FUNDAMENTALS OF REAL ANALYSISby

Dogan Comez

III. MEASURABLE FUNCTIONS AND LEBESGUE INTEGRAL

III.1. Measurable functions

Having the Lebesgue measure define, in this chapter, we will identify the collection of functionsthat are compatible with Lebesgue measure and then define Lebesgue integral on these functions.

Recall: f is continuous if and only if f−1(O) is open for all O open. Also, recall that

O =∞⋃k=1

Ik ⇔ f−1(O) =∞⋃k=1

f−1(Ik).

Continuous functions are well behaved for even Riemann-integral; hence we will naturally becompatible with measurable sets. We will consider a larger class of functions:

Definition. Let f : E → R#, where E ∈ F . f is called (Lebesgue) measurable if for alla ∈ R, the set {x ∈ E : f(x) > a} ∈ F .

Fact.1 Let f : E → R#, E ∈ F . The following are equivalent:

i) f is measurable.ii) {x ∈ E : f(x) ≥ a} ∈ F for all a ∈ R.

iii) {x ∈ E : f(x) < a} ∈ F for all a ∈ R.iv) {x ∈ E : f(x) ≤ a} ∈ F for all a ∈ R.

Furthermore i)-iv) imply

v) {x ∈ E : f(x) = a} ∈ F for all a ∈ R#.

Proof. (i) ⇒ (ii): First, note that {x ∈ E : f(x) ≥ a} = f−1([a,∞]). Also, since

[a,∞] =⋂n

(a− 1

n,∞],

it follows that

{x ∈ E : f(x) ≥ a} = f−1

(∞⋂n=1

(a− 1

n,∞]

)=∞⋂n=1

f−1(

(a− 1

n,∞]

)∈ F .

(ii) ⇒ (iii):

{x ∈ E : f(x) < a} = f−1((−∞, a)) = f−1(R \ [a,∞]) = E \ f−1([a,∞]) ∈ F .

(iii) ⇒ (iv) and (iv) ⇒ (i) are similar and left as an exercise.

For (v), if a ∈ R, then {f(x) = a} = f−1({a}) = f−1((−∞, a]) ∩ f−1([a,∞]) ∈ F . If a =∞,

{f(x) =∞} =∞⋂n=1

{f(x) > n} ∈ F . �

1

2

Examples. 1. Let f : [0, 1]→ R be a function such that

f(x) =

{0 if x ∈ Q1 if x ∈ Qc .

Then,

f−1(12,∞) = {x ∈ [0, 1] : f(x) > 1

2} = Qc ∩ [0, 1] ∈ F ,

if a = 1, f−1(1,∞) = ∅ ∈ F ,if a = −1, then f−1(−1,∞) = [0, 1] ∈ F .

It follows that, for all a ∈ R, {x ∈ [0, 1] : f(x) > a} ∈ F ; hence, f is measurable.

2. Let f(x) be the piecewise function defined as

f(x) =

0 if x ≤ −1x2 if − 1 < x < 12 if x = 12− x if x > 1

If a < 0, then f−1(a,∞) = {x : f(x) > a} = (−∞, 2− a) ∈ F .If a = 0, then f−1(0,∞) = {x : f(x) > 0} = (−1, 0) ∪ (0, 2) ∈ F .If 0 < a < 1, then f−1(−1,

√a) = (−1,−

√a) ∪ (

√a, 2− a) ∈ F .

If 2 ≤ a, then f−1(a,∞) = ∅ ∈ F .

Hence, f is measurable.

Fact.2 f : E → R, E ∈ F , is measurable if and only if for all O ∈ B, f−1(O) ∈ F .

Proof. Exercise.

Corollary.1 Continuous functions are measurable.

Proof. Exercise.

Corollary.2 Let f : E → R be a measurable function on E.

i) If D ⊂ E, D ∈ F , then f |D and f |E\D are measurable on E.ii) If f = g almost everywhere on E, then g is measurable on E.

Proof. (i) Since f : E → R be measurable on E,

{x ∈ D : f(x) > a} = (f |D)−1(a,∞) = D ∩ f−1(a,∞) ∈ F

(ii) Let A = {x ∈ E : f(x) 6= g(x)}. Note that m(A) = 0; hence A ∈ F . Then,

{x ∈ E : g(x) > a} = {x ∈ A : g(x) > a} ∪ {x ∈ E \ A : g(x) > a}= {x ∈ A : g(x) > a}︸ ︷︷ ︸

∈F

∪{x ∈ E \ A : f(x) > a}︸ ︷︷ ︸(f |E\A)−1(a,∞)∈F

∈ F . �

Fact.3 Let E ∈ F and f, g : E → R be measurable functions and finite almost everywhere.Then:

i) αf + βg is measurable on E for all α, β ∈ R.ii) f 2 is measurable on E (hence, fg is measurable on E).

iii) 1f

is measurable on E.

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iv) |f | is measurable on E.

Proof. (i) First, we show that αf is measurable on E for all α ∈ R. If α = 0, then αf = 0.Hence αf is measurable. If α > 0, then

{x ∈ E : (αf)(x) > a} ={x ∈ E : f(x) >

a

α

}∈ F .

Similarly for α < 0. Hence αf is measurable, and so is βg. Therefore, it is enough to show thatf + g is measurable on E. Notice that,

{x ∈ E : f(x) + g(x) > a} = {x ∈ E : f(x) > a− g(x)}Now, there exists rx ∈ Q such that f(x) > rx > a− g(x). Hence

{x ∈ E : f(x) > a− g(x)} =⋃r∈Q

({x ∈ E : f(x) > r}︸ ︷︷ ︸∈F

∩{x ∈ E : r > a− g(x)}︸ ︷︷ ︸∈F

) ∈ F .

(ii) First, observe that fg = 12[(f + g)2 − f 2 − g2]. Therefore it is enough to show that f

measurable implies f 2 is measurable. Now, we have

{x ∈ E : (f 2)(x) > a} = {x ∈ E : [f(x)]2 > a}= {x ∈ E : f(x) >

√a}︸ ︷︷ ︸

∈F

∪{x ∈ E : f(x) < −√a}︸ ︷︷ ︸

∈F

∈ F .

(iii) If a = 0, then {x ∈ E : 1f(x) > a} = {x ∈ E : f(x) > 0} ∈ F . So assume that a > 0. Then{

x ∈ E :1

f(x) > a

}=

{x ∈ E : f(x) <

1

a

}∈ F .

(iv) Follows from

{x ∈ E : |f |(x) > a} = {x ∈ E : |f(x)| > a}= {x ∈ E : f(x) > a} ∪ {x ∈ E : f(x) < −a} ∈ F . �

Definition. For functions f, g : E → R, define

(f ∨ g)(x) = max{f(x), g(x)}(f ∧ g)(x) = min{f(x), g(x)}

f+(x) = f(x) ∨ 0

−f−(x) = f(x) ∧ 0

Notice that, with the definitions above, we have that |f | = f+ + f− and f = f+ − f−.

