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Math 641 Lecture #2 ¶1.9,1.11,1.12 Consequences (1.9) (of Theorems 1.7 and 1.8). a) The function Φ : R2 → C defined by Φ(u, v) = u + iv is continuous, where we take the usual topology for C. So if u, v : X → R are measurable, then f (x) = Φ(u(x), v(x)) = u(x) + iv(x) is a complex measurable function on X. b) The functions Re : C → R defined by u + iv 7→ u, Im : C → R defined by u + iv 7→ √ v, and | · | : C → R defined by u + iv 7→ u2 + v 2 are continuous. For a complex measurable f : X → C defined by f = u + iv, the compositions Re(f ) = u, Im(f )√= v, and |f | = u2 + v 2 are measurable. c) The functions Φ : R2 → R defined by Φ(s, t) = s + t and Ψ : R2 → R defined by Ψ(s, t) = st are continuous. For complex measurable functions f, g on X, the compositions Φ(f (x), g(x)) = f (x) + g(x) = (f + g)(x) and Ψ(f (x), g(x)) = f (x)g(x) = (f g)(x) are complex measurable. d) The characteristic function of a measurable set E in X is defined by ( 1 if x ∈ E χE (x) = 0 if x ∈ /E It is a complex measurable function because for any open V in C, ∅ if 1 ∈ / V and 0 ∈ / V, E if 1 ∈ V but 0 ∈ / V, χ−1 E (V ) = {x ∈ X : χE (x) ∈ V } = X if 0, 1 ∈ V, C E if 0 ∈ V but 1 ∈ / V, is measurable in X. e) If f is a complex measurable function on X, then there is a complex measurable function α on X such that |α| = 1 and f = α|f |. Proof: Let Y = C \ {0}, and set ϕ(z) = z/|z| , z = u + iv. Let E = {x ∈ X : f (x) = 0}, and set α(x) = ϕ f (x) + χE (x) , x ∈ X. If x ∈ E, then α(x) = 0+1 = 1; |0 + 1| If x ∈ / E, then α(x) = f (x) + 0 f (x) = . |f (x) + 0| |f (x)| Thus |α(x)| = 1 for all x ∈ X, and ( 0 = α(x)|f (x)| if x ∈ E f (x) = α(x)|f (x)| if x ∈ / E. The function ϕ : Y → C is continuous, and the set E = f −1 ({0}) ! ∞ \ 1 = f −1 {z ∈ C : |z| < } n n=1 ∞ \ 1 −1 = f {z ∈ C : |z| < } n n=1 [open disks centered at origin with radius 1/n] [set theoretic property of preimages and intersections] is measurable. Thus, since the sum of two measurable functions is measurable, we have α(x) is measurable. Definition (1.11). Let τ be a topology on X. The smallest σ-algebra in X that contains τ is the Borel σ-algebra in X, and is denoted by B or BX . The members of B are the Borel sets of X. Consequences. a) Since each open set in X is a Borel set, the countable intersection of open sets is a Borel set and is called a Gδ . b) Each closed set in X is a Borel set because the complement of a closed set is open, and so the countable union of closed sets is a Borel set and is called an Fσ . Definition. Let X, Y be topological spaces. A function f : X → Y is a Borel function (or Borel measurable) if f −1 (V ) ∈ BX for all open V in Y. Observation. A continuous f : X → Y is Borel measurable because for every open V in Y , f −1 (V ) is open in X, hence f −1 (V ) ∈ BX . Lemma (1.12a). If M is a σ-algebra in X, and f : X → Y for a topological space Y , then Ω = {E ⊂ Y : f −1 (E) ∈ M} is a σ-algebra in Y . Proof: The set Y belongs to Ω because f −1 (Y ) = X ∈ M. For A ∈ Ω, f −1 (A) ∈ M. So AC ∈ Ω because f −1 (AC ) = [f −1 (A)]C ∈ M. For An ∈ Ω, n = 1, 2, 3, . . ., we have f −1 (An ) ∈ M, n = 1, 2, 3, . . .. −1 ∞ −1 (An ) ∈ M. (∪∞ So ∪∞ n=1 An ) = ∪n=1 f n=1 An ∈ Ω because f Thus, Ω is a σ-algebra in Y . Theorem (1.12b,c). Let M be a σ-algebra in X, let Y be a topological space, and let f : X → Y. i) If f is measurable and E is a Borel set in Y , then f −1 (E) ∈ M. ii) If Y = [−∞ , ∞] and f −1 (α , ∞] ∈ M for all α ∈ R, then f is measurable. Proof: i) Let Ω = {E ⊂ Y : f −1 (E) ∈ M} Measurability of f implies that Ω contains all the open sets in Y . (f −1 (V ) ∈ M for all open V in Y ) By the Lemma, Ω is a σ-algebra; hence BY ⊂ Ω. (BY is the smallest σ-algebra that contains the open sets in Y .) In other words, f −1 (E) ∈ M for every E ∈ BY . ii) Let Ω = {E ⊂ [−∞, ∞] : f −1 (E) ∈ M}. By the Lemma, Ω is a σ-algebra in [−∞, ∞]. Choose α ∈ R, and a sequence {αn } ∈ R such that αn < α and αn → α. Since f −1 (αn , ∞] ∈ M, it follows that [−∞, α) = ∞ [ [−∞, αn ] = n=1 ∞ [ (αn , ∞]C ∈ Ω n=1 because Ω is a σ-algebra and because (αn , ∞] ∈ Ω. Hence, (α, β) = [−∞, β) ∩ (α, ∞] ∈ Ω. Every open set in [−∞, ∞] is a countable union of intervals of the types (α, ∞], [−∞, β), and (α, β). Thus, B[−∞,∞] ⊂ Ω, implying that f is measurable, i.e, f −1 (V ) ∈ M for all open V in [−∞, ∞]. Corollary (1.12(d)). If f : X → Y is measurable and g : Y → Z is Borel measurable, then h = g ◦ f : X → Z is measurable. Proof: Let V be open in Z. Then g −1 (V ) is a Borel set in Y . f −1 g −1 (V ) , the Theorem (part i) shows that h−1 (V ) ∈ M. Since h−1 (V ) =