function f is an interval (which may be open, half-closed, orclosed) whose length is < αr. Hence there is the following inequality for themeasure of the image f (Vright,ε(Eα,β,A+(α,O))) of Vright,ε(Eα,β,A+(α,O))under the map f .

m (f (Vright,ε(Eα,β,A+(α,O)))) < αm (Vright,ε(Eα,β,A+(α,O))) .

Likewise, for x ∈ Eα,β and r ∈ A−(x, β,O′) the image of (x − r, x) underthe non-decreasing function f is an interval (which may be open, half-closed,or closed) whose length is > βr. Hence there is the following inequality for


the measure of the image f (Vleft,ε(Eα,β,A−(β,O′))) of Vleft,ε(Eα,β,A−(β,O′))under the map f .

m(f(Vleft,ε(Eα,β,A−(β,O′))

))> βm

(Vleft,ε(Eα,β,A−(β,O′))

).

SinceVleft,ε(Eα,β,A−(β,O′)) ⊂ O′ = Vright,ε(Eα,β,A+(α,O)),


f(Vleft,ε(Eα,β,A−(β,O′))

)⊂ f (Vright,ε(Eα,β,A+(α,O)))

holds and

m(f(Vleft,ε(Eα,β,A−(β,O′))

))≤ m (f (Vright,ε(Eα,β,A+(α,O)))) ,

which implies that

βm(Vleft,ε(Eα,β,A−(β,O′))

)< αm (Vright,ε(Eα,β,A+(α,O))) .

From (∗)′ and (∗∗)′ we have

βm (m∗(Eα,β)− 2ε) < αm (m∗(Eα,β) + ε) ,


β (m∗(Eα,β)− 2ε) > α (m∗(Eα,β) + ε))


Verification of Step (iii). Let Eα,β be the set of points x ∈ (a, b) withD−f(x) < α < β < D+f(x). Assume that m∗ (Eα,β) > 0 and we are goingto derive a contradiction. Choose an arbitrary ε > 0 on which we will addsome conditions later. Let O be an open subset of R which contains Eα,β

such that m(O) < m∗(Eα,β) + ε. Consider Vleft,ε(Eα,β,A−(α,O)). Let E ′α,β

be E ∩ Vleft,ε(Eα,β,A−(α,O)). Then

(∗)′′ m∗(Eα,β)−ε < m∗(E ′

α,β

)≤ m (Vleft,ε(Eα,β,A−(α,O))) ≤ m∗(Eα,β)+ε.

LetO′ = Vleft,ε(Eα,β,A−(α,O)). Now consider Vright,ε(E′α,β,A+(β,O′)). Then

(∗∗)′′ m∗(E′α,β)− ε < m

(Vright,ε(E

′α,β,A+(β,O′))

).


For x ∈ Eα,β and r ∈ A−(x, α,O) the image of (x − r, x) under the non-decreasing function f is an interval (which may be open, half-closed, orclosed) whose length is < αr. Hence there is the following inequality forthe measure of the image f (Vleft,ε(Eα,β,A−(α,O))) of Vleft,ε(Eα,β,A−(α,O))under the map f .

m (f (Vleft,ε(Eα,β,A−(α,O)))) < αm (Vleft,ε(Eα,β,A−(α,O))) .

Likewise, for x ∈ Eα,β and r ∈ A+(x, β,O′) the image of (x, x+ r) under thenon-decreasing function f is an interval (which may be open, half-closed, orclosed) whose length is > βr. Hence there is the following inequality for themeasure of the image f (Vright,ε(Eα,β,A+(β,O′))) of Vright,ε(Eα,β,A+(β,O′))under the map f .


))> βm

(Vright,ε(Eα,β,A+(β,O′))

).