Corollary. If f, g : E → R, E ∈ F is measurable on E, then:

i) f+, f− are measurable on E.ii) (f ∨ g)(x) and (f ∧ g)(x) are measurable on E.

Proof. Exercise.

Convention. For the rest of this chapter f : E → R. E will always be measurable.

Definition. Let A ⊂ R be a set, then then the function

χA(x) =

{1 if x ∈ A0 otherwise

4

is called the characteristic function of the set A.

Easy Fact. E ∈ F if and only if χE is a measurable function.

For, observe that,

χ−1E (a,∞) =

{∅ if a ≥ 1E otherwise

Hence χE is measurable (on E). If E 6∈ F , E ∈ R, then χE is not measurable.

Question: If f and g are measurable functions. What about f ◦ g?

The answer is “No” in general.

Example. Recall that f : [0, 1] → [0, 2] defined by f(x) = x + ϕ(x), where φ is the Cantor-Lebesgue function, is strictly increasing. Then, by Fact 9 in Chapter.II, there exists A ∈ Fsuch that f(A) 6∈ F .

Extend f continuously in a strictly increasing manner onto R. Say g : R → R such thatg|[0,1] = f . Then g−1 is continuous. Next, consider χA and define h(x) := (χA ◦ g−1)(x). Now

h−1(

1

2,∞)

= (χA ◦ g−1)−1(

1

2,∞)

= (g ◦ χ−1A )

(1

2,∞)

= g

(χ−1A

(1

2,∞))

= g(A) = f(A) 6∈ F .

Theorem.1 Let f, g be measurable functions on their respective domains. If g is continuous,then g ◦ f is measurable.

Proof. Define h(x) := (g◦f)(x). If O ⊂ R is a Borel set, then we have h−1(O) = (g◦f)−1(O) =f−1(g−1(O)). Since g is continuous, g−1(O) ∈ B. Hence f−1(g−1(O)) ∈ F . �

Recall that, for a given collection of functions {fn}, f : E → R, we say that fn → fpointwise if and only if for all x ∈ E, for all ε > 0, there exists N(ε, x) such that whenevern ≥ N we have |fn(x)− f(x)| < ε.

Also, fn → f uniformly if and only if for all ε > 0, there exists N(ε) such that whenevern ≥ N we have |fn(x)− f(x)| < ε for all x ∈ E.

Definition. Let {fn} be a sequence of real-valued functions on a measurable set E. We saythat {fn} converges almost everywhere (a.e.) to f : E → R if there exists N ⊂ E withm(N) = 0 such that fn → f pointwise on E \N .

Remarks. Uniform convergence ⇒ pointwise convergence ⇒ almost everywhere convergence.Note that the converse is not always true (Exercise: Give examples).

Observation. Let fn → f pointwise, where each fn is measurable (on E). For x ∈ E, if f(x) > a,then ∃m ≥ 1 such that f(x) > a+ 1

m. Thus, since fn → f pointwise, for large enough n, we will

have fn(x) > a+ 1m. This means that,

∃m ∃N such that, ∀n ≥ N, fn(x) > a+1

m. (∗)

5

Conversely, if (*) is satisfied, then f(x) = limn fn(x) ≥ a+ 1m. Hence, we have

{x : f(x) > a} = {x ∈ E : (∃m ≥ 1)(∃N ≥ 1)(∀n ≥ N)(fn(x) > a+ 1/m)}

=⋃m≥1

(⋃N≥1

(⋂n≥N

{x : fn(x) > a+

1

m

}))∈ F .

Theorem.2 Let E ∈ F and fn : E → R be a measurable function for all n ≥ 1. If fn → falmost everywhere on E, then f is measurable on E.

Proof. Since the set of x ∈ E such that {fn(x)} fails to converge to f(x) has measure zero,which is a measurable set, without loss of generality by restricting fn’s to a subset of E onwhich fn → f pointwise, we can assume that fn → f pointwise on E. Then by the observationabove, for all a ∈ R

{x ∈ E : f(x) > a} =⋃m≥1

(⋃N≥1

(⋂n≥N

{x : fn(x) > a+

1

m

}))∈ F . �

Corollary. Let E ∈ F , {fn} be a family of measurable functions on E. If fn → f uniformly orpointwise on E, then f is measurable on E.

Definition. Let fn be a sequence of real-valued measurable functions on E ∈ F . Assume eachfn is bounded (i.e. there exits M > 0 such that |fn(x)| < M for all x ∈ E). Fix x ∈ E anddefine:

f ∗(x) = supn≥1

fn(x)

f∗(x) = infn≥1

fn(x)

f(x) = lim supn≥1

fn(x)

f(x) = lim infn≥1

fn(x)

Then f ∗, f∗, f , and f are all well-defined real valued functions on E.

Exercise. Let fn be a sequence of real-valued measurable functions on E ∈ F .

1. Show that f ∗, f∗, f , and f are all measurable on E.2. Show that fn → f almost everywhere on E if and only if

m({x ∈ E : lim sup fn(x) > lim inf fn(x)}) = 0.

Question: If |f | is measurable, does it imply that f is measurable?

The answer is “NO” in general.

Exercise. Find an example of a non-measurable function f for which |f | is measurable.

Definition. A function f : E → R which is measurable and takes only finitely many values iscalled a simple function.

6

Fact.4 A function f : E → R is simple if and only if there exists a1, a2, . . . , an ∈ R andE1, E2, . . . En ∈ F such that

E =∞⋃k=1

Ek and f(x) =n∑k=1

akχEk(x).

Proof. (⇐) follows from the definition.(⇒) In order to show that f is measurable, assume a1, a2, . . . an are its values. Then we havef−1({ak}) = Ek ∈ F . If i 6= j, then ai 6= aj and hence Ei ∩ Ej = ∅, and

⋃nk=1Ek = E. Let

f(x) =n∑k=1

akχEk(x).

It is left as an exercise to verify that both sides agree for all x ∈ E. �

Note that the formula in the Fact.4 is a simple function in canonical form. Of course, thesame function can be expressed in many different ways involving characteristic functions ofdifferent sets.

Fact.5 Let f and g be simple functions on E ∈ F , then

i) f + g is simple.ii) fg is simple.

iii) fg

is simple (provided g 6= 0).

Proof. Exercise.

Fact.6 Let E ∈ F and f : E → R be a bounded measurable function. Then for all ε > 0, thereexists simple functions φε and ψε on E such that

φε(x) ≤ f(x) ≤ ψε(x) on E and 0 ≤ (ψε − φε)(x) < ε on E.

Proof. Since f is bounded there exists [a, b] ⊃ f(E). Partition [a, b] into subintervals of length` < ε, say

a < y0 < y1 < · · · < yn−1 < yn = b.

For i = 0, 1, . . . n− 1 let Ei = f−1((yi, yi+1]) ∈ F . Then

φε(x) =n−1∑i=0

yiχEi(x) ≤ f(x) ≤n−1∑i=0

yi+1χEi(x) = ψε(x).

