Real Analysis Homework: #1

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

∞ ∞ Question: Let (xn) ⊂ X be a Banach space, and Pn=1 kxnk is convergent. Proof that Pn=1 xn is convergent in X. ∞ ∞ Proof: Suppose Pn=1 kxnk = M < ∞, then ∀ ε> 0, ∃N, s.t. Pn=N kxnk < ε. n Let sn = Pi=1 xi, then ∀n>m>N, n ∞ ∞ ksn − smk = k X xik 6 X kxik < X kxik < ε. i=m i=m i=N

Hence, (sn) is a Cauchy sequence and must converge to an element in X.

2 Continuous function

Question: f : (X,τX ) → (Y,τY ) is continuous ⇔∀x0 ∈ X and any neighborhood V of f(x0), there is a neighborhood U of x0 such that f(U) ⊂ V . Proof: “⇒”: Let x0 ∈ X f(x0) ∈ Y . For each open set V containing f(x0), since f is −1 continuous, f (V ) which containing x0 is open. Then, there is a neighborhood U of x0 such that −1 x0 ∈ U ⊂ f (V ), that is to say f(U) ⊂ V . “⇐”: Let V ∈ Y . ∀y ∈ V , choose Vy satisfy y ∈ Vy ⊂ V, Vy ∈ τY . Then V = S Vy. −1 Let x = f (y), then ∀Vy, ∃Ux ∈ τX , s.t. x ∈ Ux and f(Ux) ⊂ Vy. Let U = S Ux, then U ∈ τX and f(U)= V . Therefore, f : (X,τX ) → (Y,τY ) is continuous.

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3 Closure

3.1 ¯ Question: Topological space (X,τ), x ∈ X, E ⊂ X, x ∈ E ⇔ ∀ neighborhoodV of x, E T V 6= ∅. Proof: Actually, this question is equal to the following one: ¯ (*) Topological space (X,τ), x∈ / E ⇔ There exist an open set V containing x that E T V = ∅. We just need to prove (*). ¯ ¯ “⇒ ” If x∈ / E, then V = X\E is an open set containing x and E T V = ∅. “⇐” If there exist an open set V containing x that E T V = ∅, then X\V is a closed set containing E. By the deﬁnition of closer E¯, that is the intersection of all closed sets containing E, the set X\V must contain E. Thus, x∈ / E¯.

3.2 ¯ ∞ Question: Metric space (X, d), x ∈ X, E ⊂ X, x ∈ E ⇔ there is a sequence (en)n=1 in E such that limn→∞ d(en,x) = 0. Proof: Similarly as Section 3.1, the above question is equal to the following one: (**) Metric space (X, d), x∈ / E¯ ⇔ ∀e ∈ E, ∃ ε0, s.t. d(e, x) > ε0. We just need to prove (**). “⇒ ” Suppose x∈ / E¯. Since X\E¯ is an open set, then ∃B(x, δ) ⊂ X\E¯. That is to say ∀e ∈ E, d(e, x) > δ. ε0 “⇐” If ∀e ∈ E, ∃ ε0, s.t. d(e, x) > ε0, then B(x, 2 ) is an open set that containing x and ε0 ε0 B(x, 2 ) T E = ∅. So X\B(x, 2 ) is a closed set containing E. ε0 By the deﬁnition of E¯, E¯ ⊂ X\B(x, 2 ). ε0 ¯ Since x ∈ B(x, 2 ), x∈ / E.

4 Distance between point and set

If E is a nonempty subset of a metric space X, deﬁne the distance from x ∈ X to E by

d(x, E)= inf d(x, z). (4.1) z∈E

4.1

(a) Prove that d(x, E) = 0 ⇔ x ∈ E¯.

Proof: If infz∈E d(x, z) = 0, then ∀ε, ∃ zε, s.t. d(x, zε) < ε.

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1 ∞ Choosing ε = n , then we can get a sequence (zn)n=1 satisfying limn→∞ d(x, zn) = 0. According to Section 3.2, x ∈ E¯. ¯ ∞ “⇐” If x ∈ E, by the previously Section 3.2, there is a sequence (zn)n=1 in E such that limn→∞ d(zn,x) = 0. Then d(x, E) = infz∈E d(x, z) = limn→∞ d(zn,x) = 0.

4.2

(b) Prove that x 7→ d(x, E) is a uniformly continuous function on X, by showing that

|d(x, E) − d(y, E)|≤ d(x,y). for all x ∈ X,y ∈ X. Proof: By the deﬁnition of (4.1), we can get:

∀x,y ∈ X, ∀ε> 0, ∃zx ∈ E, s.t. d(x, E)= d(x, zx)+ ε, ∃zy ∈ E, s.t. d(y, E)= d(y, zy)+ ε. Without loss of generality, we suppose that d(x, E) ≤ d(y, E). Then when ε is suﬃcient small, d(y, zy) ≤ d(y, zx). According to triangle inequality,

d(x,y) ≥ d(y, zx) − d(x, zx) ≥ d(y, zy) − d(x, zx)= d(x, E) − d(y, E).

Besides, in the case that d(x, E) ≥ d(y, E), we can get d(x,y) ≥ d(y, E) − d(x, E). So we can conclude that |d(x, E) − d(y, E)|≤ d(x,y). ε ∀x0 ∈ X, ∀ε > 0, just choose δ = 2 , then ∀x ∈ B(x0, δ), we have |d(x, E) − d(x0, E)| ≤ d(x,x0) <δ<ε. That is to say x 7→ d(x, E) is a uniformly continuous function on X.

III Real Analysis Homework: #2

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Banach space

Question: Let C([a, b]) denote the linear space of continuous function f : [a, b] → R. Show that C[a, b] is a Banach space with respect to the norm

kfk = max{|f(t)| : t ∈ [a, b]}.

∞ Proof: Let (fn)n=1 be a Cauchy sequence in C([a, b]). For ∀t ∈ [a, b], ∀m,n ∈ N, |fm(t)−fn(t)|≤ ∞ kfn − fmk, which means (fn(t))n=1 is a Cauchy sequence in R and must converge to an element in R. So we can deﬁne a function f : [a, b] → R that

f(t) = lim fn(t), ∀t ∈ [a, b]. n→∞

First, we will show that fn → f when n →∞. ∀ε> 0, there is an N s.t. ∀n,m ≥ N we have kfn − fmk < ε. For each x ∈ [a, b] we have

|fn(x) − f(x)| = lim |fn(x) − fm(x)| m→∞ ≤ lim kfn − fmk m→∞ ≤ ε, for all n ≥ N. Then for all n ≥ N,

kfn − fk = max {|fn(x) − f(x)|} ≤ ε. x∈[a,b]

Thus fn → f.