SinceVright,ε(Eα,β,A+(β,O′)) ⊂ O′ = Vleft,ε(Eα,β,A−(α,O)),


f(Vright,ε(Eα,β,A+(β,O′))

)⊂ f (Vleft,ε(Eα,β,A−(α,O)))

holds and


))≤ m (f (Vleft,ε(Eα,β,A−(α,O)))) ,

which implies that

βm(Vright,ε(Eα,β,A+(β,O′))

)< αm (Vleft,ε(Eα,β,A−(α,O))) .

From (∗)′′ and (∗∗)′′ we have

βm (m∗(Eα,β)− 2ε) < αm (m∗(Eα,β) + ε) ,


β (m∗(Eα,β)− 2ε) > α (m∗(Eα,β) + ε))


This finishes the proof that the derivative f ′(x) of a non-decreasing func-tion f(x) on (a, b) exists almost everywhere on (a, b).


Differentiation of Indefinite Integral in the Fundamental Theorem of Calcu-lus. Now we would like to show that for an integrable function on (a, b) witha < x0 ≤ a < b < b the derivative of

x 7→∫ x

x0

f

is equal to f(x) almost everywhere on (a, b). This is one half of the funda-mental theorem of calculus in Lebesgue’s theory of integration. By writingf = f+ − f− with f+(x) = max(f(x), 0) and f−(x) = max(−f(x), 0), we canassume without loss of generality that f ≥ 0 on (a, b). Let

F (x) =

∫ x

x0

f.

Then F (x) is continuous. As a matter of fact F (x) is absolutely continuousin the sense that for any ε > 0 there exists δε > 0 such that∣∣∣∣∫

E

f

∣∣∣∣ < ε

whenever E is a measurable subset of (a, b) with m(E) < δε. In particular,

k∑j=1

|F (bj)− F (aj)| < ε

whenever∪k

j=1(aj, bj) is a disjoint union of intervals in (a, b), because

k∑j=1

|F (bj)− F (aj)| =∫∪

1≤j≤k(aj ,bj)

f.

Since F is nondecreasing, for any εn ↘ 0 as n→ ∞, when Fatou’s lemma isapplied to the sequence of nonnegative functions

x 7→ F (x+ εn)− F (x)

εn

for n ∈ N, we get∫ b

a

lim infn→∞

F (x+ εn)− F (x)

εn≤ lim inf

n→∞

∫ b

a

F (x+ εn)− F (x)

εn.


Also from the nondecreasing property of F (x) we know that F ′(x) existsalmost everywhere on (a, b) and

lim infn→∞

F (x+ εn)− F (x)

εn= F ′(x)

for almost all x ∈ (a, b). We compute∫ b

a

F (x+ εn)− F (x)

εn

by shifting the domain of integration for the first term to (a+εn, b+εn) from(a, b) to get∫ b

a

F (x+ εn)− F (x)

εn=

1

εn

(∫ b

a

F (x+ εn)−∫ b

a

F (x)

)=

1

εn

(∫ b+εn

a+εn

F (x)−∫ b

a

F (x)

)=

1

εn

(∫ b+εn

b

F (x)−∫ a+εn

a

F (x)

)which approaches F (b) − F (a) as n → ∞ due to the continuity of F (x) atx = b and at x = a. Note that this identity is simply the trivial statementthat the difference of two integrals computed over two intervals is equal tothe integral over the difference of two intervals. Namely,∫

I

g −∫J

g =

∫I−J

g

for intervals I and J when the appropriate interpretation of I − J is used.Thus, the final conclusion from the application of Fatou’s lemma is∫ b

a

F ′ ≤ F (b)− F (a).

This conclusion needs only the assumption that F is continuous and non-decreasing. In particular, the derivative of any continuous nondecreasingfunction is integrable.