Also,

ψε(x)− φε(x) =n−1∑i=0

(yi+1 − yi)χEi(x) < ε. �

Theorem.3 (Approximation by Simple Functions) Let E ∈ F and f : E → R#. Thenf is measurable on E if and only if there exists a sequence {φn} of simple functions on E suchthat φn → f almost everywhere and |φn| ≤ f on E. Furthermore, if f ≥ 0, then φn’s can bechosen as φn ↑ f almost everywhere on E.

Proof. Since f = f+ − f− and |f | = f+ + f− without loss of generality, we can assume f isnon-negative (i.e. f ≥ 0). Therefore let f ≥ 0. Note that (⇐) part is Theorem.2 above.

7

(⇒) Given n ≥ 1, let En = {x ∈ E : f(x) ≤ n}. Apply the previous fact with ε = 1n. So there

exists φn and ψn simple functions on En such that φn ≤ f ≤ ψn on En and ψn− φn < 1n

on En.

Now φn ≤ f on En and f − φn < 1n

on En.

Extend φn to the rest of E by letting φn(x) = n on E \ En. Now, we’ll show that φn → fpointwise on E.

If x is such that f(x) <∞, then we have f(x) < N for some N. Then, for any n ≥ N, sincex ∈ En for some n, we have 0 ≤ f(x)− φn(x) ≤ 1

n; hence, were done.

If x is such that f(x) =∞, then take φn(x) = n; it follows that

φn(x)→ f(x)

pointwise. Hence φn ↑ f pointwise. �

Corollary. Let E ∈ F , a function f : E → R is bounded and measurable if and only if thereexists {φn}, simple functions on E, such that φn → f uniformly.

Proof. Exercise.

Exercises

1. Let E ∈ F and f : E → R be a function. f is a measurable function if and only iff−1(O) ∈ F , ∀ O ∈ B.

2. Let {fn} be a sequence of R-valued measurable functions on E ∈ F and f : E → R bemeasurable. fn → f almost everywhere on E if and only if

m({x ∈ E : lim supn

fn(x) > lim infn

fn(x)}) = 0.

3. Find a function f : E → R, E ∈ F , such that |f | is measurable whereas f is not.

III.2 Lebesgue Integration of Bounded Measurable Functions

Now that we sorted out those functions compatible with the (Lebesgue) measurable sets, wewill proceed to define the Lebesgue integral (of measurable functions). Throughout we willassume that: E ∈ F with m(E) < ∞, all functions f : E → R are bounded, and all simplefunctions will be in canonical form.

Definition. Let φ : E → R be a simple function defined as φ(x) =∑n

i=1 aiχEi(x). The(Lebesgue) integral of φ, denoted by

∫Eφ(x) dm, is defined as∫

E

φ(x) dm =n∑i=1

aim(Ei).

Notation:∫Eφ(x) dm =

∫Eφ(x) dx =

∫Eφ dm =

∫Eφ.

Remarks. 1. If we extend φ to R by letting φ(x) = 0 for all x ∈ R \ E, then∫E

φ(x) dm =

∫R(φ · χE)(x) dm.

2. The Lebesgue integral of a step function is the same as the Riemann integral of a stepfunction.

8

Example. Let E = [0, 1] and let φ(x) = χQ∩[0,1]. Then∫E

φ(x) dx = 1 ·m(Q ∩ [0, 1]) = 0.

Hence φ is Lebesgue integrable. Recall φ is not Riemann integrable!

Exercise. Calculate∫[0,1]

φ, where φ(x) = χC∩Q + 2χC∩Qc + 3χ[0,1]\C.

Fact.7 Let φ, ψ : E → R be simple functions. Then

i)∫E

(aφ+ bψ) = a∫Eφ+ b

∫Eψ for all a, b ∈ R.

ii)∫Eφ ≤

∫Eψ if φ ≤ ψ.

iii)∫Eφ =

∫E1φ+

∫E2φ where E1, E2 ∈ F , E = E1 ∪ E2, and E1 ∩ E2 = ∅.

Proof. i) Let φ(x) =∑n

i=1 aiχEi(x) and ψ(x) =∑m

j=1 bjχFj(x) be in canonical form. Let

{Ak}Nk=1 be defined as Ak = Ei ∩ Fj with an appropriate ordering. Then

{Ak}Nk=1 = {Ei ∩ Fj}n,mi,j=1

is a disjoint collection with⋃Nk=1Ak = E. Hence

aφ+ bψ =N∑k=1

(aai + bbj)χAk

where i, j corresponds to the set Ak with Ak = Ei ∩ Fj. Therefore, we have∫E

(aφ+ bψ) =N∑k=1

(aai + bbj)m(Ak) =N∑k=1

aaim(Ak) +N∑k=1

bbjm(Ak)

= aN∑k=1

aim(Ak) + bN∑k=1

bjm(Ak) = an∑i=1

aim(Ei) + bm∑j=1

bjm(Fj)

= a

∫E

φ+ b

∫E

ψ.

ii) Since φ ≤ ψ on E, ψ − φ ≥ 0 on E. Also since ψ − φ is simple, by i) we have∫E

(ψ − φ) ≥ 0⇒∫E

ψ ≥∫E

φ.

Proof of iii) is left as an exercise. �

Recall: Given any f : E → R bounded, we can always find (construct) simple functions φ andψ such that φ ≤ f ≤ ψ on E and 0 ≤ (ψ − φ)(x) can be made arbitrarily small on E.

Definition. Let E ∈ F and f : E → R be a bounded function. We say that the function f is(Lebesgue) integrable over E if

sup

{∫E

φ : φ ≤ f on E

}= inf

{∫E

ψ : ψ ≥ f on E

},

where φ and ψ are simple functions. The common value is called the (Lebesgue) integral off over E and is denoted by ∫

E

f(x) dx =

∫E

f =

∫E

f dm.

9

For the simplicity of the arguments below, we will denote:

inff≤ψ

∫E

ψ = Uf and supf≥φ

∫E

φ = Lf .

Theorem.4 Let E ∈ F , m(E) < ∞, and f : E → R be a bounded function. Then f ismeasurable on E if and only f is Lebesgue integrable over E

Proof. (⇒) Assume f is measurable and |f | ≤ M . Partition the interval [−M,M ] into 2nsubintervals of equal length. Let

−M = y−n < y−n+1 < · · · < yn−1 < yn = M

be the endpoints of these subintervals. Define Ek = f−1([yk, yk+1]). Then, Ek ∈ F for allk = −n, . . . n with Ek ∩ E` = ∅ for k 6= `, and

n∑k=−n

m(Ek) = m(E).

Now define φn =∑n−1

k=−n ykχEk and ψn =∑n−1

k=−n yk+1χEk . Therefore, on the set E,

φn(x) ≤ f(x) ≤ ψn(x),

where {ψn} and {φn} are integrable. Then

inff≤ψ

∫E

ψ ≤∫E

ψn and supf≥φ

∫E

φ ≥∫E

φn.

