FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez III

Home , Almost everywhere, Measurable function, Simple function

Do˘ganC¸¨omez

III. MEASURABLE FUNCTIONS AND LEBESGUE INTEGRAL

III.1. Measurable functions

Having the Lebesgue measure deﬁne, in this chapter, we will identify the collection of functions that 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 ∞ ∞ [ −1 [ −1 O = Ik ⇔ f (O) = f (Ik). k=1 k=1

Continuous functions are well behaved for even Riemann-integral; hence we will naturally be compatible 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 all a ∈ 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 \ 1 [a, ∞] = (a − , ∞], n n it follows that ∞ ! ∞ \ 1 \ 1 {x ∈ E : f(x) ≥ a} = f −1 (a − , ∞] = f −1 (a − , ∞] ∈ F. n n n=1 n=1 (ii) ⇒ (iii):

−1 −1 −1 {x ∈ E : f(x) < a} = f ((−∞, a)) = f (R \ [a, ∞]) = E \ f ([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) = ∞} = {f(x) > n} ∈ F. n=1 1 2

Examples. 1. Let f : [0, 1] → R be a function such that 0 if x ∈ f(x) = Q . 1 if x ∈ Qc Then,

−1 1 1 c f ( 2 , ∞) = {x ∈ [0, 1] : f(x) > 2 } = Q ∩ [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   0 if x ≤ −1  x2 if − 1 < x < 1 f(x) =  2 if x = 1  2 − x if x > 1

If a < 0, then f −1(a, ∞) = {x : f(x) > a} = (−∞, 2 − a) ∈ F. −1 If a = 0, then f (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, −1 −1 {x ∈ D : f(x) > a} = (f|D) (a, ∞) = D ∩ f (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} ∪ {x ∈ E \ A : f(x) > a} ∈ F. | {z } | −{z1 } ∈F (f|E\A) (a,∞)∈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). 1 iii) f is measurable on E. 3

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 n a o {x ∈ E :(αf)(x) > a} = x ∈ E : f(x) > ∈ F. α Similarly for α < 0. Hence αf is measurable, and so is βg. Therefore, it is enough to show that f + 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)} = ({x ∈ E : f(x) > r} ∩ {x ∈ E : r > a − g(x)}) ∈ F. | {z } | {z } r∈Q ∈F ∈F

1 2 2 2 (ii) First, observe that fg = 2 [(f + g) − f − g ]. 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} ∪ {x ∈ E : f(x) < − a} ∈ F. | {z } | {z } ∈F ∈F 1 (iii) If a = 0, then {x ∈ E : f (x) > a} = {x ∈ E : f(x) > 0} ∈ F. So assume that a > 0. Then 1 1 x ∈ E : (x) > a = x ∈ E : f(x) < ∈ F. f a (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 1 if x ∈ A χ (x) = A 0 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,

∅ if a ≥ 1 χ−1(a, ∞) = E E 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 ∈ F such that f(A) 6∈ F.

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

1 1 1 h−1 , ∞ = (χ ◦ g−1)−1 , ∞ = (g ◦ χ−1) , ∞ 2 A 2 A 2 1 = g χ−1 , ∞ = g(A) = f(A) 6∈ F. A 2

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) = −1 −1 −1 −1 −1 f (g (O)). Since g is continuous, g (O) ∈ B. Hence f (g (O)) ∈ F.

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

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

Deﬁnition. Let {fn} be a sequence of real-valued functions on a measurable set E. We say that {fn} converges almost everywhere (a.e.) to f : E → R if there exists N ⊂ E with m(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, 1 then ∃m ≥ 1 such that f(x) > a + m . Thus, since fn → f pointwise, for large enough n, we will 1 have fn(x) > a + m . This means that,

1 ∃m ∃N such that, ∀n ≥ N, f (x) > a + . (∗) n m 5

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

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

Theorem.2 Let E ∈ F and fn : E → R be a measurable function for all n ≥ 1. If fn → f almost 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 on which fn → f pointwise, we can assume that fn → f pointwise on E. Then by the observation above, for all a ∈ R !! [ [ \ 1 {x ∈ E : f(x) > a} = x : f (x) > a + ∈ F. n m m≥1 N≥1 n≥N

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

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

∗ f (x) = sup fn(x) n≥1

f∗(x) = inf fn(x) n≥1

f(x) = lim sup fn(x) n≥1

f(x) = lim inf fn(x) n≥1

∗ 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 is called a simple function. 6