For the special case of f being bounded by some positive number M on[a, b], we can apply the dominated convergence theorem instead of Fatou’slemma to get∫ b

a

limn→∞

F (x+ εn)− F (x)

εn= lim

n→∞

∫ b

x0

F (x+ εn)− F (x)

εn,

that is, ∫ b

a

F ′(x) = F (b)− F (a),

because ∣∣∣∣F (x+ εn)− F (x)

εn

∣∣∣∣ = 1

ε

∣∣∣∣∫ x+ε

x

f

∣∣∣∣ ≤M

for x ∈ [a, b] and, as computed above,∫ b

a

F (x+ εn)− F (x)

εn=

1

εn

(∫ b

a

F (x+ εn)−∫ b

a

F (x)

)=

1

εn

(∫ b+εn

a+εn

F (x)−∫ b

a

F (x)

)=

1

εn

(∫ b+εn

b

F (x)−∫ a+εn

a

F (x)

)approaches F (b)− F (a) as n → ∞ due to the continuity of F at x = b andx = a.

Now we look at the general case where f ≥ 0 may not be boundedon (a, b). We are going to use the tool of truncation in the target space.For n ∈ N we let fn be the maximum of f and n and let Fn(x) =

∫ x

x0fn.

Then F ′n(x) = fn(x) almost everywhere on [a, b] for every n ∈ N from the

nondecreasing property of Fn due to the nonnegativity of fn. Since

F (x)− Fn(x) =

∫ x

x0

(f − fn)

is a nondecreasing function, we know that (F −Fn)′ ≥ 0 and F ′ ≥ F ′

n almosteverywhere on [a, b]. Thus∫ b

a

F ′ ≥∫ b

a

F ′n = Fn(b)− Fn(a).


Passing to limit as n→ ∞ yields∫ b

a

F ′ ≥ F (b)− F (a),

because

F (b) =

∫ b

x0

f = limn→∞

∫ b

x0

fn = limn→∞

Fn(b)

and

F (a) =

∫ a

x0

f = limn→∞

∫ a

x0

fn = limn→∞

Fn(a)

by fn ↗ f as n→ ∞ and the monotone convergence theorem. Since we havealready the inequality in the other direction∫ b

a

F ′ ≤ F (b)− F (a)

from Fatou’s lemma, it follows that∫ b

a

F ′ = F (b)− F (a).

On the other hand, from the definition of F (x) =∫ x

x0f we have∫ b

a

f = F (b)− F (a)

so that ∫ b

a

(F ′ − f) = 0.

Since the same argument works with (a, b) replaced by any open interval in(a, b), we know that the integration of F ′ − f over any disjoint union of afinite number of open intervals in (a, b) vanishes. Since F ′ − f is integrableover (a, b), by the dominated convergence theorem applied to the product ofF ′ − f with the characteristic function of an open subset O of (a, b) as thelimit of the product of F ′ − f with the characteristic function of a disjointunion of a finite number of open intervals in (a, b), it follows that∫

O(F ′ − f) = 0


for any open subset O of (a, b). Hence∫E

(F ′ − f) = 0

for any closed subset E of [a, b]. If F ′ − f is not zero almost everywhere on[a, b], there is a closed subset E of positive measure either in

[a, b] ∩ {F ′ − f ≥ c > 0}

or in[a, b] ∩ {F ′ − f ≤ −c < 0}

for some positive rational number c, which would yield∫E

(F ′ − f) ≤ cm(E) > 0

or ∫E

(F ′ − f) ≤ −cm(E) < 0

and contradict ∫E

(F ′ − f) = 0.

This shows that F ′ = f almost everywhere on [a, b] and finishes the discussionof one half of the fundamental theorem of calculus in Lebesgue’s theory ofintegration. Note that, to prove that an integrable function g vanishes almosteverywhere without knowing where the exceptional set of measure zero lies,a very useful technique is to show that

∫ β

αg = 0 for any α < β. The last

step of our proof of the first half of the fundamental theorem of calculusin Lebesgue’s theory of integration is an example of using this technique toshow that the integrable function F ′−f vanishes almost everywhere. We nowlook at the second half of the fundamental theorem of calculus in Lebesgue’stheory of integration.