Thus,

inff≤ψ

∫E

ψ − supf≥φ

∫E

φ ≤∫E

(ψn − φn) =

∫E

[n−1∑k=−n

(yk+1 − yk)χEk

]<M

nm(E).

Since n is arbitrary, we have Lf = Uf and f is Lebesgue integrable.

(⇐) Given that f is bounded on E and Lf = Uf , let 0 < εn = 1n

for n ≥ 1. From theseassumptions and the definitions of inf / sup, we can pick a sequence of simple functions {ψn}and {φn} on E such that φn ≤ f ≤ ψn on E, and

0 ≤(∫

E

ψn −∫E

φn

)<

1

n.

Let ψn = infn ψn and φn = supn φn. Since each ψn and φn are measurable, we have that both

ψn and φn are measurable on E, and φn ≤ f ≤ ψn on E. We need to show ψn = φn almosteverywhere on E. To show this let

A = {x ∈ E : φ(x) < ψ(x)}.

We will show that m(A) = 0. To do this let Ak = {x ∈ E : ψ(x) − φ(x) > 1/k}, then⋃k≥1Ak = A. Now let

Bnk = {x ∈ E : ψn(x)− φn(x) > 1/k}.

Then Ak ⊂ Bnk ∈ F and, on E, k(ψn − φn) > 1. So

m(Bnk ) =

∫E

χBnk ≤∫E

k(ψn − φn) = k

∫E

(ψn − φn) <k

n.

It follows that m(Ak) = 0. Hence, we have m(A) = 0. �

10

Fact.8 Let f : [a, b]→ R be a bounded function. If f is Riemann integrable, then f is Lebesgueintegrable with ∫

[a,b]

f dm =

∫ b

a

f(x) dx.

Proof. Exercise.

Fact.9 Let E ∈ F , m(E) <∞, f, g : E → R be bounded functions. Then we have:

a)∫E

(αf + βg) = α∫Ef + β

∫Eg for all α, β ∈ R.

b) If f = g almost everywhere on E, then∫Ef =

∫Eg.

c) If f ≤ g almost everywhere, then∫Ef ≤

∫Eg (Hence |

∫Ef | ≤

∫E|f |).

d) If there exists a, b ∈ R such that a ≤ f(x) ≤ b on E, then

am(E) ≤∫E

f ≤ bm(E).

e) If E1, E2 ∈ F , and E = E1 ∪ E2 with E1 ∩ E2 = ∅, then∫E

f =

∫E1

f +

∫E2

f.

Proof. Exercise.

Exercise. Let E ∈ F with m(E) < ∞, and {fn} be a sequence of real valued, boundedmeasurable functions on E such that fn → f uniformly. Then∫

E

fn →∫E

f.

The uniform convergence condition in the exercise above cannot be relaxed as the followingexample shows.

Example. Let fn : [0, 1]→ R, n ≥ 1, be given by

fn(x) =

n2x if x ∈ [0,

1

n]

− n2x+ 2n if x ∈ [1

n,

2

n]

0 otherwise,

and f ≡ 0 on [0, 1]. Then, ∫[0,1]

fn = 1 6= 0 =

∫[0,1]

f !

Theorem.5 (Egoroff) Let E ∈ F with m(E) < ∞, and {fn} be a sequence of real valuedmeasurable functions on E such that fn → f almost everywhere. Then for all ε > 0, there existsa measurable Aε ⊂ E, m(E \ Aε) < ε, such that fn → f uniformly on Aε.

Proof. Given ε > 0 and for n, j ∈ Z+, define

Ejn =

∞⋂k=n

{x ∈ E : |fk(x)− f(x)| < 1/j}.

11

Then Ejn ∈ F for all n, j. If A ⊂ E such that fn → f pointwise on A (i.e. m(E \A) = 0), then

for all x ∈ A, x ∈ Ejn for some n, j. Hence

A ⊂∞⋃n=1

Ejn.

Therefore

A ⊂∞⋃n=1

Ejn ⊂ E ⇒ m

(∞⋃n=1

Ejn

)= m(E), since fn → f a.e. (hence m(A) = m(E)).

Observe for a fixed j, Ejn ⊂ Ej

n+1. Hence by continuity of m, m(E \Ejn) = m(E)−m(Ej

n) and

limn→∞

m(E \ Ejn) = m(E)− lim

n→∞m(Ej

n) = m(E)−m

(∞⋃n=1

Ejn

)= 0.

Consequently, for all j ≥ 1, we can find n(j) such that n ≥ n(j) and m(E \ Ejn) < ε

2j. Let

Aε =⋂∞j=1E

jn(j), then Aε ∈ F . Then

m(E \∞⋂j=1

Ejn(j)) = m

(∞⋃j=1

(E \ Ejn(j))

)≤

∞∑j=1

m(E \ Ejn(j)) <

∑j

ε

2j= ε.

Now, it is left as an exercise to show that fn → f uniformly on Aε. �

Remark. By the example above, if fn → f pointwise on E ∈ F , in general, we cannot expect∫Efn →

∫Ef. However, under somewhat mild additional conditions we have a positive answer:

Theorem.6 (Bounded Convergence Theorem) Let E ∈ F , m(E) < ∞, and {fn} be asequence of uniformly bounded measurable, real valued functions on E. If fn → f almosteverywhere, then ∫

E

fn →∫E

f.

Proof. Since {fn} is uniformly bounded, we can find a positive real number M such that|fn(x)| ≤M . Then, by Egoroff’s Theorem, given ε > 0, there exists A ⊂ E such that m(E\A) <ε

4Mand fn → f uniformly on A. Then, find N large enough that, for n ≥ N , |fn(x)− f(x)| <

ε2m(A)

for all x ∈ A. Now for such n ≥ N,∫E

fn −∫E

f =

∫A

(fn − f) +

∫E\A

(fn − f),

and hence,

|∫E

fn −∫E

f | ≤∫A

|(fn − f)|+∫E\A|(fn − f)| ≤

∫A

|fn − f |+∫E\A

2Mdm,

since |fn − f | ≤ 2M. Therefore, we have∣∣∣∣∫E

fn −∫E

f

∣∣∣∣ < ε

2m(A)m(A) +

2Mε

4M= ε. �

Remark. The BCT can be restated as saying that under the given conditions the “lim” and“∫

” can be interchanged.

12

Examples. 1. Given

f(x) =

{1 if x ∈ [1/2, 1] ∪ [1/4, 1/3] ∪ [1/6, 1/5] ∪ . . .0 otherwise

Define

f1(x) = χ[1/2,1]

f2(x) = χ[1/4,1/3] + χ[1/2,1]

...

fn(x) =n∑k=1

χ[1/(2k),1/(2k+1)]

...