Fact.4 A function f : E → R is simple if and only if there exists a1, a2, . . . , an ∈ R and E1,E2,...En ∈ F such that ∞ n [ X E = Ek and f(x) = akχEk (x). k=1 k=1

Proof. (⇐) follows from the definition. (⇒) In order to show that f is measurable, assume a1, a2, . . . an are its values. Then we have −1 Sn f ({ak}) = Ek ∈ F. If i 6= j, then ai 6= aj and hence Ei ∩ Ej = ∅, and k=1 Ek = E. Let n X f(x) = akχEk (x). k=1 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, the same function can be expressed in many different ways involving characteristic functions of different sets. Fact.5 Let f and g be simple functions on E ∈ F, then

i) f + g is simple. ii) fg is simple. f iii) g 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 > 0, there exists simple functions φ and ψ on E such that

φ(x) ≤ f(x) ≤ ψ(x) on E and 0 ≤ (ψ − φ)(x) < on E.

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

a < y0 < y1 < ··· < yn−1 < yn = b. −1 For i = 0, 1, . . . n − 1 let Ei = f ((yi, yi+1]) ∈ F. Then

n−1 n−1 X X φ(x) = yiχEi (x) ≤ f(x) ≤ yi+1χEi (x) = ψ(x). i=0 i=0 Also, n−1 X ψ(x) − φ(x) = (yi+1 − yi)χEi (x) < . i=0

Theorem.3 (Approximation by Simple Functions) Let E ∈ F and f : E → R#. Then f is measurable on E if and only if there exists a sequence {φn} of simple functions on E such that φn → f almost everywhere and |φn| ≤ f on E. Furthermore, if f ≥ 0, then φn’s can be chosen as φn ↑ f almost everywhere on E. Proof. Since f = f + − f − and |f| = f + + f − without loss of generality, we can assume f is non-negative (i.e. f ≥ 0). Therefore let f ≥ 0. Note that (⇐) part is Theorem.2 above. 7

1 (⇒) Given n ≥ 1, let En = {x ∈ E : f(x) ≤ n}. Apply the previous fact with = n . So there 1 exists φn and ψn simple functions on En such that φn ≤ f ≤ ψn on En and ψn − φn < n on En. 1 Now φn ≤ f on En and f − φn < n on En.

Extend φn to the rest of E by letting φn(x) = n on E \ En. Now, we’ll show that φn → f pointwise on E. If x is such that f(x) < ∞, then we have f(x) < N for some N. Then, for any n ≥ N, since 1 x ∈ En for some n, we have 0 ≤ f(x) − φn(x) ≤ 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 there exists {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 if f −1(O) ∈ F, ∀ O ∈ B.

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

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

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, we will proceed to define the Lebesgue integral (of measurable functions). Throughout we will assume that: E ∈ F with m(E) < ∞, all functions f : E → R are bounded, and all simple functions will be in canonical form. Pn Definition. Let φ : E → R be a simple function defined as φ(x) = aiχEi (x). The R i=1 (Lebesgue) integral of φ, denoted by E φ(x) dm, is defined as n Z X φ(x) dm = aim(Ei). E i=1 R R R R Notation: E φ(x) dm = E φ(x) dx = E φ dm = E φ. Remarks. 1. If we extend φ to R by letting φ(x) = 0 for all x ∈ R \ E, then Z Z φ(x) dm = (φ · χE)(x) dm. E R 2. The Lebesgue integral of a step function is the same as the Riemann integral of a step function. 8

Example. Let E = [0, 1] and let φ(x) = χQ∩[0,1]. Then Z φ(x) dx = 1 · m(Q ∩ [0, 1]) = 0. E Hence φ is Lebesgue integrable. Recall φ is not Riemann integrable! R c Exercise. Calculate [0,1] φ, where φ(x) = χC∩Q + 2χC∩Q + 3χ[0,1]\C.