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Second, we will show that f ∈ C([a, b]). Choose a point t0 ∈ [a, b]. Since fn ∈ C([a, b]), for the previous ε, ∃ δn, s.t. ∀t ∈ (t0 − δ, t0 + δ), |fn(t) − fn(t0)| < ε. For t ∈ (t0 − δ, t0 + δ), n>N we have

2 Banach space

Question: Let l1(N) be the linear space of all the functions f : N → R with the property that ∞ 1 |f|1 := n=1 |f(n)| < ∞. Prove that (l (N), k · k1) is a Banach space. P ∞ 1 Proof: Let (fi)i=1 be a Cauchy sequence in l (N). For ∀n ∈ N, ∀i, j ∈ N, |fi(t) − fj(t)| ≤ ∞ kfi − fjk, which means (fi)i=1(n) is a Cauchy sequence in R and must converge to an element in R. So we can deﬁne a function f : N → R that

f(n) = lim fi(n), ∀n ∈ N. i→∞

First, we will show that fi → f when i → ∞. ∀ε > 0, there is an I s.t. ∀i, j ≥ I we have kfi − fjk < ε. For each n ∈ N we have

|fi(n) − f(n)| = lim |fi(x) − fj(x)| j→∞

≤ lim kfi − fjk j→∞ ≤ ε, for all i ≥ I. Then for all i ≥ N,

kfi − fk = max{|fi(n) − f(n)|} ≤ ε. n∈N

Thus fi → f. Second, we will show that f ∈ l1(N). ∞ kfk = |f(n)| 1 X n=1 ∞ = | lim fi(n)| X i→∞ n=1 ∞ ≤ lim |fi(n)| < ∞. i→∞ X n=1

II Yingwei Wang Real Analysis

1 1 Then f ∈ l (N) and (l (N), k · k1) is a Banach space.

3 Application of Baire Theorem

Question: Let f : R → R be a smooth function (i.e. C∞). Suppose that for each t ∈ R there is (nt) nt ∈ N such that f (t) = 0. Prove that there is an interval I of positive length such that the restriction of f to I is a polynomial. (n) Proof: Deﬁne Tn := {t ∈ R : f (t) = 0}. It is easy to verify that for each n the restriction of f to Tn is a polynomial with at most (n + 1) degree. Then we just need to show that ∃n0, s.t. ◦ (Tn0 ) 6= ∅. c c (n) First, we claim that ∀n ≥ 1, Tn is a closed set. Consider the set Tn = R\Tn: ∀t ∈ Tn, f 6= 0. Since f ∈ C∞, f (n) is continuous, we know that ∃ δ, s.t. ∀x ∈ (t − δ, t + δ), f (n)(x) 6= 0. That is c to say (t − δ, t + δ) ⊂ Tn. Then R\Tn is an open set while Tn is a closed set. ∞ (nt) Second, we claim that Tn = R. ∀t ∈ R, ∃ nt ∈ N, s.t. f (t) = 0, which means t ∈ Tnt . nS=1 ∞ Thus, Tn = R. nS=1 ◦ According to the Baire Category Theorem, ∃n0, s.t. (Tn0 ) 6= ∅. That is to say ∃ interval I ⊂ ◦ (Tn) , s.t. the restriction of f to I is a polynomial.

4 Outer measure

Question: Let m∗(A) denote the outer measure of A ⊂ R. Show that if B ⊂ R and m∗(B) = 0, then m∗(A ∪ B)= m∗(A). Proof: On one hand, A ⊂ (A ∪ B) ⇒ m∗(A) ≤ m∗(A ∪ B). On the other hand, m∗(A ∪ B) ≤ m∗(A)+ m∗(B)= m∗(A)+0= m∗(A).

Thus, m∗(A ∪ B)= m∗(A).

5 Continuity of function

∞ 1 Question: Let (xn)n be an enumeration of Q. Deﬁne f : R → R by f(x) = n . where the P 2 summation is extended over all n such that xn

III Yingwei Wang Real Analysis

∞ 1 Proof: Deﬁne Ax = {i ∈ N : xi

5.1 Rational points ⇒ Discontinuity

Fix xr ∈ Q, r ∈ N. Let xn ∈ Q and xn >xr. It is obviously that r ∈ (Axn \ Axr ). Then

1 1 1 1 |f(xn) − f(xr)| = − = > . X 2p X 2q X 2i 2r p∈Axn q∈Axr i∈(Axn \Axr )

which means lim f(x) 6= f(xr). That is to say f is discontinuous at each rational number. + x∈Q, x→xr

5.2 Irrational points ⇒ Continuity ∞ ∞ 1 1 Since 2n = 1, we can know that ∀ε > 0, ∃N ∈ N, s.t. 2n < ε. Let the set Sε = nP=1 n=PN+1 {x1,x2, · · · ,xN }. We can rearrange the elements of the set Sε such that x1 0 as the following way:

1 2 (x1 − α) if αxN . 

Then (Ax+δ \ Ax−δ) ∩ ASε = ∅. That is to say ∀n ∈ (Ax+δ \ Ax−δ), n ≥ N + 1. ∞ 1 1 So ∀x ∈ (x − δ, x + δ), |f(x) − f(α)| ≤ |f(x + δ) − f(x − δ)| = 2n < 2n < ε, n∈(Ax+Pδ\Ax−δ) n=PN+1 which means f is continuous at the point α ∈ (R \ Q).

IV Real Analysis Homework: #3

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Measure inequality

∞ Question: Let (X, A,µ) be a measure space. Let (Ak)k=1 be a sequence of sets in A. Prove that ∞ ∞

µ Ak ≤ lim inf µ(Ak). k→∞ n=1 = ! [ k\n ∞ Proof: Let Bn = Ak, then Bn = Bn+1 ∩ An, Bn ⊂ Bn+1. So k=n T m ∞ µ( Bn)= µ(Bm)= Ak ⊂ Ak, ∀k ≥ m n=1 k=m [m \ ⇒ µ Bn ≤ µ(Ak), ∀k ≥ m n=1 ! [∞

⇒ µ Bn ≤ lim inf µ(Ak). k→∞ n=1 ! [ 2 Example of measure

Question: Let µ be a measure on (R, B), where B are the Borel sets, such that µ([0, 1)) = 1 and µ(x + B)= µ(B) for all x ∈ R and B ∈B. Prove that (1) µ([0, 1/n)) = 1/n for all integers n ≥ 1 and (2) µ([a, b)) = b − a for all real numbers a < b.

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Proof: (1) By the assumption that µ(x + B) = µ(B), we can know that µ([0, 1/n)) = µ([1/n, 2/n)) = · · · = µ([(n − 1)/n, 1)). Then

n 1= µ([0, 1]) = µ([(k − 1)/n, k/n)) = nµ([0, 1/n)) =1 Xk ⇒ µ([0, 1/n)) = 1/n.

(2) Since for any a < b, µ([a, b)) = µ([0, b − a)), we only need to consider the cases that a = 0,b> 0. p If b ∈ Q, then b = q where p,q ∈ N,q 6= 0.

p−1 p k k + 1 1 p µ([0, b)) = µ([0, )) = µ([ , )) = p µ([0, )) = = b. q =0 q q q q Xk ∞ If b ∈ R\Q, then ∃ (bn) =1 ⊂ Q s.t. lim bn = b. Then n n→∞

µ([0, b)) = lim µ([0, bn)) = lim bn = b. n→∞ n→∞

3 Lebesgue outer measure

Question: Let m∗ be the Lebegue outer measure on R. For any two sets A, B ⊂ R, prove the inequality: ∗ ∗ ∗ ∗ m (A)+ m (B) ≤ 2m (A△B) + 2m (A ∩ B). (3.1) Proof: For the set A, we have

∗ ∗ ∗ m (A) ≤ m (A\B)+ m (A ∩ B) (since A = (A\B) ∪ (A ∩ B)) ∗ ∗ ≤ m (A△B)+ m (A ∩ B). (since (A\B) ⊂ (A△B)) (3.2)

Similarly, for the set B, we have

∗ ∗ ∗ m (B) ≤ m (A△B)+ m (A ∩ B). (3.3)

By (3.2)-(3.3), we can get (3.1).

4 Zero measure set

Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebesgue measurable set. Suppose that m(A ∩ [a, b]) < (b − a)/2 for all a < b real numbers. Show that m(A) = 0.