It is straightforward to show that fn → f pointwise (hence, is left as an exercise). Then {fn} isa uniformly bounded sequence of measurable functions. By the Bounded Convergence Theorem,∫

[0,1]

f = limn→∞

∫[0,1]

fn = limn→∞

[n∑k=1

m

([1

2k,

1

2k − 1

])]

= limn→∞

[n∑k=1

(1

2k − 1− 1

2k

)]= lim

n→∞

[2n∑k=1

(−1)k+1 1

k

]

=∞∑k=1

(−1)k+1

k= ln(2).

2. Let

f(x) =

{2x if x ∈ [0, 1] ∩Q1− x if x ∈ [0, 1] ∩Qc

If we define g(x) = 1− x on [0, 1], then f = g almost everywhere and g is Riemann Integrable= Lebesgue Integrable. Hence,∫

[0,1]

f(x) dx =

∫[0,1]∩Q

f(x) dx+

∫[0,1]∩Qc

f(x) dx =

∫ 1

0

g(x) dx =1

2.

Exercises

1. Using only the definition, evaluate the integrals:∫[0,1]

√x dx and

∫[0,1]

x1/3 dx.

2. Let C denote the Cantor Set. Find∫[0,1]

f(x) dx if

f(x) =

x2 if x ∈ C ∩Qx if x ∈ C ∩Qc

1 if x ∈ [0, 1] \ C.

3. Prove that fn → f uniformly on Aε in Egoroff’s Theorem.

III.3. Lebesgue Integral of Non-negative Functions

In the previous section we defined the Lebesgue integral for bounded measurable functions.Now, we will continue with not necessarily bounded functions. We will deal with this case in

13

two steps: first we will define the Lebesgue integral of the functions f : E → R+, and thenextend it to more general functions f : E → R.

Definition. Let E ∈ F (m(E) =∞ is allowed) and f : E → R+ be measurable. We define∫E

fdm = sup{∫E

hdm : h : E → R is bounded, measurable, h ≤ f, m({x : h(x) 6= 0}) <∞}.

Remark.

(1) It is possible that∫Ef =∞

(2) Functions h : E → R such that m({x ∈ E : h(x) 6= 0}) < ∞ are called functions withfinite support. The set {x ∈ E : h(x) 6= 0} is called the support of h.

Exercise. Find ∫[0,1]

1√xdm.

Fact.10 Let E ∈ F , f, g : E → E+ be measurable functons. Then

(a)∫E

(αf + βg) = α∫Ef + β

∫Eg.

(b) If f ≤ g almost everywhere on E, then∫Ef ≤

∫Eg.

(c)∫Ef =

∫E1f +

∫E2f where E = E1 ∪ E2 and E1 ∩ E2 = ∅.

Proof. The result follows from the the same fact on bounded functions; hence, it is left as anexercise. �

Theorem.7 (Fatou’s Lemma) Let E ∈ F and {fn} be a sequence of R+-valued, measurablefunctions on E. If fn → f almost everywhere on E, then∫

E

f dm ≤ lim infn→∞

∫E

fn dm.

Proof. Let φ be a bounded, measurable, non-negative function with m({φ 6= 0}) < ∞ suchthat φ ≤ f almost everywhere on E. For each n, define hn = fn ∧ φ = min{fn, φ}. Then:

i) hn is bounded and measurable for all n.ii) hn ≤ fn.

iii) hn’s vanish outside A = supp(φ) = {x : φ(x) 6= 0}.iv) hn → φ almost everywhere on A.

Then, by the Bounded Convergence Theorem,∫E

φ dm =

∫A

φ dm = limn→∞

∫A

hn dm ≤ lim infn→∞

∫E

fn dm.

(Last inequality follows since limn

∫Efn dm may not exist.) Therefore,∫

E

f dm = supφ≤f

∫E

φ dm ≤ lim infn→∞

∫E

fn dm. �

Remarks. 1. Strict inequality in Fatou’s Lemma is possible.

2. If non-negativity of fn’s is removed, the conclusion of Fatou’s Lemma is not valid anymore.

14

Exercise. Provide examples for each of the remarks made above.

Theorem.8 (Monotone Convergence Theorem) Let E ∈ F , and {fn} be a sequence ofR+-valued measurable functions on E such that fn ≤ fn+1 a.e. (on E) for all n ≥ 1. If fn → fa.e., then

limn→∞

∫E

fn dm =

∫E

f dm.

Proof. From hypothesis, fn ≤ fn+1 for all n almost everywhere on E. Hence∫E

fn dm ≤∫E

f dm

for all n. Therefore by Fatou’s Lemma

lim sup

∫E

fn dm ≤∫E

f dm ≤ lim inf

∫E

fn dm.

Hence, it follows that limn→∞∫Efn dm =

∫Ef dm. �

Remarks.

1. MCT ⇒ Fatou’s Lemma.2. MCT does not hold for Riemann Integral.3. Monotonicity in MCT is essential.

Example. Let E = R ([0,∞]) and fn(x) = χ[n,∞)(x), n ≥ 1. Then

fn → 0 pointwise (but not monotonically), whereas, ∞ =

∫Rfn 6→ 0 =

∫Rf.

Corollary.1 Let E ∈ F and {un} be a sequence of positive real valued functions on E. If

f =∞∑n=1

un (pointwise), then

∫E

f dm =∞∑n=1

∫E

un dm.

Proof. Define fn =∑n

k=1 uk, for n ≥ 1. Then fn ↑ f a.e. on E. Apply MCT:

limn→∞

∫E

fn dm =

∫E

f dm⇒ limn→∞

n∑k=1

∫E

uk dm =∞∑k=1

∫E

uk dm =

∫E

f dm. �

Corollary.2 Let E ∈ F and {En} ⊂ F such that E =⋃nEn and Ei ∩ Ej = ∅ if i 6= j. Then

for any f : E → R+ measurable,∞∑n=1

∫En

f dm =

∫E

f dm

Proof. Let un = fχEn and apply Corollary.1. �

Theorem.9 (Chebychev’s Inequality) Let E ∈ F and f : E → R+ be a measurable function.Then for all λ > 0,

m({x ∈ E : f(x) > λ}) ≤ 1

λ

∫E

f dm.

15

Proof. Let Eλ = {x ∈ E : f(x) > λ}. Assume first that m(Eλ) <∞ and let φ = λχEλ . Thenon Eλ, φ ≤ f . Hence ∫

E

φ dm ≤∫E

f dm

‖∫E

λχEλ dm = λm(Eλ)⇒ m(Eλ) ≤1

λ

∫E

f dm.

Next we will assume that m(Eλ) = ∞. If we let Enλ = Eλ ∩ [−n, n], then En

λ ↑ Eλ andm(En

λ ) <∞ for each n. Now, applying above case to Enλ and φn = λχEnλ , we obtain

λm(Enλ ) ≤

∫E

f dm.

Then, by continuity of m,

λm(Eλ) = λ limn→∞

m(Enλ ) ≤

∫E

f dm.