Fact.7 Let φ, ψ : E → R be simple functions. Then R R R i) E(aφ + bψ) = a E φ + b E ψ for all a, b ∈ R. ii) R φ ≤ R ψ if φ ≤ ψ. RE RE R iii) φ = φ + φ where E1,E2 ∈ F, E = E1 ∪ E2, and E1 ∩ E2 = ∅. E E1 E2 Pn Pm Proof. i) Let φ(x) = i=1 aiχEi (x) and ψ(x) = j=1 bjχFj (x) be in canonical form. Let N {Ak}k=1 be deﬁned as Ak = Ei ∩ Fj with an appropriate ordering. Then N n,m {Ak}k=1 = {Ei ∩ Fj}i,j=1 SN is a disjoint collection with k=1 Ak = E. Hence N X aφ + bψ = (aai + bbj)χAk k=1 where i, j corresponds to the set Ak with Ak = Ei ∩ Fj. Therefore, we have N N N Z X X X (aφ + bψ) = (aai + bbj)m(Ak) = aaim(Ak) + bbjm(Ak) E k=1 k=1 k=1 N N n m X X X X = a aim(Ak) + b bjm(Ak) = a aim(Ei) + b bjm(Fj) k=1 k=1 i=1 j=1 Z Z = a φ + b ψ. E E ii) Since φ ≤ ψ on E, ψ − φ ≥ 0 on E. Also since ψ − φ is simple, by i) we have Z Z Z (ψ − φ) ≥ 0 ⇒ ψ ≥ φ. E E E Proof of iii) is left as an exercise.

Recall: Given any f : E → R bounded, we can always ﬁnd (construct) simple functions φ and ψ 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 Z Z sup φ : φ ≤ f on E = inf ψ : ψ ≥ f on E , E E where φ and ψ are simple functions. The common value is called the (Lebesgue) integral of f over E and is denoted by Z Z Z f(x) dx = f = f dm. E E E 9

For the simplicity of the arguments below, we will denote: Z Z inf ψ = Uf and sup φ = Lf . f≤ψ E f≥φ E

Theorem.4 Let E ∈ F, m(E) < ∞, and f : E → R be a bounded function. Then f is measurable 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 2n subintervals of equal length. Let

−M = y−n < y−n+1 < ··· < yn−1 < yn = M −1 be the endpoints of these subintervals. Deﬁne Ek = f ([yk, yk+1]). Then, Ek ∈ F for all k = −n, . . . n with Ek ∩ E` = ∅ for k 6= `, and n X m(Ek) = m(E). k=−n Pn−1 Pn−1 Now deﬁne φn = k=−n ykχEk and ψn = k=−n yk+1χEk . Therefore, on the set E,

φn(x) ≤ f(x) ≤ ψn(x), where {ψn} and {φn} are integrable. Then Z Z Z Z inf ψ ≤ ψn and sup φ ≥ φn. f≤ψ E E f≥φ E E Thus,

" n−1 # Z Z Z Z X M inf ψ − sup φ ≤ (ψn − φn) = (yk+1 − yk)χEk < m(E). f≤ψ f≥φ n E E E E k=−n

Since n is arbitrary, we have Lf = Uf and f is Lebesgue integrable. 1 (⇐) Given that f is bounded on E and Lf = Uf , let 0 < n = n for n ≥ 1. From these assumptions 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 Z Z 1 0 ≤ ψn − φn < . E E n

Let ψen = infn ψn and φen = supn φn. Since each ψn and φn are measurable, we have that both ψen and φen are measurable on E, and φen ≤ f ≤ ψen on E. We need to show ψen = φen almost everywhere on E. To show this let

A = {x ∈ E : φe(x) < ψe(x)}.

We will show that m(A) = 0. To do this let Ak = {x ∈ E : ψe(x) − φe(x) > 1/k}, then S k≥1 Ak = A. Now let n Bk = {x ∈ E : ψn(x) − φn(x) > 1/k}. n Then Ak ⊂ Bk ∈ F and, on E, k(ψn − φn) > 1. So Z Z Z n k m(B ) = χ n ≤ k(ψ − φ ) = k (ψ − φ ) < . k Bk n n n n E E E 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 Lebesgue integrable with Z Z b f dm = f(x) dx. [a,b] a

Proof. Exercise.

Fact.9 Let E ∈ F, m(E) < ∞, f, g : E → R be bounded functions. Then we have: R R R a) E(αf + βg) = α E f + β E g for all α, β ∈ R. b) If f = g almost everywhere on E, then R f = R g. R ER E R R c) If f ≤ g almost everywhere, then E f ≤ E g (Hence | E f| ≤ E |f|). d) If there exists a, b ∈ R such that a ≤ f(x) ≤ b on E, then Z am(E) ≤ f ≤ bm(E). E

e) If E1,E2 ∈ F, and E = E1 ∪ E2 with E1 ∩ E2 = ∅, then Z Z Z f = f + f. E E1 E2 Proof. Exercise.

