REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS Sample Exam Problems: (1) Be able to do the problems on the two previous tests. OK (2) Prove that a function f : X → R is measurable if and only if f −1 (−∞, t) is a measurable set for every t ∈ R. Proof: If f is measurable, then f −1 ([−∞, y]) = f −1 ((−∞, y]) is measurable for each y ∈ R. Given any t ∈ R, let 1 −1 An = f −∞, t − , n which is measurable. Thus, [∞ n=1 An S 1 is measurable. If x ∈ ∞ n=1 An , then f (x) < t − k for some k ∈ N, so then f (x) < t, so x ∈ f −1 (−∞, t). On the other hand, if x ∈ f −1 (−∞, t), then f (x) < t, and since lim n1 = 0,Sthere exists m ∈ N such that m1 < t − f (x). Thus f (x) < t − m1 , so x ∈ Am ⊂ ∞ n=1 An . Thus, [∞ An f −1 (−∞, t) = n=1 is measurable. Conversely, if f −1 (−∞, t) is measurable for every t ∈ R, then for every y ∈ R, the set 1 −1 Bn = f −∞, y + n is measurable, so that \∞ Bn n=1 T 1 is measurable. If x ∈ ∞ n=1 Bn , then f (x) < y + n for every n ∈ N, so by the order limit theorem f (x) ≤ y, which implies that x ∈ f −1 ((−∞, y]) = f −1 ([−∞, y]). On 1 −1 the other T∞ hand, if x ∈ f ((−∞, y]), then f (x) ≤ y < y + n for every n ∈ N, so x ∈ n=1 Bn . Thus \∞ f −1 ([−∞, y]) = Bn n=1 is measurable for each y ∈ R, so f is a measurable function. QED 1 2 REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS (3) Prove that every continuous function on R is Borel measurable. If f is a continous function on R, then for each y ∈ R, f −1 ([−∞, y]) = f −1 ((−∞, y]) is the inverse image of a closed set and is thus closed, and so it is a Borel set. QED (4) Let (fn )n≥1 be a sequence of measurable functions on R. Let A = {x | lim fn (x) n→∞ exists}. Prove that A is measurable. Proof: First observe that if g : R → R is measurable, and (a, b) is any open interval, then g −1 (a, b) is measurable. The reason is that g −1 (a, b) = { x | a < g (x) < b} = { x | g (x) ∈ (−∞, b) and g (x) ∈ / (−∞, a]} c = g −1 (−∞, b) ∩ g −1 (−∞, a] , which is measurable by definition and by Problem 2 above. Also, for any n, m ∈ N, the function (fn − fm ) is measurable. Next, note that by the Cauchy Criterion, A = {x | ∀ε > 0, ∃N ∈ N such that n, m ≥ N implies |fn (x) − fm (x)| < ε} = {x | ∀ε > 0, ∃N ∈ N such that n, m ≥ N implies − ε < fn (x) − fm (x) < ε} 1 1 = x | ∀k ∈ N, ∃N ∈ N such that n, m ≥ N implies − < fn (x) − fm (x) < k k 1 1 = x | ∀k ∈ N, ∃N ∈ N such that n, m ≥ N implies x ∈ (fn − fm )−1 − , k k \ [ \ \ 1 1 (fn − fm )−1 − , = k∈N N ∈N n≥N m≥N k k Since countable unions and countable intersections of measurable sets are measurable, A is measurable. QED (5) Define ∞ X xn . g (x) = ln (n) n=2 Prove that g is continuous on the interval [−1, 1). Proof: Observe that g (−1) is (after n = 2) an alternating series that converges by the Alternating Series Test. By Abel’s theorem, the series converges uniformly on [−1, 0]. Further, given any fixed x ∈ (0, 1), the series converges at x, because it is comparable to a convergent geometric series (for n ≥ 3), since n x 1 n ln (n) ≤ ln 3 x . Again, by Abel’s theorem (or by Weierstrass M -test), the series converges uniformly on [0, x]. Thus, the series converges uniformly on [−1, x] (choosing the maximum of the N ’s for the two intervals in the definition of uniform convergence). Thus, the REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS 3 series converges to a continuous function on [−1, x]. Since x ∈ (0, 1) was chosen arbitrarily, we conclude that the series g (x) is continuous on [−1, 1). QED (6) For a fixed p ∈ (0, ∞), for which m ∈ R is the function hm ∈ Lp (1, ∞), where hm (x) = xm for all x ∈ R? Proof: ( mp+1 Z ∞ b 1 lim − if mp 6= −1 p b→∞ mp+1 mp+1 (xm ) dx = limb→∞ (log (b) − log (1)) if mp = −1 1 is finite if and only if mp + 1 < 0, so that m < −1 p in order