Note that this gives equality on both sides since m(Eλ) =∞. Hence,

m(Eλ) =1

λ

∫E

f dm. �

Theorem.10 Let E ∈ F and f : E → R+ be measurable. Then∫E

f dm = 0⇔ f = 0 almost everywhere on E.

Proof. (⇒) Assume∫Ef dm = 0 and consider, for any n, the set

S = {x ∈ E : f(x) > 1/n}.Then by Chebychev’s Inequality,

m(S) ≤ n

∫E

f dm = 0.

Now, since

{x ∈ E : f(x) > 0} =⋃n

{x ∈ E : f(x) > 1/n},

it follows that f = 0 a.e. on E.

(⇐) Next, assume that f = 0 almost everywhere on E. Let, for all n, En = E ∩ [−n, n]. Oneach En let h be a bounded, measurable, non-negative function such that h ≤ f, and φ be anybounded simple non-negative function with φ ≤ h (i.e. 0 ≤ φ ≤ h ≤ f a.e.). Hence φ must be0 almost everywhere on En. Then ∫

En

φ = 0⇒∫En

h = 0.

Since h is arbitrary,∫Ef = 0 follows from the definition. �

Remark. Theorem.10 implies that f = g a.e. on E ∈ F ⇐⇒∫Ef =

∫Eg.

Definition. Let E ∈ F . A function f : E → R+ is called (Lebesgue) integrable over E if∫E

f dm <∞.

16

Example. Let’s show that the function f(x) = 1x2

is Lebesgue-integrable on [1,∞) and calculate∫[1,∞)

1x2. For, we will use MCT. First, let, for n ≥ 1,

fn(x) =

1

x2if 1 ≤ x ≤ n

0 if otherwise

Hence, each fn is bounded and measurable on [1,∞). Thus,∫[1,∞)

fn =

∫[1,n]

fn =

∫ n

1

1

x2.

Since 1x2

is Riemann-integrable on [1, n], it is also Lebesgue integrable on [1, n] (with both

integrals being equal). Hence,∫[1,∞)

fn =∫ n1

1x2

= 1− 1n. Therefore, by MCT,∫

[1,∞)

1

x2= lim

n

∫[1,∞)

fn = 1.

Fact.11 Let E ∈ F and f : E → R+ be integrable over E. Then f(x) <∞ a.e. on E.

Proof. It is enough to show that m({x ∈ E : f(x) = ∞}) = 0. For, first observe that{x ∈ E : f(x) =∞} =

⋂n≥1{x ∈ E : f(x) > n}. Now, by Chebychev’s Inequality,

m({x ∈ E : f(x) =∞}) ≤ m({x ∈ E : f(x) > n}) ≤ 1

n

∫E

f.

Since∫Ef <∞, letting n→∞, we obtain the desired result. �

Theorem.11 (Beppo Levi’s Lemma) Let {fn} be an increasing sequence of nonnegativemeasurable functions on E ∈ F . If the sequence {

∫Efn} is bounded, then

i) fn →p f on E for some non-negative measurable function f : E → R#,ii) limn

∫Efn =

∫Ef <∞, and

iii) f is finite a.e.

Proof. For each x ∈ E, the sequence {fn(x)} is an increasing sequence of real numbers; hence,it converges to an extended real number. Therefore, we can define f pointwise by

f(x) = limnfn(x), for all x ∈ E.

Since fn ↑ f, by MCT, we have∫Efn →

∫Ef. Therefore, by the fact that {

∫Efn} is bounded,

we have have∫Ef <∞; and by Fact.11 above we must also have that f is finite a.e. �

The following fact is another important property of integrable functions.

Theorem.12 Let f : E → R+ be an integrable function on E ∈ F . Then, for any ε > 0 thereexists δ > 0 such that ∀A ⊂ E, A ∈ F , with m(A) < δ, we have

∫Af < ε.

Proof. If |f | < M for some M on E, take δ = εM. The the assertion follows.

If f is arbitrary integrable function, let

fn(x) =

{f(x) if x ≤ n

n if x > n.

17

Then {fn} is a bounded sequence and fn →p f. on E. So, by MCT,∫Efn →

∫Ef. Hence, given

ε > 0, there exists N ≥ 1 such that if n ≥ N, then [∫Ef −

∫Efn] < ε

2. In particular, if n = N,

we have∫E

(f − fN) < ε2. Now, pick δ < ε

2N, then for any A ⊂ E with m(A) < δ we have∫

A

f =

∫A

(f − fN) +

∫A

fN ≤∫E

(f − fN) +

∫A

N <ε

2+ε

2= ε. �

Exercise. Let f : R→ R+ be an integrable function. Define a set function ν : F → [0,∞] byν(A) =

∫Afdm, A ∈ F . Show that

i) 0 ≤ ν(A) ≤ ∞ for all A ∈ F .ii) ν(A) ≤ ν(B) if A ⊂ B, A,B ∈ F .

iii) ν(∅) = 0 = ν({a}) for any a ∈ R.iv) ν(∪∞i=1Ai) =

∑∞i=1 ν(Ai) if {Ai} ⊂ F is a disjoint family of sets.

v) ν(A) = 0 if m(A) = 0.

Question. For f and ν as in the exercise above, what can you say about

i) ν(I) = `(I) for any interval I ⊂ R?ii) ν(A+ α) = ν(A) for any A ∈ F and α ∈ R?iii) m(A) = 0 if ν(A) = 0 for A ∈ F?

III.4. Lebesgue Integral of Arbitrary Functions

In this section, having defined the (Lebesgue) integral of a non-negative measurable function,we will extend it to arbitrary measurable functions. First, recall that

(i) f = f+ − f−, and(ii) if f is measurable, then so are f+ and f−.

Definition. A function f : E → R where E ∈ F , is called (Lebesgue) integrable over E ifboth f+ and f− are integrable over E. In that case,∫

E

f =

∫E

f+ −∫E

f−.

Remark. Since f+ and f− are non-negative and integrable, the set on which they take ∞ asa value is of measure zero; hence, the set on which f may take ∞+∞ or ∞−∞ values is alsoof measure zero.

Fact.12 If E ∈ F , f : E → R is integrable and A ⊂ E, A ∈ F with m(A) = 0, then∫E

f =

∫E\A

f.

Proof. Exercise.

Fact.13 Let E ∈ F , f : E → R be measurable and g : E → R+ be integrable with |f | ≤ g onE. Then f is integrable with ∫

E

f ≤∫E

g.

In particular, |∫Ef | ≤

∫E|f |.

18

Proof. Exercise.

Fact.14 If E ∈ F and f, g : E → R are integrable over E, then

(a) αf is integrable over E and∫Eαf = α

∫Ef .

(b) f + g is integrable over E and∫E

(f + g) =∫Ef +

∫Eg.

(c) f ≤ g on E, then∫Ef ≤

∫Eg.

(d) If E = E1 ∪ E2, where E1 and E2 are measurable and disjoint, then fχE1 and fχE2 areintegrable with ∫

E

f =

∫E1

f +

∫E2

f.