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Proof: On one hand, by the deﬁnition of Lebesgue measure,

∞ ∞

m(A) = inf{ l(In) : In ⊂ R are open intervals, A ⊂ In}, n=1 n=1 X [ ∞ we know that ∀ε> 0, ∃ {In}n=1, s.t. ∞

m(A) ≥ l(In) − ε, n=1 X ∞ where A ⊂ In and Ii ∩ Ij = ∅, i 6= j. n=1 1 We can chooseS ε = 4 m(A) in the above inequality, then we can get ∞ 4 m(A) ≥ l(I ). (4.1) 5 n n=1 X 1 + On the other hand, by the assumption, m(A ∩ In) < 2 l(In), ∀n ∈ N . So we have ∞ ∞ 1 m(A)= m(A ∩ I ) ≤ l(I ). (4.2) n 2 n n=1 n=1 X X From (4.1)-(4.2) we can conclude that m(A) = 0.

5 Measure function

Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebegue measurable set with m(A) < ∞. Show that the function f : R → [0, ∞), f(x) = m(A ∩ (−∞,x)) is continuous. Deduce that for every β ∈ [0,m(A)] there is a Lebesgue measurable set B ⊂ A such that m(B)= β. Proof:

5.1 The continuity of the auxiliary function g

Since m(A) < ∞, by the Proposition 15 on Page 63 in Royden’s book, given ∀ε> 0, there is a ﬁnite union U of open intervals such that

∗ m (U△A) < ε. (5.1)

III Yingwei Wang Real Analysis

n Suppose U = (ai, bi), and a = a1, b = bn, then U ⊂ [a, b]. i=1 On the intervalS [a, b], we can deﬁne the function g such that

g(x)= m(A ∩ [a, x)), ∀x ∈ [a, b].

Then for ∀∆x> 0,

A ∩ [a, x +∆x) = (A ∩ [a, x)) (A ∩ [x,x +∆x)) ⇒ g(x +∆x) ≤ g(x)+∆x [ ⇒ g(x +∆x) − g(x) ≤ ∆x

For ∆x< 0, we can get the similar result:

g(x +∆x) − g(x) ≥−∆x

Then we have |g(x +∆x) − g(x)| ≤ |∆x| which means g ∈ C([a, b]).

5.2 The continuity of f

For ∀x ∈ [a, b], choose ∆x < ε, then

5.3 The Intermediate Value Theorem

Since g(a) = 0, g(b) = m(A), by the Intermediate Value Theorem of continuous functions, for ∀β ∈ [0,m(A)] there is an x0 ∈ [a, b] such that g(x0)= β. Then we can choose B = A ∩ [a, x0].

IV Real Analysis Homework: #4

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Measurable function

∞ Question: Let (X, A,µ) be a measure space. Let (fn)n=1 be a sequence of measurable functions: fn : X → R. Show that the set of points x where the limit lim fn exists n→∞ (ﬁnite or inﬁnite) is a measurable set. 1 Proof: Let Ak = {x ∈ X : |fn(x) − fm(x)| < , ∀m,n > k}, A = lim Ak, where k k→∞ m,n,k ∈ N. It is easy to know that each Ak is measurable set since fn is a measurable function. We have this observation: Ak+1 ⊂ Ak ⇒ A ⊂ Ak, ∀k. Thus, A is a measurable set.

Let B = {x ∈ R : lim fn(x) exists}. I will show that A = B. n→∞ N 1 On one hand, let x ∈ A, then for ∀k ∈ , |fn(x) − fm(x)| < k , ∀m,n > k, which mean fn(x) is a Cauchy sequence in R. So lim fn(x) exists and then x ∈ B. n→∞ On the other hand, let x ∈ B, then ∀ε> 0, ∃N ∈ N s.t. |fn(x) − fm(x)| < ε, ∀m,n > N, which means x ∈ AN . Then if ε → 0, N →∞, x ∈ A. Now we can conclude that B is a measurable set.

2 Diﬀerential function

Question: Let f : R → R be a diﬀerential function. Prove that f ′ is Lebesgue measurable.

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f(x+1/n)−f(x) R N Proof: Deﬁne a sequence of functions: gn(x)= 1/n , x ∈ , n ∈ . Then f is measurable

⇒ gn(x) is measurable , ∀n, ′ ⇒ f (x) = lim gn is measurable. n→∞

3 The sets with measure zero

Question: Let f : R → R be deﬁned by f(x)= x5 +sin x. Suppose that a set A ⊂ R has Lebesgue measure zero. Show that f(A) has Lebesgue measure zero. Proof: I want to prove that for any function f ∈ C1(R), the image of the set with Lebesgue measure zero has also Lebesgue measure zero. ∞ Let AN = [−N, N] ∩ A, then A = AN , m(A) = 0 ⇒ m(AN ) = 0, ∀N ∈ N. We NS=1 will focus the problem on each AN . By the deﬁnition of Lebesgue measure, for ∀ε> 0, there exists a sequence of intervals ∞ (In)n=1, Ii ∩ Ij = ∅, i 6= j and AN ⊂∪In, s.t.

l(In) < ε, since m(AN ) = 0. (3.1) Xn

Let In = (an, bn), I¯n = [an, bn], then m(In)= m(I¯n). Then we have

m(f(AN )) < m(f(In)) = m(f(I¯n)). (3.2) Xn Xn Since f is continuous, we can know that

m(f(I¯n)) = max f(x) − min f(x). x∈I¯n x∈I¯n

1 Since I¯n is a closed interval and f ∈ C (I¯n), we can ﬁnd x1,x2 ∈ I¯n s.t. f(x1) = max f(x), f(x2) = min f(x). Without loss of generality, we can assume that x1 ≤ x2. x∈I¯n x∈I¯n Then

m(f(I¯n)) = f(x1) − f(x2) ′ ≤ |f (ξ)|(x2 − x1), ξ ∈ [x1,x2], ′ ≤ max f (x) l(In) x∈I¯n ≤ max f ′(x) ε. x∈I¯n

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′ Since maxx∈I¯n f (x) < ∞ and let ε → 0, we can get m(f(In)) = 0. By (3.2) we have m(f(AN )) = 0. So

m(f(A)) = m(f(∪AN )) ≤ m(f(AN )) = 0. XN

4 Measurable function

Question: f R R g R R Let :n → be a Lebesgue measurable function. Deﬁne : → by ∞ f(x) g(x)= n=1 n!(n+3) . Prove that g is Lebesgue measurable. P R n ak Proof: First, we claim that ∀a ∈ , the series sn = k=1 k!(k+3) is convergent. ∀a ∈ R, ∃N ∈ N, s.t. |a| < N. Then for ∀k> 2N, we have P ak |a|N |a|N |a|k−2N 1 ≤ · · · . k!(k + 3) 1 · · · N (N + 1) · · · (2N) (2N + 1) · · · k k + 3

N N N |a| |a| 1 |a| Let M = 1···N , we can choose k > M, s.t. 1···N · k+3 < 1. Since (N+1)···(2N) < 1, k−2N |a| 1 k−2N (2N+1)···k < 2 , we have ak 1 k−2N < . k!(k + 3) 2

∞ 1 k−2N ∞ ak Since for ﬁxed N, k=1 2 is convergent, we have for ﬁxed a, k=1 k!(k+3) is also convergent. P P n n f(x) Second, let fn = k=1 n!(n+3) which is measurable and P g(x) = lim fn(x), for ∀x ∈ R. n→∞ Hence, we know that g(x) is also measurable.