Integration of Derivative in the Fundamental Theorem of Calculus. Let F bean absolutely continuous function on [a, b] in the sense that for any positivenumber ε > 0 there exists some δε > 0 such that

k∑j=1

|F (bj)− F (aj)| < ε


whenever∪k

j=1(aj, bj) is a disjoint union of a finite number of open intervalsin [a, b] whose measure is < δε. We are going to prove that∫ b

a

F ′ = F (b)− F (a).

As a preparatory step, we introduce the definition of a function of boundedvariation and its characterization as the difference of two nondecreasing func-tion.

Definition of Function of Bounded Variation. The total variation Varba(f) ofa bounded function f on [a, b] is defined as

Varba(f) = supP

n∑j=1

|f(xj)− f(xj−1)| ,

where the supremum is taken over all partitions P

a = x0 < x1 < · · · < xn−1 < xn = b

of the interval [a, b]. A function f on [a, b] is said to be of bounded variationif its total variation Varba(f) on [a, b] is finite.

Characterization of Function of Bounded Variation as Difference of TwoNondecreasing Functions. A function f of bounded variation on [a, b] canbe written as the difference f+ − f− of two nondecreasing functions f+ andf− on [a, b]. Moreover, the nondecreasing functions f+ and f− can be chosento be continuous on [a, b] when f is continuous on [a, b].

The verification of the statement is as follows. Let a < ξ ≤ b. For anypartition Pξ

a = x0 < x1 < · · · < xn−1 < xn = ξ

of the interval [a, ξ], let

gPξ=

∑1≤j≤nwith f(xj)>f(xj−1)

(f(xj)− f(xj−1)) ≥ 0

andhPξ

=∑

1≤j≤nwith f(xj)<f(xj−1)

(f(xj−1)− f(xj)) ≥ 0.


Then

gPξ− hPξ

=n∑

j=1

(f(xj)− f(xj−1)) = f(ξ)− f(a)

and

gPξ+ hPξ

=n∑

j=1

|f(xj)− f(xj−1)| .

Solving for gPξand hPξ

, we get

2gPξ=

n∑j=1

|f(xj)− f(xj−1)|+ (f(ξ)− f(a))

and

2hPξ=

n∑j=1

|f(xj)− f(xj−1)| − (f(ξ)− f(a)) .

Taking the supremum over the partitions P of [a, ξ], we get

2 supPξ

gPξ= Varξa(f) + (f(ξ)− f(a))

and2 sup

Pξ

hPξ= Varξa(f)− (f(ξ)− f(a))

so thatsupPξ

gPξ− sup

Pξ

hPξ= f(ξ)− f(a).

We can now setf+(ξ) = sup

Pξ

gPξ+ f(a)

andf−(ξ) = sup

Pξ

hPξ.

Thenf+(ξ)− f−(ξ) = sup

Pξ

gPξ+ f(a)− sup

Pξ

hPξ= f(ξ).

Clearlyf+(ξ) = sup

Pξ

gPξ+ f(a)


andf−(ξ) = sup

Pξ

hPξ

are nondecreasing functions of ξ.

Assume that the function f is continuous. We have to verify that supPξgPξ

is a continuous function of ξ. Fix ξ and let η approach ξ from the left. Wewould like to show that

limη→ξ−

supPη

gPη = supPξ

gPξ.

For a given ε > 0, we have to show that when η is sufficiently close to ξ wehave

supPη

gPη > supPξ

gPξ− ε.

There exists a partition P ′ξ

a = x′0 < x′1 < · · · < x′m−1 < x′m = ξ

of [a, ξ] such that

gP ′ξ> sup

Pξ

gPξ− ε

2.

Choose η ∈ (x′m−1, ξ) to define a partition Pη

a = x′0 < x′1 < · · · < x′m−2 < x′m−1 < η

of [a, η]. When η is sufficiently close to ξ, we have

|f(ξ)− f(η)| < ε

2.

Since ∣∣(f(ξ)− f(x′m−1))−

(f(η)− f(x′m−1)

)∣∣ = |f(ξ)− f(η)| < ε

2,

it follows thatgPη > gP ′

ξ− ε

2

andsupPη

gPη ≥ gPη > gP ′ξ− ε

2> sup

Pξ

gPξ− ε.