Proof. The proof of (a) and (c) are left as exercises. For (b) without loss of generality, ifnecessary, by removing a larger set of measure zero, we can assume that f + g is finite on E.Thus

(f + g)+ − (f + g)− = f + g = (f+ − f−) + (g+ − g−) and

(f + g)+ + f− + g− = (f + g)− + f+ + g+.

Since both sides are non-negative, we have∫E

(f + g)+ +

∫E

f− +

∫E

g− =

∫E

(f + g)− +

∫E

f+ +

∫E

g+.

If we reorganize we have∫E

[(f + g)+ − (f + g)−

]=

∫E

(f+ + f−) +

∫E

(g+ − g−).

Hence f + g is integrable.

For the proof of (d), since f is integrable we know that |fχE1| ≤ |f | and |fχE2| ≤ |f |. Nowboth fχE1 and fχE2 are integrable with χE = χE1 + χE2 . �

Theorem.13 (Dominated Convergence Theorem) Let {fn} be a sequence of real valuedmeasurable functions on E ∈ F . Let g : E → R be an integrable function over E such that|fn| ≤ g for all n on E. If fn → f almost everywhere on E, then fn and f are integrable overE and ∫

E

fn −→∫E

f.

Proof. Since g is integrable over E and |fn| ≤ g, then each fn is integrable over E. But then,since fn → f almost everywhere, we also have |f | ≤ g on E. Hence f is integrable over E.

Observe that {g − fn} is a sequence of non-negative integrable functions on E converging tog − f . Similarly {g + fn} is a sequence of non-negative functions converging to g + f . Now byapplying Fatou’s Lemma to {g − fn} we have∫

E

g −∫E

f =

∫E

(g − f) ≤ lim inf

∫E

(g − fn) =

∫E

g − lim sup

∫E

fn

⇒∫E

f ≥ lim sup

∫E

fn.

Next, by applying Fatou’s Lemma to {g + fn} we have∫E

f +

∫E

g =

∫E

(g + f) ≤ lim inf

∫E

(g + fn) =

∫E

g + lim inf

∫E

fn

⇒∫E

f ≤ lim inf

∫E

fn.

19

These two inequalities imply that∫Efn →

∫Ef. �

Exercise. For each n ≥ 1, let fn(x) = n3/2x1+n2x2

, x ∈ [0, 1]. Show that

(a) fn → 0 almost everywhere on [0, 1].(b) |fn(x)| ≤ g(x) on [0, 1] for all n where g(x) = 1√

x.

(c) limn→∞∫[0,1]

fn = 0.

Remark. Converse of the DCT is not valid!

Examples. Let E = [0, 1].

1. Let fn(x) = 1− x for all n and f(x) = x. Then,∫E

fn −→∫E

f, but fn 6→ f.

2. Let fn(x) = n−1n− x and f(x) = x. Then∫

E

fn −→∫E

f, but fn 6→ f.

3. Define

f0 = χ[0,1]

f1 = χ[0, 12], f2 = χ[ 1

2,1]

f3 = χ[0, 13], f4 = χ[ 1

3, 23], f5 = χ[ 2

3,1]

f6 = χ[0, 14], f7 = χ[ 1

4, 12], f8 = χ[ 1

2, 34], f9 = χ[ 3

4,1]

f10 = χ[0, 15], . . .

...

Then we have∫Efn −→ 0, but fn 6→ 0 almost everywhere.

Theorem.14 (Generalized Dominated Convergence Theorem) Let {fn} be a sequenceof real valued measurable functions on E ∈ F such that fn → f almost everywhere on E.Let {gn} be a sequence of real valued non-negative functions on E such that gn → g almosteverywhere on E. Also let |fn| ≤ gn for all n on E. If∫

E

gn −→∫E

g, then

∫E

fn −→∫E

f.

Proof. Exercise.

Corollary.1 Let E ∈ F and f : E → R be integrable over E.

(a) If E =∞⋃n=1

En (disjoint), En ∈ F . Then

∫E

f =∞∑n=1

∫En

f.

20

(b) Given {En} ⊂ F , En ⊂ E and En ↑ . If A = ∪nEn, then∫A

f = limn→∞

∫En

f.

(c) Given {En} ⊂ F , En ⊂ E, and En ↓. If A = ∩nEn, then∫A

f = limn→∞

∫En

f.

Proof. Exercise. [Hint: Use Corollary.1 before Chebychev’s Inequality.]

Corollary.2 Let E ∈ R and f : E → R be integrable over E. Then given ε > 0 there existsδ > 0 such that for all A ⊂ F , A ⊂ E and m(A) < δ, then∫

A

f < ε.

Proof. Exercise.

The results stated above have some important and interesting ramifications. Let ν(A) =∫Af

for all A ∈ F , where f : R→ R is integrable. Then by Corollary.1(a) above,

ν

∞⋃k=1

disjoint

Ak

=∞∑k=1

ν(Ak).

Also ν(∅) = 0, 0 ≤ ν(A), A ⊂ B ⇒ ν(A) ≤ ν(B). Notice that ν is not a measure; however,it acts like a measure. Such set functions are called signed measures.

It turns out that the converse of Corollary.2 is also valid if m(E) <∞.

Fact.15 Let E ∈ F , m(E) <∞, and f : E → R be a measurable function on E. If for all ε > 0there exists δ > 0 such that for all A ∈ F , A ⊂ E and m(A) < δ, we have that

∫Af < ε, then

f is integrable over E.

Proof. WLOG we can assume that f ≥ 0. Take ε = 1 and let δ0 be the corresponding valuesuch that whenever A ⊂ E, A ∈ F , with m(A) < δ0, we have

∫Af < 1.

Claim: E =n⋃k=1

Ek (disjoint) such that m(Ek) < δ0 for all 1 ≤ k ≤ n.

(The proof of this claim is left as an exercise.) Hence we now have∫A

fχEk =

∫Ek

f < 1 for all 1 ≤ k ≤ n.

So each fχEk is integrable. Therefore, f =∑n

k=1 fχEk is integrable. �

Exercises

1. Let f : [a, b] → R be a bounded function. If f is Riemann-integrable, then f is Lebesgue-integrable and ∫

[a,b]

f(x)dm =

∫ b

a

f(x)dx.

21

2. Let E ∈ F with m(E) < ∞ and {fn} be a sequence of R-valued, bounded measurablefunctions on E such that fn → f uniformly on E. Then∫

E

fndm→∫E

fdm.

[Do not use BCT!]

3. The function f(x) = 1√x

is measurable on [0, 1]. Calculate∫[0,1]

1√xdm. [Notice: f is not

Riemann-integrable on [0, 1].]

4. Find∫[0,1]

f(x) dm if

f(x) =

x2 if x ∈ C ∩Qx if x ∈ C ∩Qc

1 if x ∈ [0, 1] \ Cwhere C is the standard Cantor Set.