5 Measurable function

Question: Let f : R → R be a Lebesgue measurable function. Suppose that A ⊂ R is a Borel set. Show that the set {x ∈ A : f(x) >x} is Lebesgue measurable. Proof: Deﬁne a function g : R → R by g(x)= f(x) − x, then g(x) is measurable (by the Theorem 6 in the page 259 of Royden’s book). So the set B = {x ∈ R : g(x) > 0} is measurable. Then, {x ∈ A : f(x) >x} = A ∩ B is measurable.

III Real Analysis Homework: #5

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

Note: In this paper, {f(x) satisﬁes some property} = {x : f(x) satisﬁes some property}

1 Integrable function

1.1 a

Question: Show that if f is integrable then the set {f(x) 6= 0} is of σ-ﬁnite measure. Proof: On one hand, ∞ 1 {f(x) 6= 0} = {|f(x)| > 0} = {|f(x)|≥ }. (1.1) n n=1 [ On the other hand, 1 1 ∞ > |f|dµ ≥ |f|dµ ≥ µ{|f(x)|≥ }, ∀n ∈ N, (1.2) 1 {|f(x)|≥ } n n Z Z n which means 1 µ{|f(x)|≥ } < ∞, ∀n ∈ N. (1.3) n From (1.1) and (1.3), we can know that the set {f(x) 6= 0} is of σ-ﬁnite measure.

1.2 b

Question: Show that if f is integrable, f ≥ 0, then f = lim ϕn for some increasing sequence of simple functions each of which vanishes outside a set of finite measure. Proof: By proposition 7 on Page 260 in Royden’s book, since the set {f(x) 6= 0} is of σ-finite measure, we can find a sequence (ϕn) of simple functions defined on {f(x) 6= 0} with ϕn+1 ≥ ϕn such that f = lim ϕn and each ϕn vanishes outside a set of finite measure. Then we can define ϕn(x) = 0 on the set {f(x) = 0}, and get the conclusion. ∗E-mail address: [email protected]; Tel: 765 237 7149

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1.3 c

Question: Show that if f is integrable with respect to µ, then given ǫ > 0 there is a simple function ϕ such that |f − ϕ|dµ<ǫ. Proof: By the assumption, f + and f − are nonnegative integrable functions. By (b), there R + − are increasing sequence (φn) and (ϕm) such that f = lim φn and f = lim ϕm. By the Monotone Convergence Theorem, we have

+ f dµ = lim φn, Z Z − f dµ = lim ϕm. Z Z

So given ǫ> 0, there are φN and ϕM such that ǫ f +dµ − φ dµ< , N 2 Z Z ǫ f −dµ − ϕ dµ< . M 2 Z Z

Let ϕ = φN − ϕM . Then ϕ is also a simple function and satisﬁes

|f − ϕ|dµ Z + − ≤ |f − φN |dµ + |f − ϕM |dµ Z Z + − = f dµ − φN dµ + f dµ − ϕM dµ Z Z Z Z < ǫ.

2 Measurable set

Question: Let f : R → R be deﬁned by f(x) = x3 − 2x2 + x − 1. Show that if A ⊂ R is Lebesgue measurable, then f(A) ⊂ R is Lebesgue measurable. Proof: Since f(x) is a polynomial, then it is measurable and there exists a sequence of step functions (ϕn) s.t. ϕn(x) → f(x), ∀x ∈ R. By assumption, A ⊂ R is Lebesgue measurable, which means ∀E ⊂ R,

m(E)= m(E ∩ A)+ m(E ∩ Ac).

II Yingwei Wang Real Analysis

For each step function ϕn(x), if A is measurable, then ϕn(A) is just a set of points so it is measurable. Thus f(A) = lim ϕn(A) is measurable. n→∞

3 Limits

Question: Let (X, A,µ) be a measure space with µ(X) < ∞. Let f : X → (−1, 1) be a measurable function. Consider the sequence

n an = (1 + f + · · · + f )dµ. ZX Show that either (an) converges to a ﬁnite number or otherwise lim an = ∞. (In other words n→∞ (an) cannot have two distinct limit points.) ∞ n Proof: Since ∀x ∈ X, |f(x)| < 1, the series n=0 |f(x)| is convergent for ∀x ∈ X. It ∞ n ∞ 2n follows that all of the functions g(x) = n=0(f(x)) and h0(x) = n=0(f(x)) , h1(x) = ∞ P (−f(x))2n+1 are absolutely convergent. n=0 P P Let A = {f ≥ 0}, B = {f < 0}, then X = A ∪ B. P n + j an = (f(x)) dµ A Z Xj=0 n k f 2jdµ − k (−f)2j+1dµ if n = 2k + 1, a− = (f(x))jdµ = B j=0 B j=0 n k f 2jdµ − k (−f)2j−1dµ if n = 2k. B j=0 ( RB j=0 RB j=0 Z X P P R P R P Hence,

+ − lim an = lim (a + a )= g(x)dµ + h0(x)dµ − h1(x)dµ. n→∞ n→∞ n n ZA ZA ZB which means lim an can not have two distinct limit points. n→∞

4 Convergence

Question: Show that the following sequence is convergent in R.

1 cos(x + 1/n) − cos x an = n dx. x3/2 Z1/n Proof: Consider the function

cos(x + 1/n) − cos x X[1/n,1](x) fn(x)= , x ∈ [0, 1], 1/n x3/2

III Yingwei Wang Real Analysis

where XA(x) is the characteristic function. It is obvious that sin x lim fn = − , x ∈ [0, 1]. n→∞ x3/2 sin x x 1 Let f(x)= − x3/2 , then |f(x)|≤ x3/2 = x1/2 since sin x ≤ x, x ∈ [0, 1]. 1 Let g(x)= 1 2 , then ∃N ∈ N, s.t. ∀n>N, |f (x)| < g(x). Then by Lebesgue dominated x / n convergence theorem, we have 1 1 1 sin x 1 1 lim an = lim fn(x)dx = lim fn(x)dx = − dx< dx = 2. n→∞ n→∞ n→∞ x3/2 x1/2 Z0 Z0 Z0 Z0 So an is convergent.

5 Inﬁnitive sum

∞ Question: Let (X, A,µ) be a complete measure space. Let (fn)n=1 be a sequence of A- measurable functions, fn : X → R. Suppose that ∞ |fn|dµ< ∞. (5.1) n=1 X X Z ∞ Show that the series n=1 fn(x) is absolutely convergent µ-almost everywhere on X. Proof: First, claim that P ∞ ∞ |fn| dµ = |fn|dµ. (5.2) X n=1 ! n=1 X Z X X Z n ∞ Let Fn = |fi|, F∞ = |fi|, ∀x ∈ X. Then the statement (5.2) becomes i=1 i=1 P P n F∞dµ = lim |fi|dµ = lim Fndµ. (5.3) X n→∞ X n→∞ X Z Xi=1 Z Z Since Fn(x) ≤ Fn+1(x) and Fn → F∞, by the Monotone Convergence Theorem, we can get (5.3). Second, by the assumption (5.1) and the claim (5.2), we can get ∞ ∞ |fn| dµ = |fn|dµ< ∞, (5.4) X n=1 ! n=1 X Z X X Z ∞ ∞ which means the measure µ ({ n=1 |fn| = ∞}) = 0. That is to say the series n=1 fn(x) is absolutely convergent µ-almost everywhere on X. P P

IV Real Analysis Homework: #6

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Integral function

Question: Let f : R → R be a Lebesgue integrable function. Prove that g(x) is also Lebesgue integrable. Proof: First, if f(x) is a non-negative simple function, then

f(x)= ckχEi (x), ∀x ∈ R, Xk=1 n R g(x)= ckχEi+{1}(x), ∀x ∈ . Xk=1

Hence, g(x) is also a non-negative simple function and Lebesgue integrable. Second, if f(x) is a non-negative measurable function, then there exists a sequence ∞ of non-negative simple monotonic increasing functions {ϕk(x)}k=1 s.t.

lim ϕk(x)= f(x), ∀x ∈ R. k→∞

∞ It is obvious that {ϕk(x − 1)}k=1 is also a monotonic increasing sequence and

lim ϕk(x − 1) = f(x − 1), ∀x ∈ R. k→∞

∗E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis

Hence,

g(x) dm ZR = f(x − 1) dm ZR

= lim ϕk(x − 1) dm k→∞ ZR

= lim ϕk(x) dm k→∞ ZR = f(x)dm< ∞. ZR

So we have g(x) is Lebesgue integrable on R. Finally, if f(x) is an arbitrary function on R, we can rewrite f as f = f + − f −, where both f + and f − are non-negative integrable function, and then get the same conclusion about g(x).

