Math 245C Homework 1
Yunbai Cao 904066974 Apr 11, 2014

Exercise 1.10.3 Proof. Since R equipped with the usual topology F is Hausdorff, (R, F ) is Haudorff as F is stronger than F. Given any x ∈ R, for any y ∈ R\{x}, there exists By open such that By ∩ Byx = ∅. Then R\{x} = is open. Let K = [0, 1]\Q and L = { 1 }. Then K c = (−∞, 0) ∪ (1, ∞) ∪ Q ∈ F , so K 2 is closed. And L is closed by previous paragraph. Let U ⊃ K, V ⊃ L be two open neighbourhoods of K, L. Claim: U ∩ V = ∅. As F is generated by open intervals of R and {Q}, there exists
1 2 y∈R\{x} By

y open and Byx

x

is open. Therefore {x}

∈ V ⊂ V where V is a finite intersection of open

intervals (thus an open interval) and {Q}. So L ⊂ I ∩ Q ⊂ V where I is an open interval. On the other hand, as K ∩ Q = ∅, we may assume U ∈ F ( since we can always find an open subset U of U such that K ⊂ U ∈ F). As I
1 2

is an open
1 2

interval, U ∩ I = ∅ since otherwise there will be some irrational number close to in (R, F). Thus U ∩ V = ∅ as claimed. Therefore (R, F ) is not normal.

that is not contained in U . Therefore U ∩ I ∩ Q = ∅ as U ∩ I ∈ F and Q is dense

Exercise 1.10.6 Proof. Since X is locally compact, for any x ∈ K there exists Ux that U x is compact. Then K ⊂ cover of K, we may write K ⊂ then O = n i=1 U i x∈K Ux . Since K is compact, there n i=1 Ui with Ui ∈ {Ux : x ∈ K}. Let

x open such exists a finite O= n i=1 Ui ,

is compact since finite union of compact sets is compact (Exercise

1.8.1). Let U0 = U ∩ O, then U0 ⊃ K is open and U 0 is compact.

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Now U 0 equipped with the subspace topology induced from X is both compact and Hausdorff thus is normal. By Urysohn’s lemma there exists a continuous function f : U 0 → [0, 1] such that 1K ≤ f ≤ 1U0 . Define f : X → [0, 1] by f (x) = f (x) for x ∈ U 0 and f (x) = 0 otherwise. Claim: f is continuous. It suffices to show f −1 ((a, ∞)) and f −1 ((−∞, b)) is open for any a, b ∈ R. If a < 0, f −1 ((a, ∞)) = X, if a ≥ 1, f −1 ((a, ∞)) = ∅. If 0 ≤ a < 1, f −1 ((a, ∞)) ⊂ U0 and is open in U 0 with subspace topology, i.e. there exists H1 ⊂ X open such that f −1 ((a, ∞)) = H1 ∩ U 0 = H1 ∩ U0 . So f −1 ((a, ∞)) is open in X. If b > 1, f −1 ((−∞, b)) = X and if b ≤ 0, f −1 ((−∞, b)) = ∅. If 0 < b ≤ 1, f −1 ((−∞, b)) = U0 ∪ (H2 ∩ U 0 ) for some open set H2 ⊂ X. Thus f −1 ((−∞, b)) = U0 ∪ H2 is open in X. Therefore f is continuous with supp(f ) ⊂ U 0 which is compact, and by construction we have 1K ≤ f ≤ 1U . c c

Exercise 1.10.7 Proof. Let me first show that C0 (X → R) ⊂ Cc (X → R): For any f ∈ C0 (X → R) and any such that |f (x)| ≤ > 0, there exists a compact set K ⊂ X for all x = K. As X is locally compact Hausdorff, by Exercise

1.10.6 there exists g ∈ Cc (X → R) such that 1K ≤ g ≤ 1X . Let f = f g, then f ∈ Cc (X → R) and f (x) = f (x) for all x ∈ K and |f (x)| = |f (x)g(x)| ≤ |f (x)| ≤ for all x ∈ K. Thus |f (x) − f (x)| ≤ 2 for all x ∈ X. So C0 (X → R) ⊂ Cc (X → R). / Therefore it suffices to show that C0 (X → R) is closed. Let’s first work with the additional hypothesis that every non-empty open set has positive measure. We want to show that if fn ∈ C0 (X → R) such that fn → f in L∞ , then f ∈ C0 (X → R). I.e. there exists a f that is continuous and vanishes at infinity that differs from f with a set of meausre 0. We can assume fn are continuous and vanish at infinity for all n. Claim: if fn → f in L∞ and f is continuous, then for any compact set K such that |f (x)| ≤ If fn → f in L∞ , for all x ∈ K. /
2 2

> 0 there exists a

then there exists N large enough such that |f (x) − fN (x)| <

for µ-almost everywhere x. And there exists a compact set K such that |fN (x)| < for all x ∈ K. Therefore |f (x)| < / |f (x)| ≤

for µ-almost everywhere x ∈ K. But then /

for all x ∈ K: suppose not, i.e. there is an x ∈ K such that |f (x)| = / /

δ > . Since f is continuous there exists an open neighbourhood of U of x such 2

that U ⊂ X\K satisfies |f (y) − f (x)| < δ − for all y ∈ U . Then |f (y)| > for all y ∈ U , but µ(U ) > 0. Contradiction, so we have proved the claim. Thus it suffices to show if fn → f in L∞ , then there exists a continuous function f such that f differs from f on a set of measure 0. fn → f in L∞ means fn → f uniformly outside of a set E with µ(E) = 0. Since {fn } ⊂ C0 (X → R), fn ’s are bounded. So {fn } is Cauchy on E c with the supnorm. Thus for any |fn (y) − fm (y)| < there exists Un > 0, there exists N ∈ N such that m, n > N implies x both open such that |fn (y) − fn (x)| < for any for all y ∈ E c . Now for any x ∈ E, since fn , fm are continuous, x and Um