Since2 sup

Pξ

gPξ= Varξa(f) + (f(ξ)− f(a)) ,

what we have just proved is equivalent to

limη→ξ−

Varηa(f) = Varξa(f).

By a slightly modified argument we are going to prove that

limη→ξ+

Varηa(f) = Varξa(f),

which is equivalent tolimη→ξ+

supPη

gPη = supPξ

gPξ.

Instead of adding one partition point just before ξ to a partition Pξ of [a, ξ],we are adding one partition point just after ξ to a partition Qξ of [ξ, b]. Wecan show in the same way as before that

limη→ξ+

Varbη(f) = Varbξ(f)

given ξ ∈ [a, b). We now use

Varba(f) = Varηa(f) + Varbη(f)

which one can verify as follows. For the verification of the direction

Varba(f) > Varηa(f) + Varbη(f),

we take a sequence of partitions of [a, η] which can be used to give Varηa(f) aslimit and we take a sequence of partitions of [η, b] which can be used to giveVarbη(f) as limit and then concatenate the two sequences to get a sequenceof partitions of [η, b]. For the verification of the direction

Varba(f) 6 Varηa(f) + Varbη(f),

we take a sequence of partitions of [a, b] and add, if necessary, the partitionpoint η to break it up into a sequence of partitions of [a, η] and a sequenceof partitions of [η, b]. From

limη→ξ+

Varbη(f) = Varbξ(f)

andVarba(f) = Varηa(f) + Varbη(f)

it follows thatlimη→ξ+

Varηa(f) = Varξa(f).


Conclusion of Proof of Second Half of Fundamental Theorem of Calculus. Anabsolutely continuous function is of bounded variation, because for any ε > 0there exists some δε > 0 such that its total variation over an interval of length< δε is < ε and the interval [a, b] can be divided up into at most k intervals of

length < ε with k ≤ b−aδε

+ 1 to give a total variation < ε(

b−aδε

+ 1)on [a, b].

Thus an absolutely continuous function can be written as the difference oftwo nondecreasing functions and hence is differentiable almost everywhereand its derivative is integrable (from Fatou’s lemma applied to the differencequotient of a continuous nondecreasing function).

Let G(x) =∫ x

aF ′. From the first part of the fundamental theorem of

calculus in Lebesgue’s theory of integration, we have already seen that Gis absolutely continuous and G′ = F ′ almost everywhere on [a, b]. To showthat F (x) = G(x) it suffices to show that if the derivative of an absolutelycontinuous function is almost everywhere zero, then the absolutely continuousfunction is identically constant, because we can apply the statement to thedifference function F (x)−G(x) whose value at x = a is zero.

Let φ be an absolutely continuous function on [a, b] such that φ′ is zeroalmost everywhere on [a, b]. We are going to use Vitali’s covering argument toshow that φ(a) = φ(b). Since b can be replaced by any point in (a, b), we canthus conclude that φ is equal to the constant value φ(a). From the absolutecontinuity of φ it follows that for any ε > 0 there exists some δε > 0 with theproperty that for any disjoint union O of a finite number of open intervalsin [a, b] which cover [a, b] up to measure < δε, the sum of the absolute valueof the difference of the values of φ at the end-points of the intervals in thecomplement is < ε. Let E be the set of points in (a, b) where φ′ is zero. Fixany η > 0. For x ∈ E, from φ′(x) = 0 it follows that there exists a nonemptysubset Aη,x of R+ such that |φ(x + r) − φ(x)| < ηr and (x, x + r) ⊂ (a, b)for r ∈ Aη,x. By Vitali’s covering argument, we can cover E up to measure< δε by a disjoint union of a finite number of such intervals (x, x+ r) (withx ∈ E and r ∈ Aη,x). The complement of such a disjoint union of a finitenumber of open intervals is of length < δε so that is the sum of the absolutevalue of the difference of the values of φ at the end-points of the intervalsin the complement is < ε. This means that |φ(b) − φ(a)| is < η(b − a) + ε.By letting η > 0 and ε > 0 approach 0, we conclude that |φ(b)− φ(a)| = 0.This finishes the discussion on the second half of the fundamental theoremof calculus in Lebesgue’s theory of integration.