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

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

Example. Let fn : [0, 1] → R, n ≥ 1, be given by  1  n2x if x ∈ [0, ]   n f (x) = 1 2 n − n2x + 2n if x ∈ [ , ]   n n  0 otherwise, and f ≡ 0 on [0, 1]. Then, Z Z fn = 1 6= 0 = f ! [0,1] [0,1]

Theorem.5 (Egoroﬀ) Let E ∈ F with m(E) < ∞, and {fn} be a sequence of real valued measurable functions on E such that fn → f almost everywhere. Then for all > 0, there exists a measurable A ⊂ E, m(E \ A) < , such that fn → f uniformly on A. Proof. Given > 0 and for n, j ∈ Z+, deﬁne ∞ j \ En = {x ∈ E : |fk(x) − f(x)| < 1/j}. k=n 11

j Then En ∈ F for all n, j. If A ⊂ E such that fn → f pointwise on A (i.e. m(E \ A) = 0), then j for all x ∈ A, x ∈ En for some n, j. Hence ∞ [ j A ⊂ En. n=1 Therefore ∞ ∞ ! [ j [ j A ⊂ En ⊂ E ⇒ m En = m(E), since fn → f a.e. (hence m(A) = m(E)). n=1 n=1 j j j j Observe for a ﬁxed j, En ⊂ En+1. Hence by continuity of m, m(E \ En) = m(E) − m(En) and ∞ ! j j [ j lim m(E \ En) = m(E) − lim m(En) = m(E) − m En = 0. n→∞ n→∞ n=1 j Consequently, for all j ≥ 1, we can ﬁnd n(j) such that n ≥ n(j) and m(E \ En) < 2j . Let T∞ j A = j=1 En(j), then A ∈ F. Then ∞ ∞ ! ∞ \ [ X X m(E \ Ej ) = m (E \ Ej ) ≤ m(E \ Ej ) < = . n(j) n(j) n(j) 2j j=1 j=1 j=1 j

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

Remark. By the example above, if fn → f pointwise on E ∈ F, in general, we cannot expect R R E fn → E f. However, under somewhat mild additional conditions we have a positive answer:

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

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 > 0, there exists A ⊂ E such that m(E\A) < 4M and fn → f uniformly on A. Then, find N large enough that, for n ≥ N, |fn(x) − f(x)| < 2m(A) for all x ∈ A. Now for such n ≥ N, Z Z Z Z fn − f = (fn − f) + (fn − f), E E A E\A and hence, Z Z Z Z Z Z | fn − f| ≤ |(fn − f)| + |(fn − f)| ≤ |fn − f| + 2Mdm, E E A E\A A E\A since |fn − f| ≤ 2M. Therefore, we have Z Z 2M fn − f < m(A) + = . E E 2m(A) 4M

Remark. The BCT can be restated as saying that under the given conditions the “lim” and “R ” can be interchanged. 12

Examples. 1. Given 1 if x ∈ [1/2, 1] ∪ [1/4, 1/3] ∪ [1/6, 1/5] ∪ ... f(x) = 0 otherwise Deﬁne

f1(x) = χ[1/2,1]

f2(x) = χ[1/4,1/3] + χ[1/2,1] . . n X fn(x) = χ[1/(2k),1/(2k+1)] k=1 . .

It is straightforward to show that fn → f pointwise (hence, is left as an exercise). Then {fn} is a uniformly bounded sequence of measurable functions. By the Bounded Convergence Theorem, " n # Z Z X 1 1 f = lim fn = lim m , n→∞ n→∞ 2k 2k − 1 [0,1] [0,1] k=1 " n # " 2n # X 1 1 X 1 = lim − = lim (−1)k+1 n→∞ 2k − 1 2k n→∞ k k=1 k=1 ∞ X (−1)k+1 = = ln(2). k k=1

2. Let 2x if x ∈ [0, 1] ∩ f(x) = Q 1 − 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, Z Z Z Z 1 1 f(x) dx = f(x) dx + f(x) dx = g(x) dx = . [0,1] [0,1]∩Q [0,1]∩Qc 0 2

Exercises

R √ R 1/3 1. Using only the deﬁnition, evaluate the integrals: [0,1] x dx and [0,1] x dx. R 2. Let C denote the Cantor Set. Find [0,1] f(x) dx if  2  x if x ∈ C ∩ Q f(x) = x if x ∈ C ∩ Qc  1 if x ∈ [0, 1] \ C.

