Homework 3 Solutions, Real Analysis I, Fall, 2010. (9) Let f : (, ) [, ] be a function whose restriction to (, 0) (0, ) is continuous. Assume the one-sided limits p = lim x 0 f(x), q = lim x 0 + f(x) exist as extended real numbers. Show that f is upper semicontinuous on (, ) if and only if f(0) max{p, q}. Solution: To prove =, assume f(0) max{p, q}. To prove f is upper semicontinuous, we need to show B α = {x : f(x) < α} is open for all α R. By assumption, we know C α = B α {0} is open for all real α. Then we have a few cases If f(0) =, then for each α R, 0 / B α. Therefore, B α = C α is open. If f(0) =, then that implies p = q = as well, and since both one-sided limits are equal, we must have lim x 0 f(x) = as well. Therefore, for all real numbers α, there is an ɛ > 0 so that x < ɛ = f(x) < α. This shows ( ɛ, ɛ) B α. Thus B α = (B α {0}) ( ɛ, ɛ) is the union of two open sets, and so is open. Finally, if f(0) R, and f(0) max{p, q}, then f(0) lim x 0 + f(x), f(0) lim f(x). x 0 In particular, if ɛ > 0, then there are δ +, δ > 0 so that and δ < x < 0 = f(x) < f(0) + ɛ, 0 < x < δ + = f(x) < f(0) + ɛ. So for δ = min{δ, δ + } > 0, we have x < δ = f(x) < f(0) + ɛ. Now choose α R. If α f(0), then B α = C α is open. On the other hand, if α > f(0), then for ɛ = α f(0), there is a δ > 0 so that B α = C α ( δ, δ), which is open. 1
Thus f is upper We will prove = by contradiction. If f(0) < max{p, q}, then without loss of generality, assume f(0) < p = lim x 0 f(x). Then for α (f(0), p), there is a δ > 0 so that δ < x < 0 implies f(x) > α. Therefore, B α = {x : f(x) < α} includes 0 but does not intersect the interval ( δ, 0). Thus no neighborhood of 0 is contained in B α, and B α is not open. Thus f is not upper semicontinuous, and = is proved by contradiction. (10) Let (X, d) be a compact metric space which contains at least 2 points. Consider a point p X. Show that X {p} is compact if and only if inf{d(p, x) x p} > 0. Solution: To prove =. Note that since X is a metric space, it is a Hausdorff topological space. So if X {p} is compact, then X {p} is a closed subset of X. Therefore, the complement {p} is an open subset of X. In a metric space, this means that there is an ɛ > 0 so that B ɛ (p) {p}. In other words, if y X and d(y, p) < ɛ, then y = p. This shows that inf{d(p, x) x p} ɛ > 0. To prove =, if L = inf{d(p, x) x p} > 0, then the open ball B L (p) = {p}. Since {p} is open, X {p} is a closed subset of X, and thus it is compact. (11) Let X be a Hausdorff topological space, and let K, K be two disjoint compact subsets of X. Prove that there are disjoint open sets U, U so that K U and K U. (Hint: Use Theorem 2.5 and mimic its proof.) Solution: For p K, since K K =, p / K. Thus we can apply Theorem 2.5 to show that there are open sets U p and V p so that p U p, K V p, U p V p =. Therefore, we have an open cover of K given by K U p. p K Since K is compact, there is a finite subcover K n U pi. 2
Now if we let U = n U pi, V = n V pi, then U, as a union of open sets, is open; V, as a finite intersection of open sets, is open; K U; K V ; and U V = since U pi V pi = for each i = 1,..., n. (12) Rudin, chapter 2, problem 1. Let f n be a sequence of real nonnegative functions on R 1, and consider the following four statements: (a) If f 1 and f 2 are upper semicontinuous, then f 1 +f 2 is upper (b) If f 1 and f 2 are lower semicontinuous, then f 1 + f 2 is lower (c) If each f n is upper semicontinuous, then 1 f n is upper (d) If each f n is lower semicontinuous, then 1 f n is lower Show that three of these are true and that one if false. What happens in the word nonnegative is omitted? Is the truth of the statements affected if R 1 is replaced by a general topological space? Proof. (c) is false. Consider the sequence of continuous functions 1 for x < 0 h n (x) = 1 x n for 0 x 1 0 for x > 1 Then {h n (x)} is an increasing sequence of nonnegative continuous functions. Then if we define f 1 = h 1 and f n = h n h n 1, then the f i are continuous and nonnegative. n h n = f i, lim h n = f i. n So we can compute lim h n(x) = n { 1 for x < 1 0 for x 1 This function is not upper semicontinuous even though it is the sum of a series of nonnegative continuous (and thus upper semicontinuous) functions. 3
