Midterm - 6 Questions on Real Analysis I with Solutions | MATH 533, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Midterm - 6 Questions on Real Analysis I with Solutions | MATH 533 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Math 533 Midterm Exam Part II, Fall 2010 Solutions 1. Let (X,M, µ) be a complete measure space. Suppose that a set E ⊂ X satisfies: for any  > 0 there exist A,B ∈M so that A ⊂ E ⊂ B µ(B \A) < . Prove that E ∈M. Solution. It is given, that for every n ∈ N there exist An, Bn ∈M with An ⊂ E ⊂ Bn and µ(Bn\An) < 1n . Let A∗ = ⋃∞ n=1An and B∗ = ⋂∞ n=1Bn. These are countable unions and intersections of sets in M, hence A∗, B∗ ∈ M. We also have A∗ ⊂ E ⊂ B∗, and since B∗ \ A∗ ⊂ Bn \ An it follows that µ(B∗ \ A∗) < 1/n for each n ∈ N. So µ(B∗ \ A∗) = 0. Since E \ A∗ ⊂ B∗ \ A∗ and M is complete, it follows that E \ A∗ ∈ M. Since A∗ ∈ M we proved E = A∗ ∪ (E \ A∗) ∈ M as required. 2. Let f ∈ L1(X,M, µ). Prove that for any  > 0 there is E ∈M with µ(E) <∞ so that∣∣∣∣∫ E f dµ− ∫ f dµ ∣∣∣∣ < . Solution. For n ∈ N let En = {x ∈ X : |f(x) > 1/n}. Then 1n · χEn(x) ≤ |f(x)| for all x ∈ X, Therefore µ(En) ≤ n · ∫ 1 n χEn dµ ≤ n · ∫ |f | dµ <∞. Hence we have an increasing sequence of finite measure sets E1 ⊂ E2 ⊂ · · ·, µ(En) < ∞. Let fn = f · χEn . Then for every x ∈ X: |f1(x)| ≤ |f2(x)| ≤ · · · ≤ |fn(x)| → |f(x)|. By Monotone Convergence Theorem: ∫ |fn| dµ→ ∫ |f | dµ. Therefore∣∣∣∣∫ X f dµ− ∫ En f dµ ∣∣∣∣ = ∣∣∣∣∣ ∫ Ecn f dµ ∣∣∣∣∣ ≤ ∫ Ecn |f | dµ = ∫ |f | dµ− ∫ |fn| dµ→ 0. Given  > 0, for n large enough, the right hand side is <  while µ(En) <∞. 3. Define λ : BR → [0,∞) by λ(E) = ∫ E 1 1 + x2 dm(x). (a) Prove that λ is a finite measure on (R,BR). (b) Find a monotonic F : R→ R so that λ is given by the Lebesgue-Stieltjes measure mF . Math 533 Midterm 1 Solution. (a). Since φ(x) = 1 1+x2 is a continuous function, the Lebesgue integral of φ over any finite closed interval [a, b] agrees with the Riemann integral which can be computed using the Fundamental Theorem of Calculus and the fact that ddx arctan(x) = 1 1+x2 . We have λ([a, b]) = ∫ [a,b] φdm = ∫ b a φ(x) dx = arctan(b)− arctan(a). Using the monotonicity from below we have λ(R) = lim n→∞ λ([−n, n]) = lim n→∞ (arctan(n)− arctan(−n)) = π 2 − −π 2 = π <∞. (b). Define F : R→ [0,∞] by F (x) = λ((∞, x]). Then F (x) = lim n∞ λ([−n, x]) = arctan(x)− −π 2 . Thus F is a continuous non-decreasing function. Denoting by mF the corresponding Lebesgue- Stiltjes measure we have λ((a, b]) = arctan(b)− arctan(a) = F (b)− F (a) = mF ((a, b]). Since λ and mF assign the same measure to all intervals (a, b]. Part of Caratheodory theorem states that since R is σ-finite for mF , the Lebesgue-Stieltjes measure mF is the unique extension of the premeasure ρ((a, b]) = arctan(b)−arctan(a). Hence λ = mF where F (x) = arctan(x)−C for some/any constant C. 4. Let f, g : X → R be measurable functions on a measure space (X,M, µ). For n ∈ N, k ∈ Z define Ak,n = {x ∈ X | f(x) < k n }, Bk,n = {x ∈ X | g(x) < k n }. Prove that f(x) = g(x) for µ-a.e. x ∈ X if and only if ∞∑ n=1 ∞∑ k∈Z µ(Ak,n4Bk,n) = 0. Solution. ”if”. For any y < z there is n ∈ N and k ∈ Z with y < kn < z. Hence {x : f(x) 6= g(x)} = {x : f(x) < g(x)} ∪ {x : f(x) > g(x)} ⊂ ⋃ n∈N, k∈Z Ak,n \Bk,n ∪Bk,n \Ak,n. Hence, since the index set n ∈ N, k ∈ Z is countable, we have µ{x : f(x) 6= g(x)} ≤ µ  ⋃ n∈N, k∈Z Ak,n4Bk,n  ≤ ∞∑ n=1 ∞∑ k∈Z µ(Ak,n4Bk,n) = 0. Page 2