y ∈ Un , and |fm (z) − fm (x)| < for any z ∈ Um . Let U = Un ∩ Um . Then U is open thus has positive measure, so U ∩ E c = ∅. Take y ∈ U ∩ E c . Then |fn (x) − fm (x)| ≤ |fn (x) − fn (y)| + |fn (y) − fm (y)| + |fm (y) − fm (x)| < 3 Thus {fn } is also Cauchy on E with the supnorm, so fn also converges uniformly to some function on E as BC(E → R) is complete. Since fn converges uniformly both on E and E c , we conclude that fn → f uniformly on X for some f . Therefore f is continuous and f (x) = f (x) for all x ∈ E c . This finishes the proof of the case with the addtional hypothesis that every non-empty open set has postivie measure. Now we drop the hypothesis and work with the general case. Let A := {O : O ⊂ X open and µ(O) = 0}. Let U :=
O∈A O.

Then µ(U ) = 0 since for any compact

K ⊂ U , K can be covered by finitely many O, thus µ(K) = 0; so µ(U ) = 0 by inner regularity. Now U c ⊂ X is closed and is LCH and σ-compact with the relative topology, and µ restricted to U c is still a Radon measure. Then if fn : X → R are continuous and fn → f for some f : X → R in L∞ , then fn → f on U c in L∞ , so by previous case we can find a continuous f : U c → R such that f differs from f on a set of measure 0 on U c . Now I want to apply the Tietze extension theorem to extend f to X then we are done because f only differs from f on a set of measure 0 as µ(A) = 0. But we need either X to be compact to apply the normal version or U c to be compact to apply the LCH version. But we only have X is σ-compact and U c is merely closed. I tried a lot of approach, like decompse X into increasing unions of compact sets and trying to glut functions, or trying to use inner regularity of µ to come up with a function. But none of them works. I have to give up in the end. (feel really bad...)

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Exercise 1.10.8 ˜ ˜ Proof. For any x ∈ X\K, WTS there exists f (x) such that |f (x) − f (y)| ≤ Ad(x, y) ˜ for all y ∈ K, i.e. f (y) − Ad(x, y) ≤ f (x) ≤ f (y) + Ad(x, y) for all y ∈ K. But for any y, y ∈ K, f (y) − f (y ) ≤ Ad(y, y ) ≤ Ad(x, y) + Ad(x, y ). Thus f (y) − Ad(x, y) ≤ f (y ) + Ad(x, y ) for all y, y ∈ K. Therefore supy∈K (f (y) − ˜ Ad(x, y)) ≤ inf y ∈K (f (y ) + Ad(x, y )), so pick f (x) to be any value in between will ˜ make the job. Thus we can find a f : K ∪ {x} → R that is lipschitz with constant A that extends f . Now consider the set of all pairs (E, f ) where K ⊂ E ⊂ X and f : E → R extending f and is lipschitz with constant A. Then the set is partially ordered by declaring (F, f ) ≥ (E, f ) if E ⊂ F and f extends f . Then it is easy to see that every chain of partial extensions has an upper bound; hence, by Zorn’s lemma, there must be a maximal partial extension (E∗ , f∗ ). If E∗ = X then we are done; otherwise, one can find x ∈ X\E∗ and by the first paragraph, we can then extend f∗ further to the larger set E∗ ∪ {x}, a contradiction.

Exercise 1.10.9 Proof. Since f : K → R is continuous, arctan(f ) : K → (− π , π ) ⊂ [− π , π ] is also 2 2 2 2 continuous. As X is normal and K is closed, by Tietze extension theorem we can find a g : X → [− π , π ] such that g = arctan(f ) on K. 2 2 Let E := g −1 ({− π }, { π }), then E ⊂ X is closed as single points are closed in R 2 2 and g is continuous. E ∩ K = ∅ since g(x) = arctan(f )(x) ∈ (− π , π ) for all x ∈ K. 2 2 Thus by Urysohn’s lemma there exists a continuous function h : X → [0, 1] such that h = 1 on K and h = 0 on E. Let g := gh. Then g is continuous, − π < g < π , and g (x) = g(x) = 2 2 ˜ := tan g : X → R is continuous, and f (x) = ˜ arctan(f )(x) for all x ∈ K. Then f tan g(x) = tan arctan(f )(x) = f (x) for all x ∈ K.

Exercise 1.10.13 Proof. Let X = E =
∞ n=1 En ∞ n=1 Xn

with each Xn compact. For any E ⊂ X borel, write

with En = E ∩ Xn . So µ(En ) ≤ µ(Xn ) < ∞. Now for any En , 4

by inner regularity, for any i ∈ N, there exists a Kni ⊂ En that is compact and µ(Kni ) + 1 > µ(En ). Let Kn = i thus µ(En \Kn ) = 0. Let K =
∞ i=1 µ(En \Kn ) ∞ i=1 Kni . ∞ n=1 Kn ∈

Then Kn ∈ Fσ and µ(Kn ) = µ(En ) < ∞, Fσ . Then µ(E\K) ≤ µ(
∞ i=1 (En \Kn ))

≤

= 0. This proves that every borel set E can be expressed as the

union of an Fσ set and a null set. On the other hand, E c is also Borel. Let F ∈ Fσ be such that F ⊂ E c and µ(E c \F ) = 0. Then F c ∈ Gδ , E ⊂ F c , and µ(F c \E) = µ(E c \F ) = 0.

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