2 Zero function

Question: Let f : [0, 1] → R be a Lebesgue measurable function. Suppose that for R any function g : [0, 1] → , fg is Lebesgue integrable and [0,1] fg dm = 0. Prove that f = 0 almost everywhere with respect to the Lebesgue measure.R Proof: Suppose that there exists a set A ⊂ [0, 1], m(A) > 0, s.t. ∀x ∈ A, f(x) 6= 0. Without loss of generality, we can assume that f(x) > 0, ∀x ∈ A.

∀ε > 0, we can ﬁnd a sequence of intervals In = (a, b) ⊂ [0, 1] such that A ⊂ ∪In and m(A\∪ In) < ε. Choose g(x) as follows:

−(x − a )(x − b ), x ∈ [a , b ], g(x)= n n n n 0, x ∈ [0, 1]\∪ I¯n.

It is easy to verify that g(x) is continuous and satisﬁes: g(x) > 0,x ∈ ∪In, and g(x) = 0,x ∈ [0, 1]\∪ I¯n. Hence, fg ≥ 0,x ∈ [0, 1] and fg > 0,x ∈ A. So we have

fg dm = fgdm = fgdm + fgdm, Z[0,1] Z∪In ZA Z∪In\A

II Yingwei Wang Real Analysis

On one hand, A fgdm > 0. On the other hand, since m(A\∪ In) < ε, we can R choose ε such that ∪In\A fgdm < A fgdm. It follows that [0,1] fg dm> 0, which contradicts that R fg dm = 0, ∀g ∈R C1([0, 1]). R [0 ,1] Thus, f = 0,R a.e. x ∈ [0, 1]. Note: Another method is by the statement that “Any measurable function can be approximated by a sequence of continuous functions”.

3 σ-ﬁnite

Question: Let (X, A),µ be a measure space and let f : X → R be A-measurable. Suppose that µ is σ-ﬁnite. Suppose also that there is c ≥ 0 such that for all E ∈ A with µ(E) < ∞ one has | E f dµ|≤ c. Show that f is Lebesgue integrable on X. Proof: Since µ is σ-ﬁnite,R we can choose a sequence of sets (Xn) such that X = ∪Xn, X1 ⊂ X2 ···⊂ Xn ⊂···⊂ X, lim Xn = X and n→∞

µ(Xn) < ∞, ∀n ∈ N.

Deﬁne fn(x)= f(x)χXn , ∀x ∈ X, then

lim fn(x)= f(x) and |fn(x)| ≤ |f(x)|, ∀x ∈ X. n→∞

Since µ(Xn), by assumption,

fn dµ = fdµ ≤ c, ∀n ∈ N. Z Z X Xn

By Lebesgue dominated control theorem, we have

f dµ = lim fn dµ ≤ lim fn dµ ≤ c, Z n→∞ Z n→∞ Z X X X which implies f is Lebesgue integrable on X.

III Yingwei Wang Real Analysis

4 Limits

∞ Question: Let (X, A),µ be a measure space. Let (fn)n=1 be a sequence of A-measurable functions, fn : X → [0, +∞]. Show that

lim inf fndµ ≤ lim inf fndµ. ZX n→∞ n→∞ ZX

Proof: Let gn(x)= inf fn(x), f(x) = lim inf fn(x), ∀x ∈ X. Then gn are nonneg- k≥n n→∞ ative and gn increases to f, lim gn(x)= f(x). Therefore, n→∞

gndµ ≤ inf fkdµ. ZX k≥n ZX

If we take the limit as n →∞, on the left side we obtain X fdµ by the monotone convergence theorem, while on the right side we obtain lim inRf fndµ. n→∞ X R

IV Real Analysis Homework: #7

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Limits

Question: Compute the following limits and justify your calculation. (i) lim ∞ 1+ x −n sin x dx. n→∞ 0 n n x −n x −x Solution:R Let fn(x)= 1+ sin , then |fn(x)|≤ e , ∀x ∈ [0, ∞). Besides, lim fn(x)= n n n→∞ 0. Then we have ∞ lim fndx = 0. n→∞ Z0 (ii) lim n 1 − x n ex/2dx. n→∞ 0 n x n x/2 −x/2 Solution:RLet fn(x)= 1 − e χ[0,n], f(x) = e then lim fn(x)= f(x). n n→∞ Let gn = 2f − fn, thengn monotonic increasingly converges to f as n →∞. Besides, when n is suﬃcient large, gn ≥ 0. By Monotonic Convergence Theorem, we can see that

lim gn(x)= f(x). n→∞ So we have n n ∞ ∞ x x/2 lim 1 − e dx = lim fndx = fdx = 2. n→∞ 0 n n→∞ 0 0 Z Z Z (ii) lim n 1+ x n e−2xdx. n→∞ 0 n x n −2x −x Solution:RLet fn(x)= 1+ e χ[0,n], f(x) = e then lim fn(x)= f(x) and fn(x) ≤ n n→∞ f(x), ∀x ∈ [0, ∞). So we have n n ∞ ∞ x −2x lim 1+ e dx = lim fndx = fdx = 1. n→∞ 0 n n→∞ 0 0 Z Z Z ∗E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis

2 Limits

Question: Let (X, A,µ) be measure space and let f : X → [0, +∞) be A-measurable. Suppose that 0 < c = X fdµ< ∞. Show that R +∞, if 0 <α< 1, f α lim n ln 1+ dµ = c, if α = 1, n→∞ X n  Z  0, if 1 <α< ∞. Proof: Let  f α f (x) = n ln 1+ n n f α n = ln 1+ n 1−α α n f α n = ln 1+ nα ! α f α n = n1−α · ln 1+ nα ! If 0 <α< 1, then α lim fn(x)=+∞ · f =+∞. n→∞ By Fatou’s Lemma, we have

lim fn(x)dµ ≥ lim fn(x)dµ ≥ lim fn(x)dµ =+∞. n→∞ n→∞ ZX n→∞ ZX ZX If α = 1, then lim fn(x)= f(x). n→∞