5. Let {fn}n≥1 be a sequence of R-valued functions on [0, 1] defined as

fn(x) =

{0 if x ∈ (0, 1

n+1) ∪ [ 1

n, 1]

x−32 if x ∈ [ 1

n+1, 1n).

Show that

a) fn → 0 a.e.b)∫[0,1]

fn → 0.

c) There is no integrable function g : [0, 1] → R such that fn ≤ g a.e. for all n ≥ 1. Howwould you reconcile this result with the DCT?

III.5. Convergence in Measure

In regards to convergence of sequence of functions, in addition to uniform, pointwise and a.e.convergence, we can add another mode of convergence as follows.

Definition. Let {fn} be a sequence of real valued measurable functions on E ∈ F and let eachfn be finite almost everywhere on E. For f : E → R measurable and finite almost everywhereon E, we say that the sequence {fn} converge in measure on E to f if for all α > 0

limn→∞

m({x ∈ E : |fn(x)− f(x)| ≥ α}) = 0.

Notation: fn →m f .

Remark. It is easy to see that if fn →u f , then fn →m f . Indeed, with an additional conditionwe can replace uniform convergence with a.e. convergence by the following result.

Fact.16 Let m(E) <∞ and {fn} be a sequence of real valued measurable functions on E ∈ F ,and f : E → R be a finite almost everywhere. Then

fn →a.e. f ⇒ fn →m f.

Proof. Clearly f is measurable on E (why?). Let α > 0 and ε > 0 be arbitrary. By Egoroff’sTheorem, there exists F ⊂ E, F ∈ F , such that m(E \ F ) < ε and fn →u f on F . Hence thereexists N such that when n ≥ N we have

|fn(x)− f(x)| < α

22

for all x ∈ F . Thus if n ≥ N ,

m({x ∈ E : |fn(x)− f(x)| ≥ α}) ≤ m(E \ F ) < ε.

Hence m({x ∈ E : |fn(x)− f(x)| ≥ α})→ 0. �

Remarks. 1. The condition m(E) < ∞ is essential. (Exercise: Provide a counterexample ifm(E) =∞.)

2. The converse of Fact.16 is not true in general, either. (Exercise: provide an example).However, it’s valid along a subsequence:

Theorem.15 If fn →m f on E ∈ F , then there exists a subsequence {fnk} of {fn} such thatfnk →a.e. f .

Proof. Let εk = 12k

. Applying the hypothesis, for each k there exists Nk such that when

n ≥ Nk, m({x ∈ E : |fn(x)− f(x)| ≥ 1k}) < 1

2k. Let

Ek = {x ∈ E : |fNk(x)− f(x)| > 1

k}.

Then∞∑k=1

m(Ek) <∞.

Hence by Borel-Cantelli Lemma, for almost every x, the point x belongs to at most finitelymany Ek’s. That means for almost every x, there exists kx such that x 6∈ Ek for k ≥ kx. So, foralmost every x,

|fnk(x)− f(x)| < 1

kfor all k ≥ kx. Hence fnk →p f almost everywhere on E. �

Remark. In Fatou’s Lemma, MCT and DCT, the condition fn →a.e. f on E can be replacedby fn →m f on E.

III.6. Riemann Integral vs Lebesgue Integral

Recall: Let f : [a, b]→ R be bounded, Pn be a tagged part partition. Then

S(f, Pn) =n∑i=1

f(ti)(xi+1 − xi)

S(f, Pn) −→∫ b

a

f =: Riemann Integral of f on [a, b].

Let’s look at this process a little differently. For f : [a, b] → R a bounded function and Pn apartition of [a, b], define step functions {αn} and {βn} by

αn(x) =n∑k=1

ukχ[xk,xk+1](x), βn(x) =n∑k=1

vkχ[xk,xk+1](x), respectively,

where uk = infx∈[xk,xk+1]{f(x)} and vk = supx∈[xk,xk+1]{f(x)}. Then,

(i) αn ↑ and βn ↓ as n→∞ pointwise (or as ‖Pn‖ → 0).(ii) For all n ≥ 1, αn and βn are Lebesgue integrable (by construction).

23

(iii) For all n ≥ 1, αn and βn are Riemann-integrable since,

Ln(f) =

∫ b

a

αn =n∑k=1

mk(xk+1 − xk) and

Un(f) =

∫ b

a

βn =n∑k=1

vk(xk+1 − xk),

and that Ln(f) and Un(f) are special cases of S(f, Pn).

We also have that Ln(f) ↑ and Un(f) ↓ . Therefore, f is Riemann-integrable if and only if

limn Un = limn Ln(=∫ baf).

In general, since f is bounded, both sequences αn and βn are uniformly bounded; hence,αn(x)→ α(x) and βn(x)→ β(x) pointwise, for some α : [a, b]→ R and β : [a, b]→ R. Now, theDominated Convergence Theorem implies that

Ln(f) =

∫[a,b]

αn −→∫[a,b]

α, where α = limn→∞

αn(x) and

Un(f) =

∫[a,b]

βn −→∫[a,b]

β, where β = limn→∞

βn(x).

If∫[a,b]

α =∫[a,b]

β, then the Lebesgue Integral of f exists and is equal to the common value.

Therefore, any bounded Riemann-integrable function is Lebesgue integrable.

Fact.17 f : [a, b]→ R is continuous at x ∈ [a, b] if and only if α(x) = β(x).

Proof. Exercise.

Theorem.16 Let f : [a, b]→ R be bounded.

(a) f is Riemann Integrable on [a, b] if and only if f is almost everywhere continuous on [a, b].(b) If f is Riemann Integrable on [a, b], then f is Lebesgue Integrable and∫

[a,b]

f =

∫ b

a

f.

Proof. From above observation, if (a) is proven, (b) follows easily.

(a) (⇒) Since f is Riemann Integrable, then

limk→∞

∫ b

a

αk = limk→∞

∫ b

a

βk ⇒∫ b

a

α =

∫ b

a

β.

Since α ≤ β, we have∫[a,b]

(β − α) = 0 =

∫ b

a

(β − α)⇒ α = β almost everywhere.

Hence by Fact.17, f is continuous almost everywhere on [a, b].

(⇐) If f is continuous almost everywhere, then by Fact.17 for almost every x ∈ [a, b] α(x) =β(x)(= f(x)). Since α and β are Lebesgue Integrable,∫

[a,b]

α =

∫[a,b]

β.

24

Hence we must have that if

Ln =

∫ b

a

αn =

∫[a,b]

αn and Un =

∫ b

a

βn =

∫[a,b]

βn,

then |Un − Ln| = 0. That is f is Riemann Integrable and we must have∫ b

a

f = limn→∞

∫ b

a

αn

(= lim

n→∞

∫ b

a

βn

). �

Important Remark. All the concepts introduced and developed in this chapter can berephrased in an arbitrary σ-finite measure space (X,A, µ) and for functions f : X → R (withappropriate modifications). We will leave this as an exercise.