3. Prove that fn → f uniformly on A 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 then extend it to more general functions f : E → R.

Deﬁnition. Let E ∈ F (m(E) = ∞ is allowed) and f : E → R+ be measurable. We deﬁne Z Z fdm = sup{ hdm : h : E → R is bounded, measurable, h ≤ f, m({x : h(x) 6= 0}) < ∞}. E E

Remark. R (1) It is possible that E f = ∞ (2) Functions h : E → R such that m({x ∈ E : h(x) 6= 0}) < ∞ are called functions with ﬁnite support. The set {x ∈ E : h(x) 6= 0} is called the support of h.

Exercise. Find Z 1 √ dm. [0,1] x

Fact.10 Let E ∈ F, f, g : E → E+ be measurable functons. Then R R R (a) E(αf + βg) = α E f + β E g. (b) If f ≤ g almost everywhere on E, then R f ≤ R g. R R R E E (c) f = f + f where E = E1 ∪ E2 and E1 ∩ E2 = ∅. E E1 E2 Proof. The result follows from the the same fact on bounded functions; hence, it is left as an exercise.

+ Theorem.7 (Fatou’s Lemma) Let E ∈ F and {fn} be a sequence of R -valued, measurable functions on E. If fn → f almost everywhere on E, then Z Z f dm ≤ lim inf fn dm. E n→∞ E

Proof. Let φ be a bounded, measurable, non-negative function with m({φ 6= 0}) < ∞ such that 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, Z Z Z Z φ dm = φ dm = lim hn dm ≤ lim inf fn dm. E A n→∞ A n→∞ E R (Last inequality follows since limn E fn dm may not exist.) Therefore, Z Z Z f dm = sup φ dm ≤ lim inf fn dm. E φ≤f E n→∞ E

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 of + R -valued measurable functions on E such that fn ≤ fn+1 a.e. (on E) for all n ≥ 1. If fn → f a.e., then Z Z lim fn dm = f dm. n→∞ E E

Proof. From hypothesis, fn ≤ fn+1 for all n almost everywhere on E. Hence Z Z fn dm ≤ f dm E E for all n. Therefore by Fatou’s Lemma Z Z Z lim sup fn dm ≤ f dm ≤ lim inf fn dm. E E E R R Hence, it follows that limn→∞ E fn dm = E f 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 Z Z fn → 0 pointwise (but not monotonically), whereas, ∞ = fn 6→ 0 = f. R R

Corollary.1 Let E ∈ F and {un} be a sequence of positive real valued functions on E. If ∞ ∞ X Z X Z f = un (pointwise), then f dm = un dm. n=1 E n=1 E

Pn Proof. Deﬁne fn = k=1 uk, for n ≥ 1. Then fn ↑ f a.e. on E. Apply MCT: n ∞ Z Z X Z X Z Z lim fn dm = f dm ⇒ lim uk dm = uk dm = f dm. n→∞ n→∞ E E k=1 E k=1 E E

S Corollary.2 Let E ∈ F and {En} ⊂ F such that E = n En and Ei ∩ Ej = ∅ if i 6= j. Then for any f : E → R+ measurable, ∞ X Z Z f dm = f dm n=1 En E

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, 1 Z m({x ∈ E : f(x) > λ}) ≤ f dm. λ E 15

Proof. Let Eλ = {x ∈ E : f(x) > λ}. Assume ﬁrst that m(Eλ) < ∞ and let φ = λχEλ . Then on Eλ, φ ≤ f. Hence Z Z φ dm ≤ f dm E E k Z 1 Z λχEλ dm = λm(Eλ) ⇒ m(Eλ) ≤ f dm. E λ E

n n Next we will assume that m(Eλ) = ∞. If we let Eλ = Eλ ∩ [−n, n], then Eλ ↑ Eλ and n n m(E ) < ∞ for each n. Now, applying above case to E and φ = λχ n , we obtain λ λ n Eλ Z n λm(Eλ ) ≤ f dm. E Then, by continuity of m, Z n λm(Eλ) = λ lim m(Eλ ) ≤ f dm. n→∞ E

Note that this gives equality on both sides since m(Eλ) = ∞. Hence, 1 Z m(Eλ) = f dm. λ E