Besides, ∀x ∈ X, fn(x) ≤ f(x). By Lebesgue Dominated Control Theorem, we have

lim fn(x)dµ = f(x)dµ = c. n→∞ ZX ZX If 1 <α< ∞, then α lim fn(x) = 0 · f = 0. n→∞ By Monotonic Convergence Theorem, we have

lim fn(x)dµ = 0 dµ = 0. n→∞ ZX ZX

II Yingwei Wang Real Analysis

3 Limits

∞ Question: Let (X, A,µ) be measure space. Let (fn)n=1 be sequence of A-measurable functions, fn : X → [0, +∞). Suppose that limn→∞ fn(x) = f(x) for all x ∈ X and that limn→∞ X fn(x)dµ = X f(x)dµ < ∞. Show that limn→∞ A fn(x)dµ = A f(x)dµ for every set A ∈A. R R R R Proof: For any set A ∈A, deﬁne gn(x)= fn(x)χA, hn(x)= fn(x)χX\A, the we have

lim fn(x)= f(x), n→∞ lim gn(x)= f(x)χA, n→∞ lim hn(x)= f(x)χX\A, n→∞ gn(x)+ hn(x)= fn(x), ∀n and ∀x. On one hand, by Fatou’s Lemma, we have

fχAdµ ≤ lim gndµ, (3.1) ZX n→∞ ZX fχX\Adµ ≤ lim hndµ, (3.2) ZX n→∞ ZX ⇒ fdµ ≤ lim gndµ + lim hndµ. (3.3) ZX n→∞ ZX n→∞ ZX On the other hand,

lim gndµ + lim hndµ = lim gndµ + hndµ ≤ lim fndµ = fdµ.(3.4) n→∞ n→∞ ZX n→∞ ZX n→∞ ZX ZX ZX ZX By (3.3) and (3.4), we can see that

lim gndµ + lim hndµ = fdµ. (3.5) n→∞ ZX n→∞ ZX ZX Then by (3.1), (3.2) and (3.5), we can see that

fχAdµ = lim gndµ, (3.6) ZX n→∞ ZX fχX\Adµ = lim hndµ, (3.7) ZX n→∞ ZX So we can get

lim fn(x)χAdµ = f(x)χAdµ n→∞ ZX ZX ⇒ lim fn(x)dµ = f(x)dµ. n→∞ ZA ZA III Yingwei Wang Real Analysis

4 The absolute continuity of integral

Question: Let f : R → [0, ∞] be a Lebesgue integrable function. Show that for each α > 0 R there is β > 0 such that if B ⊂ is Lebesgue measurable and m(B) < β, then B fdm < α. Proof: f α > , Since is nonnegative and Lebesgue integrable, ∀ 0 ∃ a nonnegativeR simple function ϕ(x), such that α f − ϕdm = fdm − ϕdm< . R R R 2 Z Z Z R α Let ϕ(x) < M, then we can choose B ⊂ s.t. m(B) < 2M , then

fdm ≤ f − ϕdm + ϕdm ZB ZB ZB ≤ f − ϕdm + M · m(B) R Zα α < + M · = α. 2 2M

IV Real Analysis Homework: #8

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

Note: In this paper, {f(x) satisﬁes some property} = {x : f(x) satisﬁes some property}

1 Integral function

Question: Let f : (0, 1) → R be a Lebesgue integrable function. Suppose that

fdm for all x ∈ (0, 1). (1.1) Z(0,x) Show that f(x) = 0 a.e. 1 Proof: First I want to claim that 0 fdm = 0. Let fn = fχ(0,1−1/n), then |fχ(0,1/n)(x)|≤ |f(x)|, ∀n ∈ N and x ∈ (0, 1) By LebesgueR dominated convergence theorem, 1 fdm = lim fndm = lim fdm = 0. Z0 n→∞ Z(0,1) n→∞ Z(0,1−1/n) Second, let A = {f(x) > 0}. Suppose m(A) > 0, then ∃ closed set B ⊂ A, m(B) > 0. Let C = [0, 1]\B, then C is an open set, i.e. C = ∪(an, bn).

fdm + fdm = fdm = 0, ZB ZC Z(0,1) ⇒ fdm 6= 0, (since fdm> 0) ZC ZB ⇒ fdm 6= 0, Z∪(an,bn)

⇒ ∃ (an, bn), s.t. fdm 6= 0, Z(an,bn) ⇒ fdm − fdm 6= 0, Z(0,bn) Z(0,an)

∗E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis which contradicts with (1.1). So m(A) = 0. Similarly, we can prove that m({f(x) < 0}) = 0. That is to say f(x) = 0 a.e.

2 Convergence of L1 norm and pointwise

1 Question: Let (fn) be a convergent sequence in L (X, A,µ) with limit f. Show that (fn) has a subsequence (fnk ) such that fnk → f for almost x ∈ X. 1 Proof: Since (fn) be a convergent sequence in L (X, A,µ), then (fn) is a Cauchy sequence in L1(X, A,µ), which means

|fn − fm|dµ → 0, as n,m →∞. ZX ⇒ ∀ε> 0, ∃N, s.t. ∀n,m>N,

m({|fn − fm|= 6 0}) < ε.

⇒ ∀i, ∃ki ∈ N s.t. ∀n,m>ki, 1 1 m({|f − f |≥ }) < . n m 2i 2i

We can assume that k1 < k2 < · · · < ki < ki+1 < · · · , and deﬁne 1 E = {|f − f |≥ }, i ki ki+1 2i

1 then m(Ei) < 2i . Let ∞ ∞ E = Ei, j\=1 i[=j then m(E) = 0. ∀x ∈ X\E, ∃j s.t. ∞ x ∈ X Ei. i[=j Then ∀i ≥ j, 1 |f (x) − f (x)| < , ki+1 ki 2i

II Yingwei Wang Real Analysis

which means |fki+1(x) − fki (x)| is convergent, so fk1 + (fki+1(x) − fki (x)) is also convergent. P P

Denote lim fki (x)= g(x), ∀x ∈ X\E, then |fnk − g|→ 0, as n →∞, ∀x ∈ X\E. So ki→∞

|fnk − g|dµ = |fnk − g|dµ → 0, ZX ZX\E

⇒ kfnk − gkL1 → 0.

1 Since kfnk − fkL1 → 0, by the completeness of L (X, A,µ), f = g for almost all x ∈ X. So we have a subsequence (fki ) which is convergent to f for almost all x ∈ X.

3 Integral functions and continuous functions

Question: Let f : [0, 1] → R be a Lebesgue integrable function. Show that there is a sequence of continuous functions fn : [0, 1] → R which converges pointwise to f almost everywhere. Proof: First, I want to introduce the Lusin’s Theorem (Problem 3.31 on Page 74 in Royden’s book):

Theorem 3.1 (Lusin). Let f be a measurable real-valued function on an interval [a, b]. Then given δ, there is a continuous function ϕ on [a, b] such that m({f 6= ϕ}) < δ.

By this theorem, for f ∈ L1([0, 1]), ∀n ∈ N, ∃ a sequence of continuous functions ϕn ∈ C([0, 1]) s.t. 1 m({ϕ 6= f}) < . n 2n Then we have

|ϕn − f|dm = |ϕn − f|dm → 0, as n →∞, ZX Z{ϕn6=f} since m({δn 6= f}) → 0 as n →∞. 1 It implies that ϕn converges to f in the L norm. The conclusion of the last problem tells us that (ϕn) has a subsequence (ϕnk ) such that ϕnk → f for almost x ∈ X.

III Real Analysis Homework: #9

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Absolutely continuous

Question: Let f, g : [a, b] → R be two absolutely continuous functions. Prove or disprove that h(x) = ef(x)g(x) is absolutely continuous on [a, b]. Proof: First, I want to show that if both f and g are absolutely continuous, then fg is also absolutely continuous. There exists M > 0 such that |f(x)|≤ M and |g(x)|≤ M for ∀x ∈ [a, b]. Given ε> 0, there exits δ > 0 such that

n n ε ε |f(x ) − f(y )| < , and |g(x ) − g(y )| < , i i 2M i i 2M Xi=1 Xi=1

n for any ﬁnite collection {(xi,yi)} of disjoint intervals in [a, b] with i=1 |xi − yi| < δ. Then P n |f(xi)g(xi) − f(yi)g(yi)| Xi=1 n n ≤ |f(xi)||g(xi) − g(yi)| + |g(yi)||f(xi) − f(yi)| Xi=1 Xi=1 ε ε < + = ε. 2 2

Second, I want to show that if q : [a, b] → [a′, b′] is absolutely continuous on [a, b] and p : [a′, b′] → R is Lipschitz continuous on [a′, b′], then p(q(x)) is absolutely continuous on [a, b].

