Problems and Solutions in - Mathematics - University of Hawaii

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1.5 2001 November 26 1 REAL ANALYSIS Now, by definition, Tnx → T x, for all x ∈ X and, since the norm · Y is uniformly continuous, 27 TnxY → T xY (∀x ∈ X). (17) Taken together, (16) and (17) imply T xY ≤ (c + 1)xX, for all x ∈ X. Therefore, T is bounded. • limn→∞ Tn − T = 0: Fix ɛ > 0 and choose N ∈ N such that n, m ≥ N implies Tn − Tm < ɛ. Then, Tnx − Tmx ≤ Tn − TmxX < ɛxX holds for all n, m ≥ N, and x ∈ X. Letting m go to infinity, then, Tnx − T x = lim m→∞ Tnx − Tmx ≤ ɛxX. That is, Tnx − T x ≤ ɛxX, for all n ≥ N and x ∈ X. Whence, Tn − T ≤ ɛ for all n ≥ N. 27 Proof: | aY − bY | ≤ a − bY (∀a, b ∈ Y ). 26 ⊓⊔
1.6 2004 April 19 1 REAL ANALYSIS 1.6 2004 April 19 Instructions. Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well written solutions will count more than several partial solutions. Notation: f ∈ C(X) means that f is a real-valued, continuous function defined on X. 1. (a) Let S be a (Lebesgue) measurable subset of R and let f, g : S → R be measurable functions. Prove that (i) f + g is measurable and (ii) if φ ∈ C(R), then φ(f) is measurable. (b) Let f : [a, b] → [−∞, ∞] be a measurable function. Suppose that f takes the value ±∞ only on a set of (Lebesgue) measure zero. Prove that for any ɛ > 0 there is a positive number M such that |f| ≤ M, except on a set of measure less than ɛ. Solution: (a) Proof 1: Since f and g are real measurable functions of S, and since the mapping Φ : R × R → R defined by Φ(x, y) = x + y is continuous, theorem A.3 implies that the function f + g = Φ(f, g) is measurable. If φ ∈ C(R), then φ(f) is measurable by part (b) of theorem A.2. Proof 2: Let {qi} ∞ i=1 be an enumeration of the rationals. Then, for any α ∈ R, {x ∈ S : f(x) + g(x) < α} = ∞ {x ∈ S : f(x) < α − qi} ∩ {x ∈ S : g(x) < qi}. i=1 Since each set on the right is measurable, and since σ-algebras are closed under countable unions and intersections, {x ∈ S : f(x) + g(x) < α} is measurable. Since α was arbitrary, f + g is measurable. The function φf is measurable if and only if, for any open subset U of R, the set (φf) −1 (U) is measurable. Let U be open in R. Then φ −1 (U) is open, since φ ∈ C(R), and so (φf) −1 (U) = f −1 (φ −1 (U)) is measurable, since f is measurable. Therefore, φf is measurable. (b) Fix ɛ > 0. For n ∈ N, define An = {x ∈ [a, b] : |f(x)| ≤ n}. Then [a, b] = ∞ An ∪ A∞, (18) n=1 where 28 A∞ = {x : f(x) = ±∞}. Also, A1 ⊆ A2 ⊆ · · · and, since f is measurable, each An is measurable. Therefore, µ(An) ↑ µ(∪nAn), as n → ∞. Note that all sets are contained in [a, b] and thus have finite measure. Let M ∈ N be such that µ(∪nAn) − µ(AM ) < ɛ. Then |f| ≤ M except on [a, b] \ AM , and by (18), µ([a, b] \ AM ) = µ(∪nAn ∪ A∞ \ AM ) ≤ µ(∪nAn \ AM ) + µ(A∞) = µ(∪nAn \ AM ) < ɛ. The second equality holds since we assumed f(x) = ±∞ only on a set of measure zero; i.e., µ(A∞) = 0. ⊓⊔ 28 Since f is an extended real valued function, we must not forget to include A∞, without which the union in (18) would not be all of [a, b]. 27
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A.2 Complex Analysis A MISCELLANEOU
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B List of Symbols F an arbitrary fi
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INDEX INDEX Riemann mapping theorem
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Problems and Solutions in - Mathematics - University of Hawaii

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