Econ 508B: Lecture 3 Riemann, Lebesgue Integral and Repeated Integration

Hongyi Liu

Washington University in St. Louis

July 24, 2017

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 1 / 25 Outline

1 Review of Riemann Integral

2 Lebesgue Integral

3 Comparison between Lebesgue and Riemann Integral

4 Repeated integration

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 2 / 25 Outline

1 Review of Riemann Integral

2 Lebesgue Integral

3 Comparison between Lebesgue and Riemann Integral

4 Repeated integration

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 3 / 25 Riemann integral

Let f be defined and finite on a finite interval [a, b]. If Γ = {a = x0 < x1 < ... < xn = b}] is a partition of [a, b], let |Γ|, called the norm of Γ, be defined as the length or diameter of a longest subinterval of Γ: |Γ| = max(xi − xi−1) i n Then we arbitrarily choose intermediate points {ξi}i=1 satisfying xi−1 ≤ ξi ≤ xi. Let

mi = inf f(x),Mi = sup f(x) xi−1≤x≤xi xi−1≤x≤xi

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 4 / 25 Definition 1.1 (Riemann sums) The Riemann sum and the upper-and lower-Riemann sums of f w.r.t the partition Γ are, respectively, defined as

n X RΓ = f(ξi)(xi − xi−1) i=1 n n X X UΓ = Mi(xi − xi−1),LΓ = mi(xi − xi−1) i=1 i=1

Definition 1.2 (Riemann integral)

We then further define the Riemann integral by if lim|Γ|→0 RΓ exists R and lim|Γ|→0 RΓ = f(x)dx. And such the definition is equivalent to the statement that Z Z Z Z Z f = inf UΓ, f = sup LΓ, f = f = f Γ Γ

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 5 / 25 Outline

1 Review of Riemann Integral

2 Lebesgue Integral

3 Comparison between Lebesgue and Riemann Integral

4 Repeated integration

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 6 / 25 Lebesgue integral of simple function

Recall from the last slide that a simple function ϕ(x) is a finite sum N X ϕ(x) = akχCk (x) k=1 The canonical form of φ is the unique decomposition of the above form, where the numbers ak are distinct and non-zero values, and the sets Ck are disjoint, formulated by taking the set Ck = {x : φ(x) = ak} where ak are distinct values, it turns out the sets Ck are disjoint. Definition 2.1 PN If φ is a simple function with canonical form φ(x) = k=1 akχCk (x), then we define the Lebesgue integral of φ on a measurable set C by

N Z Z Z X ϕ(x)dx = ϕ(x)χC (x)dx = ϕ(x)dµ = akµ(Ck) C C C k=1

where C = ∪kCk Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 7 / 25 Monotone Convergence Theorem for nonnegative functions

Theorem 2.1 (Nonnegative functions)

If {fk} is a sequence of nonnegative measurable functions such that f % f on C, then k Z Z fk → f. C C This allows us to interchange the integral with limits. Theorem 2.2 Let f be nonnegative and measurable on C, then

Z X f = sup [ inf f(x)]µ(Cj) x∈Cj C j

We say, f is Lebesgue integrable or integrable if R f(x)dx < +∞.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 8 / 25 Properties

The integral of nonnegative measurable functions satisfies the following properties: (i) Linearity: If f, g ≥ 0, and a, b are positive real numbers, then Z Z Z (af + bg) = a f + b g

(ii) Additivity: If f ≥ 0 and measurable on C and C is the countable union of disjoint measurable sets Cj,C = ∪jCj. Then

Z X Z f = f. C Cj

(iii) Monotonicity: If 0 ≤ f ≤ g a.e. in C and measurable on C, then R R C f ≤ C g.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 9 / 25 continued

(iv) Let C1 and C2 be measurable and C1 ⊂ C2. If f is nonnegative and measurable on C , then R f ≤ R f. 2 C1 C2 R (v) Let f be nonnegative on C. If µ(C) = 0, then C f = 0. R (vi) Let f ≥ 0 and measurable on C. Then C f = 0 if and only if f = 0 a.e. in C.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 10 / 25 R R Is it always true that fn → f?

However, the following example provides a negative answer to this. Example 2.1 ( n if 0 < x < 1/n fn(x) = 0 o.w. R Here fn(x) → 0 for all x while fn(x)dx = 1 for all n. In particular, R R fn > f.