∗E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis

There exists N > 0 such that for ∀x,y ∈ [a′, b′], |p(x) − p(y)|≤ N|x − y|.

Given ε, there exits δ > 0 such that n ε |q(x ) − q(y )| < i i N Xi=1 n for any ﬁnite collection {(xi,yi)} of disjoint intervals in [a, b] with i=1 |xi − yi| < δ. Then P n |p(q(xi)) − p(q(yi))| Xi=1 n ≤ N |q(xi) − q(yi)| Xi=1 < ε.

Finally, since ex is Lipschitz continuous in any ﬁnite interval [a′, b′], so h = efg is absolutely continuous if both f and g are absolutely continuous.

2 Bounded variation

b b ′ Question: Let f ∈ BV [a, b]. Show that Ta f ≥ a |f (x)|dx. Proof: By Jordan Theorem, f(x) can beR written as the diﬀerence of two monotone increasing functions on [a, b], f(x)= g(x) − h(x), where 1 g(x)= (T xf + f(x)), 2 a 1 h(x)= (T xf − f(x)). 2 a It is easy to verify that both g and h are increasing, so g′ and h′ are nonnegative. By Lebesgue Theorem, b g′(x)dx ≤ g(b) − g(a), Za b h′(x)dx ≤ h(b) − h(a). Za

II Yingwei Wang Real Analysis

Since f ′ = g′ − h′, |f ′|≤ g′ + h′. Then we have

b b ′ ′ b |f (x)|dx ≤ g (x)+ h(x)dx ≤ g(b) − g(a)+ h(b) − h(a)= Ta f. Za Za

3 Bounded variation

Question: Let f ∈ BV [a, b]. Show that f has at most countably many points of discontinuity.

Proof: First, I want to show that ∀x0 ∈ [a, b], both lim + f(x) and lim − f(x) x→x0 x→x0 exist. By Jordan Theorem, f = g − h where g and h are monotone increasing functions on

[a, b]. Let G = supx∈[a,x0) g(x) and H = supx∈[a,x0) h(x). It is obvious that A,B < ∞. Given ε> 0, there exists δ > 0 such that ε G − < g(x − δ) ≤ G, 2 0 ε H − < h(x − δ) ≤ H. 2 0

Then for ∀x ∈ (x0 − δ, x0), we have ε ε G − < g(x) ≤ G ⇒ 0 ≤ G − g(x) < , 2 2 ε ε H − < h(x) ≤ H ⇒ 0 ≤ H − h(x) < . 2 2 It follows that |G − H − f(x)|≤ (G − g(x)) + (H − h(x)) < ε, for ∀x ∈ (x0 − δ, x0). So lim + f(x)= G − H. Similarly we can know that lim − f(x) x→x0 x→x0 also exists. Second, I want to show that the set of discontinuities of f is at most countably.

Let E = {x ∈ [a, b] : f(x+) 6= f(x−)}, E1 = {x ∈ [a, b] : g(x+) > g(x−)}, E2 = {x ∈ [a, b] : h(x+) > h(x−)}. Since g and h are monotone increasing and f = g − h, we have E = E1 ∪ E2. For ∀x ∈ E1, ∃rx ∈ Q such that g(x−) < rx < g(x+). Besides, if x1 < x2, then g(x1+) ≤ g(x2−) so rx1 6= rx2 . Thus x → rx is a bijection between E1 and a subset of Q, which means E1 is at most countably. Similarly, E2 is at most countably and further E = E1 ∪ E2 is also at most countably.

III Yingwei Wang Real Analysis

4 L1 space

1 Question: Let (fn) be a sequence of nonnegative functions in L[0, 1] such that 0 fn(t)dt 1 1 R and 1/n dt ≤ 1/n for all n ≥ 1. If g(x) = supn fn(x) show that g∈ / L [0, 1]. Proof:R Suppose that g ∈ L1[0, 1], I want to get a contradiction. 1 By the absolutely continuity of the integral, for ε = 3 , there exists δ such that for any measurable subset E ⊂ [0, 1], if m(E) < δ, then

1 gdm< . (4.1) ZE 3

1 1 Choosing an integer n satisfying n ≥ 3 and 1/n < δ, then since 0 fn(t)dt and 1/n dt ≤ 1/n, we have R R 1/n 1 fndt = 1 − fndt ≥ 1 − 1/n, Z0 Z1/n for all n> 3. By the deﬁnition of g, we have

1/n 1/n 2 gdt ≥ fndt ≥ 1 − 1/n ≥ 1 − 1/3 ≥ . (4.2) Z0 Z0 3

But by (4.1), we have 1/n 1 gdt< . (4.3) Z0 3 It is obvious that (4.2) contradicts with (4.3).

IV Real Analysis Homework: #10

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Convergence series

∞ Question: Let (an)n=1 be sequence of non-negative numbers such that ∞ a < . (1.1) n ∞ nX=1

Let r1, r2,... be an enumeration of the rational numbers. Show that the series ∞ a2 n (1.2) x r nX=1 | − n| converges for almost all x R. ∈ Proof: I found two methods to prove this. Hopefully, both of them are correct.

1.1 Method one: integral

Consider the new series: ∞ a n (1.3) x r nX=1 | − n| p If (1.3) converges, then (1.2) is sure to converge. If x r 1 for some n, then | − n|≥ an an. (1.4) x r ≤ | − n| ∗ p E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis

So the convergence of the series is obvious. If x r < 1 for some n, then r 1

∞ an which means n=1 is convergent almost everywhere for x (rn 1, rn + 1). √|x−rn| ∈ − P In sum, (1.3) converges for x R, a.e. and so does (1.2). ∈ 1.2 Method two: measure

Consider the set Eε = x : x r < εa , (1.6) n { | − n| n} then the measure of this set is ε m(En) = 2εan. (1.7) For x R Eε , we have ∀ ∈ \∪ n an 1 x rn ≤ ε | −2 | an an ⇒ x rn ≤ ε |∞− | ∞ a2 1 n a . (1.8) ⇒ x r ≤ ε n Xn=1 | − n| nX=1

2 As ε 0, m( Eε ) 0. Hence, ∞ an converges for almost all x R. → ∪ n → n=1 |x−rn| ∈ P

II Yingwei Wang Real Analysis

2 Bounded variation functions

Question: Find all functions f BV [0, 1] with the property that ∈ f(x) + (T xf)1/2 = 1, x [0, 1]. (2.1) 0 ∀ ∈ Solution: Let x = 0 in Eqn.(2.1), we can get

f(0) = 1.

x 1/2 Since F (x) = (T0 f) is an increasing function, f(x) should be a decreasing function. Hence, F (x) = (T xf)1/2 = (f(0) f(x))1/2. So 0 − f(x) + (f(0) f(x))1/2 = 1, − (f(0) f(x)) = (1 f(x))2, ⇒ − − (1 f(x)) = (1 f(x))2, ⇒ − − f(x) = 1, x [0, 1]. ⇒ ∀ ∈

3 Absolutely continuous function

Question: Let f : [0, 1] [0, 1],f(x)= √x. Show that f AC[0, 1]. 1→ t −1/2 ∈ Proof: Since √x = 2 0 t dt, by the Absolutely continuity of the integral (Propo- sition 4.14 in Page 88 ofR Royden’s book), it is easy to know that f(x)= √x AC[0, 1]. ∈

4 Non-absolutely continuous function

Question: Construct f : [0, 1] [0, 1],f AC[0, 1] such that √f / AC[0, 1]. → ∈ ∈ Solution: Let f(x) be as following

x2 sin2( 1 ), 0

It is easy to know that f(x) AC[0, 1] since f ′(x) = 2x sin2(1/x) 2 sin(1/x) cos(1/x) ∈ − is bounded in [0, 1].