Theorem.10 Let E ∈ F and f : E → R+ be measurable. Then Z f dm = 0 ⇔ f = 0 almost everywhere on E. E R Proof. (⇒) Assume E f dm = 0 and consider, for any n, the set S = {x ∈ E : f(x) > 1/n}. Then by Chebychev’s Inequality, Z m(S) ≤ n f dm = 0. E Now, since [ {x ∈ E : f(x) > 0} = {x ∈ E : f(x) > 1/n}, 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]. On each En let h be a bounded, measurable, non-negative function such that h ≤ f, and φ be any bounded simple non-negative function with φ ≤ h (i.e. 0 ≤ φ ≤ h ≤ f a.e.). Hence φ must be 0 almost everywhere on En. Then Z Z φ = 0 ⇒ h = 0. En En R Since h is arbitrary, E f = 0 follows from the deﬁnition. R R Remark. Theorem.10 implies that f = g a.e. on E ∈ F ⇐⇒ E f = E g.

Deﬁnition. Let E ∈ F. A function f : E → R+ is called (Lebesgue) integrable over E if Z f dm < ∞. E 16

1 Example. Let’s show that the function f(x) = x2 is Lebesgue-integrable on [1, ∞) and calculate R 1 [1,∞) x2 . For, we will use MCT. First, let, for n ≥ 1,  1  2 if 1 ≤ x ≤ n fn(x) = x  0 if otherwise

Hence, each fn is bounded and measurable on [1, ∞). Thus, Z Z Z n 1 fn = fn = 2 . [1,∞) [1,n] 1 x 1 Since x2 is Riemann-integrable on [1, n], it is also Lebesgue integrable on [1, n] (with both R R n 1 1 integrals being equal). Hence, [1,∞) fn = 1 x2 = 1 − n . Therefore, by MCT, Z 1 Z 2 = lim fn = 1. [1,∞) x n [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 T {x ∈ E : f(x) = ∞} = n≥1{x ∈ E : f(x) > n}. Now, by Chebychev’s Inequality, 1 Z m({x ∈ E : f(x) = ∞}) ≤ m({x ∈ E : f(x) > n}) ≤ f. n E R Since E f < ∞, letting n → ∞, we obtain the desired result.

Theorem.11 (Beppo Levi’s Lemma) Let {fn} be an increasing sequence of nonnegative R measurable functions on E ∈ F. If the sequence { E fn} is bounded, then

p # i) fn → f on E for some non-negative measurable function f : E → R , R R ii) limn E fn = E f < ∞, 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) = lim fn(x), for all x ∈ E. n R R R Since fn ↑ f, by MCT, we have fn → f. Therefore, by the fact that { fn} is bounded, R E E E we have have E f < ∞; 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 > 0 there R exists δ > 0 such that ∀A ⊂ E,A ∈ F, with m(A) < δ, we have A f < . Proof. If |f| < M for some M on E, take δ = M . The the assertion follows. If f is arbitrary integrable function, let f(x) if x ≤ n f (x) = n n if x > n. 17

p R R Then {fn} is a bounded sequence and fn → f. on E. So, by MCT, E fn → E f. Hence, given R R > 0, there exists N ≥ 1 such that if n ≥ N, then [ E f − E fn] < 2 . In particular, if n = N, R we have E(f − fN ) < 2 . Now, pick δ < 2N , then for any A ⊂ E with m(A) < δ we have Z Z Z Z Z f = (f − fN ) + fN ≤ (f − fN ) + N < + = . A A A E A 2 2

Exercise. Let f : R → R+ be an integrable function. Deﬁne a set function ν : F → [0, ∞] by R ν(A) = A fdm, 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. ∞ P∞ 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 if both f + and f − are integrable over E. In that case, Z Z Z f = f + − f −. E E E

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

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

Proof. Exercise.

Fact.13 Let E ∈ F, f : E → R be measurable and g : E → R+ be integrable with |f| ≤ g on E. Then f is integrable with Z Z f ≤ g. E E R R In particular, | E f| ≤ E |f|. 18

Proof. Exercise.

Fact.14 If E ∈ F and f, g : E → R are integrable over E, then R R (a) αf is integrable over E and E αf = α E f. (b) f + g is integrable over E and R (f + g) = R f + R g. R R E E E (c) f ≤ g on E, then E f ≤ E g.