Lemma 2.1 (Fatou’s Lemma)

Suppose {fk} is a sequence of measurable functions with fn ≥ 0 defined on C, then Z Z ( lim inf fk) ≤ lim inf fk C k→∞ k→∞ C

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 11 / 25 Lebesgue’s Dominatd Convergence Theorem

Theorem 2.3

Let {fk} be a sequence of nonnegative measurable functions on E such R that fk → f a.e. in C. If fk ≤ g a.e. for all k,where g < +∞ or it is integrable. Then Z Z fk → f. C C Proof: R R R By Fatou’s lemma, f = ( lim inf fk) ≤ lim inf fk. Then apply C C k→∞ k→∞ C Fatou’s lemma to the nonnegative functions g − fk, obtaining Z Z Z Z Z (g−f) = lim inf(g−fk) ≤ lim inf (g−fk) = g−lim sup fk. C C C C C Combining the above two inequalities completes the proof.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 12 / 25 The integral of a measurable f

Theorem 2.4 R Let f ≥ 0 defined on a measurable set C. Then C f exists if and only if f is measurable.

Moreover, f = f + − f −, where f + = max{f, 0} and f − = − min{f, 0}. + − R + Since f , f are nonnegative and measurable, therefore, C f and R − C f exist and are nonnegative, possibly having value +∞.Then define Z Z Z f(x)dx = f(x)+dx − f(x)−dx C C C As mentioned before, f is integrable if R f is finite. Usually we write R C f ∈ L(C), L(C) = {f : C f} is finite. Theorem 2.5 Let f be measurable on C. f is integrable over C if and only if |f| is.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 13 / 25 M.C.T & Uniform Convergence Theorem

Let {fk} be a sequence of measurable functions on C.

(i) If fk % f a.e. on C and ∃g ∈ L(C) such that fk ≥ g a.e. on C for R R all k, then C fk → C f. (ii) If fk & f a.e. on C and ∃g ∈ L(C) such that fk ≤ g a.e. on C for R R all k, then C fk → C f. Their proofs are alike to the proof of the nonnegative case. Theorem 2.6 (Uniform Convergence Theorem)

Let fk ∈ L(C) and suppose {fk} uniformly converge to f on R R C, µ(C) < ∞. Then f ∈ L(C) and C fk → C f.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 14 / 25 Fatou’s Lemma

Theorem 2.7 (Fatou’s lemma)

Let {fk} be a sequence of measurable functions on C. If ∃g ∈ L(C) such that fk ≥ g a.e. on C for all k, then Z Z lim inf fk ≤ lim inf fk. C k→∞ k→∞ C

Corollary 2.1

Let {fk} be a sequence of measurable functions on C. If ∃g ∈ L(C) such that fk ≤ g a.e. on C for all k, then Z Z lim sup fk ≥ lim sup fk. C k→∞ k→∞ C

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 15 / 25 Dominated Convergence Theorem & Bounded Convergence Theorem

Theorem 2.8 (Dominated Convergence Theorem)

Let {fk} be a sequence of measurable functions on C such that fk → f R R a.e. in C. If ∃g ∈ L(C) such that |fk| ≤ g a.e. in C, then C fk → C f.

Corollary 2.2 (Bounded Convergence Theorem)

Let {fk} be a sequence of measurable functions on C such that fk → f a.e. in C. If µ(C) < +∞ and ∃ a finite constant M such that |fk| ≤ M R R a.e. in C, then C fk → C f.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 16 / 25 Outline

1 Review of Riemann Integral

2 Lebesgue Integral

3 Comparison between Lebesgue and Riemann Integral

4 Repeated integration

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 17 / 25 Dirichlet function

The Riemann and Lebesgue integral are defined in different ways, with the latter generally perceived as the more general. The example that follows shows that functions exist that are Lebesgue integrable but not Riemann integrable.

Example 3.1 Consider the characteristic function of the rational numbers in [0, 1], ( 1, if x rational, 1 (x) = Q 0, o.w.

This function, known as the Dirichlet function, is not Riemann integrable.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 18 / 25 To shed light on that, choose an arbitrary partition Γ of the interval 1 1 [0, 1]. It turns out that mi = inf Q = 0,Mi = sup Q = 1. Z Z f = 1 6= f = 0.

Therefore, the Riemann integral cannot exist. On the other hand, 1 Q ∈ L([0, 1]) turns out to be Lebesgue integrable. Then enumerate the rationals in [0, 1] as {r0, r1, ...}. Each rational number is covered by i an open set Gi of size /2 . It means that {r0, r1, ...} contained in the ∞ set G = ∪i=0Gi. It is well-known that the countable union of open sets is a Lebesgue measurable set. It turns to consider the following indicator function: ( 1, if x ∈ O, 1G (x) =  0, o.w. O is any open set covering [0, 1]. Therefore, Z Z 1 1 0 ≤ Qdx ≤ G ≤  → 0. . Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 19 / 25 Outline