III Yingwei Wang Real Analysis

However, for √f, x sin( 1 ), 0

So T 1√f = and √f / BV [0, 1] and further √f / AC[0, 1]. 0 ∞ ∈ ∈

IV Real Analysis Homework: #11

Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Open sets

Question: Show that every open subset U of R can be written uniquely as the union of a countable family of mutually disjoint open intervals.

Proof: Let U ∈ R be open. For x1 ∈ U, consider the interval I1 = (a1, b1) where

a1 = inf{t ∈ R : (t,x) ⊂ U},

b1 = sup{t ∈ R : (x,t) ⊂ U}.

Then for x2 = U\I1, consider the interval I2 = (a2, b2) where

a2 = inf{t ∈ R : (t,x) ⊂ U\I1},

b2 = sup{t ∈ R : (x,t) ⊂ U\I1}. ∞ ∞ In this way, we get a sequence of disjoint open intervals {In}n=1 such that ∪n=1In = U.

2 Measurable sets

Question: Let f : R → R be a continuous functions with the property that m∗(f(A)) = 0 whenever m∗(A) = 0, A ⊂ R. Show that if A ⊂ R is Lebesgue measurable, then f(A) is Lebesgue measurable. ∞ Proof: That A is measurable means ∀ε > 0, ∃ disjoint open intervals {In}n=1 s.t. A ⊂∪In and m(∪In − A) < ε. ∗ ∗ ∗ ∗ Since m (A) = inf{ l(In) : A ⊂∪In}, we have m (∪In) → m (A) and m (∪In −A) → 0 as ε → 0. P ∗ By the assumption, m (f(∪In − A)) → 0, so m(∪f(In) − f(A)) < ε, for ∀ε> 0. Since ∪f(In) is an open set and f(A) ⊂∪f(In), we know that f(A) is measurable. ∗E-mail address: [email protected]; Tel: 765 237 7149

I Yingwei Wang Real Analysis

3 Absolutely continuity

Question: Let f : R → R ∈ AC[a, b] for all a < b, a, b ∈ R. Show that if A ⊂ R is Lebesgue measurable, then f(A) is Lebesgue measurable. Proof: It is suﬃces to show that f ∈ AC[a, b] for all a < b, a, b ∈ R implies that f has the property that m(f(E)) = 0 whenever m(E) = 0, A ⊂ R. That f is absolutely continuous means ∀ε > 0, ∃ δ such that for disjoint intervals n {Ii = (xi,yi)}i=1 satisfying n (yi − xi) < δ, i X=1 then n |f(yi) − f(xi)| < ε. i X=1 Suppose E ⊂ R and m(E) = 0. Let G = ∪(xi,yi) ⊃ E. Choose ci, di ∈ [xi,yi] such that

f([xi,yi]) = [f(ci),f(di)], then m(f(E)) ≤ m(f(G)) ≤ |f(di) − f(ci)| < ε.

So m(f(E)) = 0. X By the conclusion of the Problem 2, we can know that if E ⊂ R is Lebesgue measurable, then f(E) is Lebesgue measurable.

4 Lebesgue measure

Question: Let f ∈ AC[a, b] for all a < b where a, b ∈ R. Suppose that f is strictly increasing. Show that if A is Lebesgue measurable then

m(f(A)) = f ′(t)dt. (4.1) ZA

Proof: Case 1: A is an open interval A = (a0, b0). On one hand, since f is strictly increasing, m(f(A)) = f(b0)−f(a0); on the other hand, f ′(t)dt = b0 f ′(t)dt = f(b ) − f(a ). Thus, (4.1) holds. A a0 0 0 A A I I I , i j I a , b R Case 2: IfR is an open set, then = n where i ∩ j = ∅ 6= , n = ( n n). F

II Yingwei Wang Real Analysis

On one hand, ∞

m(f(A)) = (f(bn) − f(an)) = f(b∞) − f(a1). n=1 X On the other hand, ∞ bn ∞ ′ ′ f (t)dt = f (t)dt = (f(bn) − f(an)) = f(b∞) − f(a1). A n an n Z X=1 Z X=1 It implies the (4.1).

Case 3: If A is a Gσ-set, it suﬃces to show that

f ′(x)dx ≤ m∗(f(A)). (4.2) ZA −1 Choose open intervals In such that f(A) ⊂ ∪In, then E ⊂ ∪Jn, where Jn = f (In). (n) (n) Choose sequences {αk } and {βk } such that (n) lim αk = inf {x}, k→∞ x∈Jn (n) lim βk = sup {x}. k→∞ x∈Jn Then we get n β( ) ′ k ′ f (x)dx = lim f (x)dx ≤ m(In). k→∞ (n) ZJn Zαk As a sequence,

′ ′ ′ f (x)dx ≤ f (x)dx ≤ f (x)dx ≤ m(In). A ∪Jn Jn Z Z X Z By the deﬁnition of outer measure, we can get (4.2).

5 Signed measure

Question: Let µ = µ+ − µ− be the Jordan decomposition of a signed measure µ deﬁned on a measurable space (X, A). Deﬁne the total variation of µ to be the measure |µ| = µ+ +µ−. Show that for each A ∈A, n n |µ|(A) = sup |µ(Ai)| : (Ai)i is a partition of A with Ai ∈A,n ≥ 1 . (5.1) ( i ) X=1 III Yingwei Wang Real Analysis

Proof: Let E, F be a Hahn decomposition for µ. That is to say X = E ∪ F , E ∩ F = ∅, and E is positive while F is negative respect to µ. On one hand, we know that

µ+A = µ(A ∩ E), µ−A = −µ(A ∩ F ), ⇒ |µ|(A)= µ(A ∩ E) − µ(A ∩ F ). (5.2)

On the other hand,

(χA∩E − χA∩F )dµ = 1dµ − 1dµ = µ(A ∩ E) − µ(A ∩ F ). (5.3) ∩ ∩ ZA ZA E ZA F By Eqn.(5.2) and Eqn.(5.3), we know that

|µ|(A) = (χA∩E − χA∩F )dµ A Zn

= (χAi∩E − χAi∩F )dµ, i Ai X=1 Z

n where (Ai)i=1 is a partition of A with Ai ∈A. ∩ ∩ Since for each Ai, Ai (χAi E − χAi F )dµ ≥ |µ(Ai)|, so we have

R n n

|µ|(A)= (χAi∩E − χAi∩F )dµ ≥ sup |µ(Ai)| , (5.4) i Ai (i ) X=1 Z X=1 n for any partition (Ai)i=1. Besides, if we choose A1 = A ∩ E and A2 = A ∩ F , then

|µ|(A)= µ(A1) − µ(A2)= |µ(A1)| + |µ(A2)|, which means n sup |µ(Ai)| ≥ |µ|(A). (5.5) ( i ) X=1 By Eqn.(5.4) and (5.5), we can get the conclusion (5.1).