(d) If E = E1 ∪ E2, where E1 and E2 are measurable and disjoint, then fχE1 and fχE2 are integrable with Z Z Z f = f + f. E E1 E2 Proof. The proof of (a) and (c) are left as exercises. For (b) without loss of generality, if necessary, 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 Z Z Z Z Z Z (f + g)+ + f − + g− = (f + g)− + f + + g+. E E E E E E If we reorganize we have Z Z Z (f + g)+ − (f + g)− = (f + + f −) + (g+ − g−). E E E Hence f + g is integrable.

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

Theorem.13 (Dominated Convergence Theorem) Let {fn} be a sequence of real valued measurable 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 over E and Z Z fn −→ f. E E

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 to g − f. Similarly {g + fn} is a sequence of non-negative functions converging to g + f. Now by applying Fatou’s Lemma to {g − fn} we have Z Z Z Z Z Z g − f = (g − f) ≤ lim inf (g − fn) = g − lim sup fn E E E E E E Z Z ⇒ f ≥ lim sup fn. E E Next, by applying Fatou’s Lemma to {g + fn} we have Z Z Z Z Z Z f + g = (g + f) ≤ lim inf (g + fn) = g + lim inf fn E E E E E E Z Z ⇒ f ≤ lim inf fn. E E 19 R R These two inequalities imply that E fn → E f.

n3/2x Exercise. For each n ≥ 1, let fn(x) = 1+n2x2 , x ∈ [0, 1]. Show that

(a) fn → 0 almost everywhere on [0, 1]. √1 (b) |fn(x)| ≤ g(x) on [0, 1] for all n where g(x) = x . R (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, Z Z fn −→ f, but fn 6→ f. E E n−1 2. Let fn(x) = n − x and f(x) = x. Then Z Z fn −→ f, but fn 6→ f. E E 3. Deﬁne

f0 = χ[0,1]

f1 = χ 1 , f2 = χ 1 [0, 2 ] [ 2 ,1] f3 = χ 1 , f4 = χ 1 2 , f5 = χ 2 [0, 3 ] [ 3 , 3 ] [ 3 ,1] f6 = χ 1 , f7 = χ 1 1 , f8 = χ 1 3 , f9 = χ 3 [0, 4 ] [ 4 , 2 ] [ 2 , 4 ] [ 4 ,1] f10 = χ 1 ,... [0, 5 ] . . R Then we have E fn −→ 0, but fn 6→ 0 almost everywhere.

Theorem.14 (Generalized Dominated Convergence Theorem) Let {fn} be a sequence of 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 almost everywhere on E. Also let |fn| ≤ gn for all n on E. If Z Z Z Z gn −→ g, then fn −→ f. E E E E

Proof. Exercise.

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

∞ [ (a) If E = En (disjoint), En ∈ F. Then n=1 ∞ Z X Z f = f. E n=1 En 20

(b) Given {En} ⊂ F,En ⊂ E and En ↑ . If A = ∪nEn, then Z Z f = lim f. n→∞ A En

(c) Given {En} ⊂ F, En ⊂ E, and En ↓. If A = ∩nEn, then Z Z f = lim f. n→∞ A En 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 > 0 there exists δ > 0 such that for all A ⊂ F, A ⊂ E and m(A) < δ, then Z f < . A

Proof. Exercise. R The results stated above have some important and interesting ramiﬁcations. Let ν(A) = A f for all A ∈ F, where f : R → R is integrable. Then by Corollary.1(a) above,   ∞ ∞  [  X ν  Ak = ν(Ak). k=1 k=1 disjoint

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 > 0 R there exists δ > 0 such that for all A ∈ F, A ⊂ E and m(A) < δ, we have that A f < , then f is integrable over E.

Proof. WLOG we can assume that f ≥ 0. Take = 1 and let δ0 be the corresponding value R such that whenever A ⊂ E, A ∈ F, with m(A) < δ0, we have A f < 1. n [ Claim: E = Ek (disjoint) such that m(Ek) < δ0 for all 1 ≤ k ≤ n. k=1 (The proof of this claim is left as an exercise.) Hence we now have Z Z

fχEk = f < 1 for all 1 ≤ k ≤ n. A Ek Pn So each fχEk is integrable. Therefore, f = 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 Z Z b f(x)dm = f(x)dx. [a,b] a 21

2. Let E ∈ F with m(E) < ∞ and {fn} be a sequence of R-valued, bounded measurable functions on E such that fn → f uniformly on E. Then Z Z fndm → fdm. E E [Do not use BCT!]