We ll prove (a) for f 1 and f 2 upper semicontinuous on X a general topological space and we also let f 1, f 2 be real-valued and not necessarily nonnegative. Then for all real β, {x : f i (x) < β} is open, i = 1, 2. Let α be a real number. Then we claim {x : f 1 (x) + f 2 (x) < α} = N R({x : f 1 (x) < N} {x : f 2 (x) < α N}). This is an open set, and so f 1 + f 2 is upper To prove the claim, first prove. So let x satisfy f 1 (x) + f 2 (x) < α. Therefore, there is an ɛ > 0 so that f 1 (x) + f 2 (x) < α ɛ. Now if N = f 1 (x) + ɛ, then f 1 (x) < N and f 2 (x) < α ɛ f 1 (x) = α N. Thus x is in the set in the right-hand side. Now we prove. Assume for some N, x satisfy f 1 (x) < N and f 2 (x) < α N. Then f 1 (x)+f 2 (x) < N +(α N) = α. This proves the claim and thus that f 1 + f 2 is upper To prove (b) in the same setting as (a) above, note that f is upper semicontinuous if and only if f is lower Thus the result follows from writing (f 1 +f 2 ) = ( f 1 )+( f 2 ). We prove (d) for a general topological space, but we still assume f n are nonnegative and lower Then (b) applied inductively implies that each partial sum h n = n 1 f i is lower Let h = lim n h n = 1 f n. Since f n 0, we have h n is an increasing sequence of functions whose limit is h. We need to prove h is lower Let α R. Then we claim {x : h(x) > α} = {x : h n (x) > α}. n=1 This is a union of open sets, and thus is open. This shows h is lower We prove of the claim. Assume h(x) > α. Therefore, there is an ɛ > 0 so that h(x) > α + ɛ. Since lim h n (x) = h(x), there is an n so that h n (x) > α. To prove of the claim, assume that there is an n so that h n (x) > α. Then since h n increases to h, h(x) h n (x) > α. This proves the claim. Finally, (d) is false if we remove the condition that f n 0. We may simply take the negative of the counterexample for (c) above. (13) Rudin, chapter 2, problem 4. 4
(4a) If E 1 V 1 and E 2 V 2, where V 1 and V 2 are disjoint open sets, then µ(e 1 E 2 ) = µ(e 1 ) + µ(e 2 ), even if E 1 and E 2 are not in M. Solution: First of all, note that since µ is a Borel measure, then µ(u 1 U 2 ) = µ(u 1 )+µ(u 2 ) for U 1 and U 2 disjoint open sets. Step I implies µ(e 1 E 2 ) µ(e 1 ) + µ(e 2 ). Therefore, we need only show µ(e 1 E 2 ) µ(e 1 ) + µ(e 2 ) to prove the result. Choose an open set W E 1 E 2. Then U i = W V i are disjoint open supersets of E i. Therefore, we have µ(w ) µ(u 1 U 2 ) = µ(u 1 ) + µ(u 2 ) µ(e 1 ) + µ(e 2 ). (The first inequality is since µ is monotone. The equality is since U i are disjoint and measurable. The last inequality is by definition of µ(e i ).) Since this is true for arbitrary supersets W of E 1 E 2, we have µ(e 1 E 2 ) = inf{µ(w ) : E 1 E 2 W open} µ(e 1 ) + µ(e 2 ). (4b) If E M F, then E = N K 1 K 2, where {K i } is a disjoint countable collection of compact sets and µ(n) = 0. Solution: E M F means that µ(e) = sup{µ(k) : E K compact}, and is equivalent to E in the σ-algebra M and µ(e) <. Choose by induction E 1 = E, K i E i so that K i compact and µ(e i K i ) < 2 i. E i+1 = E i K i. Then induction and the fact that M is a σ-algebra containing all Borel sets show that each E i+1 is measurable. As subsets of E, they have finite measure and thus are in M F. The condition µ(e i K i ) < 2 i is possible by the definition of M F and the fact that µ is an additive measure. It s also clear by induction that the K i are disjoint, and that E i = E K 1 K i 1. Let N = E ( K i ) = E i. By construction, µ(e i ) < 2 i+1, and the E i are nested. Then the characteristic functions χ Ei χ N pointwise and χ E χ Ei for all i. Since E has finite measure, the Dominated Convergence Theorem applies, and µ(n) = χ N dµ = lim χ Ei dµ = lim µ(e i ) = 0. X i X i 5