Example of Continuous Function of Bounded Variation Yet Not AbsolutelyContinuous (Constructed from Cantor Set). For the second part of the fun-damental theorem of calculus∫ b

a

F ′ = F (b)− F (a),

in order to make sure that F ′ exists almost everywhere we assume that Fis of bounded variation so that it can be written as the difference of twonondecreasing functions. In order to get always the difference F (b) − F (a)of the values of F at the two end-points a and b for the right-hand side of∫ b

a

F ′ = F (b)− F (a)

and not just the identity ∫ x2

x1

F ′ = F (x2)− F (x1)

for x1, x2 in [a, b] not belonging to a subset of [a, b] of measure zero, weassume that F is continuous on [a, b]. The question arises whether the twoconditions of F being of bounded variation and being continuous is sufficientfor the second part of the fundamental theorem of calculus∫ b

a

F ′ = F (b)− F (a)

to hold. In other words, the question arises whether there is an example ofa continuous function of bounded variation on [a, b] which is not absolutelycontinuous. There is indeed such an example which can be constructed byusing the Cantor set.

For α < β and γ < δ in R let ψα,β;γ,δ be the continuous nondecreasingfunction on [α, β] with range [γ, δ] defined as follows. Let

α = x0 < x1 < x2 < x3 = β

be the partition of [α, β] by equally spaced points so that xj − xj−1 = β−α3

for j = 1, 2, 3. Letγ = y0 < y1 < y2 = δ


be the partition of [γ, δ] by equally spaced points so that yj − yj−1 = δ−γ2

for j = 1, 2. The graph of ψα,β,γ,δ over [x0, x1] is the line-segment joiningthe point (x0, y0) to the point (x1, y1). The function ψα,β;γ,δ over [x1, x2] isconstant (with constant value y1). The graph of ψα,β,γ,δ over [x2, x3] is theline-segment joining the point (x2, y1) to the point (x3, y2).

Let E be the Cantor set which is constructed by starting with [0, 1] anditerating the removal of the open middle third from remaining closed inter-vals. Our example will be a continuous decreasing function φ on [0, 1] withφ(0) = 0 and φ(1) = 1 whose derivative at any point of [0, 1]− E is zero sothat the second part of the fundamental theorem of calculus fails to hold forφ, because

0 =

∫ 1

0

φ′ = φ(1)− φ(0) = 1

due to the Cantor set being of measure zero. Thus φ is not absolutelycontinuous on [0, 1].

The construction of φ is as follows. We start out with the function ψ0,1;0,1.Then we perform the following operation on the graph of ψ0,1;0,1. We divide[0, 1] into three equal subintervals and we do the following replacement of thegraph over each subinterval except the middle subinterval (where ψ0,1;0,1 isconstant). We replace the graph of ψ0,1;0,1 over [0, 1

3] by the graph of ψ0, 1

3;0, 1

2

and we replace the graph of ψ0,1;0,1 over [23, 1] by the graph of ψ 2

3,1; 1

2,1. We

repeat the similar operation with [0, 1] replaced by the two sub-intervals [0, 13]

and [13, 1] of [0, 1] other than the middle subinterval [1

3, 23] of [0, 1]. Induc-

tively, at each step we repeat the similar operation of dividing each of thethree sub-intervals except the middle third from the preceding step into threeequal subintervals and do the replacement of the graph on each subintervalexcept the middle third (where the function is constant). In other words, thefunction φ maps the point

∞∑k=1

ak3k

of [0, 1] with ak = 0, 1, or 2 to the point

∞∑k=1

bk2k


of [0, 1] with

bk =

{0 if ak = 0 or 11 if ak = 2.

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