√1 R √1 3. The function f(x) = x is measurable on [0, 1]. Calculate [0,1] x dm. [Notice: f is not Riemann-integrable on [0, 1].] R 4. Find [0,1] f(x) dm if  2  x if x ∈ C ∩ Q f(x) = x if x ∈ C ∩ Qc  1 if x ∈ [0, 1] \ C where C is the standard Cantor Set.

5. Let {fn}n≥1 be a sequence of R-valued functions on [0, 1] deﬁned as 1 1 0 if x ∈ (0, n+1 ) ∪ [ n , 1] fn(x) = 3 1 1 − 2 x if x ∈ [ n+1 , n ). Show that

a) fn → 0 a.e. R 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. How would 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 each fn be finite almost everywhere on E. For f : E → R measurable and finite almost everywhere on E, we say that the sequence {fn} converge in measure on E to f if for all α > 0

lim m({x ∈ E : |fn(x) − f(x)| ≥ α}) = 0. n→∞

m Notation: fn → f. u m Remark. It is easy to see that if fn → f, then fn → f. Indeed, with an additional condition we 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 a.e. m fn → f ⇒ fn → f.

Proof. Clearly f is measurable on E (why?). Let α > 0 and > 0 be arbitrary. By Egoroﬀ’s u Theorem, there exists F ⊂ E, F ∈ F, such that m(E \ F ) < and fn → f on F . Hence there exists 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 ) < .

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

Remarks. 1. The condition m(E) < ∞ is essential. (Exercise: Provide a counterexample if m(E) = ∞.) 2. The converse of Fact.16 is not true in general, either. (Exercise: provide an example). However, it’s valid along a subsequence:

m Theorem.15 If fn → f on E ∈ F, then there exists a subsequence {fnk } of {fn} such that a.e. fnk → f. 1 Proof. Let k = 2k . Applying the hypothesis, for each k there exists Nk such that when 1 1 n ≥ Nk, m({x ∈ E : |fn(x) − f(x)| ≥ k }) < 2k . Let 1 E = {x ∈ E : |f (x) − f(x)| > }. k Nk k Then ∞ X m(Ek) < ∞. k=1 Hence by Borel-Cantelli Lemma, for almost every x, the point x belongs to at most ﬁnitely many Ek’s. That means for almost every x, there exists kx such that x 6∈ Ek for k ≥ kx. So, for almost every x, 1 |f (x) − f(x)| < nk k p for all k ≥ kx. Hence fnk → f almost everywhere on E.

a.e. Remark. In Fatou’s Lemma, MCT and DCT, the condition fn → f on E can be replaced m by fn → 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 n ˙ X S(f, Pn) = f(ti)(xi+1 − xi) i=1 Z b ˙ S(f, Pn) −→ f =: Riemann Integral of f on [a, b]. a

Let’s look at this process a little diﬀerently. For f :[a, b] → R a bounded function and Pn a partition of [a, b], deﬁne step functions {αn} and {βn} by n n X X αn(x) = ukχ[xk,xk+1](x), βn(x) = vkχ[xk,xk+1](x), respectively, k=1 k=1 where u = inf {f(x)} and v = sup {f(x)}. Then, k x∈[xk,xk+1] k x∈[xk,xk+1]

(i) αn ↑ and βn ↓ as n → ∞ pointwise (or as kPnk → 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, n Z b X Ln(f) = αn = mk(xk+1 − xk) and a k=1 n Z b X Un(f) = βn = vk(xk+1 − xk), a k=1 ˙ 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 R b limn Un = limn Ln(= a f).

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, the Dominated Convergence Theorem implies that Z Z Ln(f) = αn −→ α, where α = lim αn(x) and [a,b] [a,b] n→∞ Z Z Un(f) = βn −→ β, where β = lim βn(x). [a,b] [a,b] n→∞ R R 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 Z Z b f = f. [a,b] a

Proof. From above observation, if (a) is proven, (b) follows easily. (a) (⇒) Since f is Riemann Integrable, then Z b Z b Z b Z b lim αk = lim βk ⇒ α = β. k→∞ a k→∞ a a a Since α ≤ β, we have Z Z b (β − α) = 0 = (β − α) ⇒ α = β almost everywhere. [a,b] a 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, Z Z α = β. [a,b] [a,b] 24

Hence we must have that if Z b Z Z b Z Ln = αn = αn and Un = βn = βn, a [a,b] a [a,b] then |Un − Ln| = 0. That is f is Riemann Integrable and we must have Z b Z b Z b f = lim αn = lim βn . a n→∞ a n→∞ a

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