that hm ∈ Lp (1, ∞). QED (7) Let (fn )n≥1 be a sequence of continuous functions from R to R, and let K be a compact subset of R. Suppose that fj (x) ≥ fj+1 (x) for every j ∈ N and every x ∈ K, and suppose that lim fj (x) = 0 for all x ∈ K. Prove that j→∞ (fj (x))j≥1 converges to zero uniformly on K. Proof: For every x ∈ K, since (fj (x))j≥1 is a decreasing sequence and limj→∞ fj (x) = 0, we see that each fj (x) is nonnegative and converges to zero. Let Mj = max fj (x) , x∈K and let xj ∈ K be chosen such that fj (xj ) = Mj . (Such an xj can be chosen since a continuous function on a compact set attains its maximum and minimum.) Note that (Mj ) is a decreasing sequence since Mj = max fj (x) ≥ max fj+1 (x) = Mj+1 . x∈K x∈K Since 0 ≤ Mj ≤ M1 , (Mj ) is a bounded monotone sequence and thus converges to a number M ≥ 0. Since K is compact, there exists a subsequence (xjk )k≥1 that converges to a point a ∈ K, and limk→∞ fjk (xjk ) = M . Given any ε > 0, there exists N ∈ N such that for all k ≥ N , |fjk (a)| < 2ε . Since K is compact, fjN is uniformly continuous on K , so there exists δ > 0 such that |y − a| < δ implies |fjN (y) − fjN (a)| < 2ε . If |a − y| < δ, then for all n ≥ jN , 0 ≤ fn (y) ≤ |fjN (y)| ≤ |fjN (y) − fjN (a)| + |fjN (a)| ε ε < + = ε. 2 2 Choose m sufficiently large so that |xjm − a| < δ. Then for k ≥ max {m, N }, we have 0 ≤ fjk (xjk ) < ε. Since such a k can be found for every ε > 0, we conclude that M = = lim fjk (xjk ) k→∞ lim max fjk (x) = 0. k→∞ x∈K 4 REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS Thus, given ε > 0, there exists L ∈ N such that for all n ≥ jL , for all x ∈ K, 0 ≤ fn (x) ≤ fjL (x) ≤ max fjL (x) < ε. x∈K Thus, (fn ) converges uniformly to the zero function on K . QED (8) Prove that the measure of any countable set in Rn is zero. Proof: Let S = {x1 , x2 , ...} be any countable set in Rn . Then, given S any ε > 0, let ε Bj be the open ball centered at xj whose measure is 2j+1 . Then G = Bj is an open set containing S, and X ε λ (G) ≤ by subadditivity j+1 2 j≥1 = ε 4 1− 1 2 = ε < ε, 2 so 0 ≤ λ∗ (S) < ε. By the order limit theorem (taking ε → 0+ ), λ∗ (S) = 0, so S is measurable and has measure zero. QED (9) Prove that the Cantor set has measure zero using the dilation property of Lebesgue measure. Observe that the part C1 of the Cantor set C contained in 0, 13 has the property S that C = C1 C1 + 32 , so that λ (C) = λ (C1 ) + λ C1 + 32 = 2λ (C1 ), since it is a disjoint union. (Note that C and C1 are compact and thus measurable.) Also, C = 3C1 , so that λ (C) = 3λ (C1 ). Thus, 3λ (C1 ) = 2λ (C1 ), so λ (C1 ) = 0, and thus λ (C) = 0. QED (10) Suppose that A ⊂ Rn is measurable, and let λ (A) < ∞. Prove that there exists a sequence of compact sets K1 ⊂ K2 ⊂ K3 ⊂ ... such that [ λ A\ j≥1 Kj = 0. Is the same statement true if the assumption λ (A) < ∞ is dropped? Proof: Since A is measurable, λ (A) = λ∗ (A) = sup {λ (K) | K ⊂ A and K is cpt} . By the sup lemma, given any m ∈ N, there exists a compact set Cm such that Cm ⊂ A and 1 λ (A) − < λ (Cm ) ≤ λ (A) . m REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS 5 Next, letting [m Km = k=1 Ck ⊂ A, observe that Km is closed (finite union of closed sets) and bounded (finite union of bounded sets) and is thus compact. Then K1 ⊂ K2 ⊂ K3 ⊂ ..., and 1 λ (A) − < λ (Cm ) ≤ λ (Km ) ≤ λ (A) . m Then λ (Km ) + λ (A \ Km ) = λ (A) , so 1 . λ (A \ Km ) ≤ m The sets Bm = A \ Km satisfy λ (Bm ) ≤ λ (A) < ∞ and B1 ⊃ B2 ⊃ B3 ⊃ ... So by a property of measurable sets, \∞ λ Bm = lim λ (Bm ) = 0, m=1 m→∞ and \∞ m=1 \∞ (A \ Km ) [∞ = A\ Km . QED (first part) Bm = m=1 m=1 Surprisingly, this proof can be adjusted to work even in the case when λ (A) = ∞. The adjustment one makes is as follows. Instead of using the definition of λ∗ (A), use the approximation result that states that for every ε > 0, there exists a closed set Fε ⊂ A such that λ (A \ Fε ) < ε. And each such Fε is a countable union of nested compact sets, that is [ Fε = Fε ∩ B (0, k) . k∈N (Note that Fε ∩ B (0, k) is compact since it is the intersection of a compact set and a closed set and is thus closed and bounded.) By choosing ε = m1 as above we let F 1 m be a specific choice of the closed set in A with those properties. Then we let [m Km = F 1 ∩ B (0, m) , j=1 j and observe that for every k ∈ N we have [ F1 ⊂ Km . m∈N k Then A\ [∞ m=1 Km ⊂ A \ F 1 k 6 REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS for all k ∈ N, so 0≤λ A\ [∞ m=1 Km ≤λ A\F 1 k < 1 k for each k, so by the order limit theorem (with k → ∞), we have [∞ λ A\ m=1 Km = 0. QED (second part) (11) Let X be a set, and let M be the σ-algebra of subsets of X. Prove that the sum of two M -measurable simple functions on X is M -measurable. Proof: A simple function f : X → R is a M -measurable function such that f (X) is finite. If f1 and f2 are any two simple functions on X, let f1 (X) = {α1 , ..., αk1 } and f2 (X) = {β1 , ..., βk2 }. Observe that (f1 + f2 ) (X) = {f1 (x) + f2 (x) | x ∈ X} [k1 [k2 ⊂ {αj + βi } , j=1 i=1 which is a set containing a maximum of k1 k2 points. Thus (f1 + f2 ) (X) is finite, and since the sum of two M -measurable functions is M -measurable, (f1 + f2 ) is a simple function. QED (12) Prove the absolute continuity of the Lebesgue integral. See text. R (13) Suppose that f : (0, 1) → R is measurable, and E f dλ = 0 for every set E ⊂ (0, 1) such that 0 < λ (E) < 1. Prove that f = 0 a.e. in (0, 1). Proof: Let E+ = f −1 ([0, ∞)), E− = f −1 (−∞, 0). Then E+ and E− are measurable. Further, let E+1 = E+2 = E−1 = E−2 = 1 E+ ∩ 0, , 2 1 E+ ∩ , 1 , 2 1 E− ∩ 0, , and 2 1 E− ∩ , 1 . 2 REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS 7 All of these sets are measurable and disjoint, and by hypothesis Z Z 1 Z Z f dλ = f+ dλ = f dλ + f dλ = 0, and E+ 0 Z Z f dλ = − 1 E+ 2 E+ Z Z 1 f dλ + f− dλ = 1 E− 0 E− f dλ = 0. 2 E− Thus, it suffices to show that if a measurable function g satisfies g ≥ 0 on a measurable R set S and S g dλ = 0, then g = 0 a.e. on S. If we are able to show that, then it follows that f+ = 0 a.e. on (0, 1) and f− = 0 a.e. on (0, 1), and thus f = 0 a.e. on (0, 1). QED (modulo Lemma below) R Lemma: If a measurable function g satisfies g ≥ 0 on a measurable set S and g dλ = 0, then g = 0 a.e. on S. S 1 Proof of Lemma: For each k ∈ N, consider the set S k = x ∈ S | g (x) > k = c S ∩ g −1 −∞, k1 , which is measurable. Observe that S1 ⊂ S2 ⊂ ... Also, Z Z Z Z g dλ = g dλ + g dλ ≥ g dλ 0 = S S\Sk Sk Sk Z 1 1 ≥ dλ = λ (Sk ) ≥ 0. k Sk k Thus, λ (Sk ) = 0 for each k ∈ N. Next, let c S+ = { x ∈ S | g (x) > 0} = S ∩ g −1 (−∞, 0] [ Sk . = k∈N Then, by a property of measurable sets, [ λ (S+ ) = λ = k∈N Sk lim λ (Sk ) = 0. k→∞ So since g (x) = 0 for x ∈ S \ S+ and λ (S+ ) = 0, g = 0 a.e. on S. QED(Lemma) (14) Suppose that f is a measurable function on a measurable set A ⊂ Rn with λ (A) < ∞, and suppose that ∞ X 2k λ x ∈ A : f (x) ≥ 2k k=1 converges. Prove that Z f dλ < ∞. A 8 REAL ANALYSIS 2 FINAL EXAM SAMPLE PROBLEM SOLUTIONS R R Proof: Since A f dλ = −∞ is a possibility, it suffices to show that A f+ dλ < ∞. Consider the (simple) function g : A → R defined by k+1 2 if 2k < f+ (x) ≤ 2k+1 for k ∈ {0, 1, 2, ...} g (x) = , ie 1 otherwise ∞ X g = 2k+1 χf −1 (2k ,2k+1 ] + χf+−1 [0,1] . + k=0 Z By construction, f+ (x) ≤ g (x) for every x ∈ A. Then Z f+ dλ ≤ g dλ A A = ≤ ∞ X k=0 ∞ X 2k+1 λ x ∈ A : 2k < f+ (x) ≤ 2k+1 2k+1 λ x ∈ A : 2k ≤ f+ (x) k=0 ∞ X ≤ 2 2k λ x ∈ A : 2k ≤ f (x) + λ ({x ∈ A : 0 ≤ f+ (x) ≤ 1 }) + λ (A) + 2λ (A) + λ (A) , k=1 which is finite by the two assumptions. QED