1 Review of Riemann Integral

2 Lebesgue Integral

3 Comparison between Lebesgue and Riemann Integral

4 Repeated integration

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 20 / 25 Fubini’s Theorem

d d d1 d2 In general, write R as a product R = R × R , where d = d1 + d2. Theorem 4.1 d d d1 d2 d2 Let f(x, y) ∈ L(R ), R = R × R . Then for almost every y ∈ R : d1 (i) f(x, y) is measurable and integrable on R . (ii) The function defined by R f(x, y)dx is integrable on d2. Rd1 R (iii) R (R f(x, y)dx)dy = R f. Rd2 Rd1 Rd By symmetry, we also have that R (R f(x, y)dy)dx = R f. Rd1 Rd2 Rd In particular, Z Z Z Z Z ( f(x, y)dx)dy = ( f(x, y)dy)dx = f. Rd2 Rd1 Rd1 Rd2 Rd d It turns out that Fubini’s theorem states that the integral of f on R can be computed by iterating lower-dimensional integrals, and the iterations can be taken in any order. Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 21 / 25 Tonelli’s Theorem

Theorem 4.2 d d1 d2 Let f(x, y) be nonnegative and measurable on R = R × R . Then d2 for almost every y ∈ R , we will have the same result to Fubini’s theorem. d1 (i) f(x, y) is measurable and integrable on R . (ii) The function defined by R f(x, y)dx is integrable on d2. Rd1 R (iii) R (R f(x, y)dx)dy = R f. Rd2 Rd1 Rd By symmetry, we also have that R (R f(x, y)dy)dx = R f. Rd1 Rd2 Rd In particular, Z Z Z Z Z ( f(x, y)dx)dy = ( f(x, y)dy)dx = f. Rd2 Rd1 Rd1 Rd2 Rd

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 22 / 25 Apply Tonelli’s theorem to the function (ω, x) 7→ 1X(ω)>x. One obtains

Z Z Z Z X(ω) Z 1X(ω)≥xdxdP (ω) = dxdP (ω) = X(ω)dP (ω) = E(X) Ω R+ Ω 0 Ω

while integrating Ω firstly and defining Ax = {ω ∈ Ω|X(ω) ≥ x}, Z Z Z Z 1 1 X(ω)≥xdP (ω)dx = ω∈Ax dP (ω)dx R+ Ω R+ Ω Z Z = P (Ax)dx = P (X ≥ x)dx R+ R+

An application

We will show if X is a non-negative random variable, then E[X] = R P (X ≥ x)dx. Proof: R+

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 23 / 25 An application

We will show if X is a non-negative random variable, then E[X] = R P (X ≥ x)dx. Proof: R+ Apply Tonelli’s theorem to the function (ω, x) 7→ 1X(ω)>x. One obtains

Z Z Z Z X(ω) Z 1X(ω)≥xdxdP (ω) = dxdP (ω) = X(ω)dP (ω) = E(X) Ω R+ Ω 0 Ω

while integrating Ω firstly and defining Ax = {ω ∈ Ω|X(ω) ≥ x}, Z Z Z Z 1 1 X(ω)≥xdP (ω)dx = ω∈Ax dP (ω)dx R+ Ω R+ Ω Z Z = P (Ax)dx = P (X ≥ x)dx R+ R+

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 23 / 25 ∞ ∞ i X X X E[X] = ip(i) = p(i) i=1 i=1 k=1 ∞ ∞ ∞ X X X = p(i) = P (X ≥ k) k=1 i=k k=1 Intuitively, we could add up probabilities column-by-column, though, and get the same result by adding row-by-row.

continued

P∞ For the discrete case, E(X) = i=1 P (X ≥ i). Proof:

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 24 / 25 continued

P∞ For the discrete case, E(X) = i=1 P (X ≥ i). Proof:

∞ ∞ i X X X E[X] = ip(i) = p(i) i=1 i=1 k=1 ∞ ∞ ∞ X X X = p(i) = P (X ≥ k) k=1 i=k k=1 Intuitively, we could add up probabilities column-by-column, though, and get the same result by adding row-by-row.

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 24 / 25 Convolution

n If f and g are measurable in R , their convolution (f ∗ g)(x) is defined by Z (f ∗ g)(x) = f(x − t)g(t)dt, Rn provided the integral exists. And f ∗ g = g ∗ f simply by changes of variable. Theorem 4.3 n n If f ∈ L(R ) and g ∈ L(R ), then (f ∗ g)(x) exists for almost every n n x ∈ R and is measurable. Moreover, f ∗ g ∈ L(R ) and Z Z Z (f ∗ g)dx = ( fdx)( gdx) Rn Rn Rn Z Z Z |f ∗ g|dx ≤ ( |f|dx)( |g|dx) Rn Rn Rn proved by Tonelli’s theorem and Fubini’s theorem

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 24, 2017 25 / 25