MATH 525a SAMPLE FINAL EXAM SOLUTIONS FALL 2013 Prof. Alexander (1)(a) Clearly φ, X ∈ M. Now E ∈ M implies either (i) A ⊂ E or (ii) E ∩ A = φ which implies either (i) A ∩ E c = φ or (ii) A ⊂ E c . Either way this shows E c ∈ M. Suppose E1 , E2 , ... ∈ M. If all Ei ∩ A = φ, then (∪i Ei ) ∩ A = φ. If instead some Ej ⊃ A then (∪i Ei ) ⊃ A. Either way, ∪i Ei ∈ M. (b) f is measurable if and only if f is constant on A. Proof: If f = c on A, then for every D ⊂ R, A ⊂ f −1 (D) if c ∈ D, and A ∩ f −1 (D) = φ if c ∈ / D, so f is measurable. Conversely if f takes any two distinct values c1 , c2 somewhere on A then f −1 ({c1 }) is not measurable so f is not measurable. (2)(a) Let > 0. Then {x : |f (x)| ≥ |g(x)| + } ⊂ {x : |f (x) − fn (x)| ≥ } for all n so µ({x : |f (x)| ≥ |g(x)| + }) ≤ µ({x : |f (x) − fn (x)| ≥ }) → 0 as n → ∞. Hence µ({x : |f (x)| ≥ |g(x)| + }) = 0 for all . Take = 1/k, let k → ∞ to get µ({x : |f (x)| > |g(x)|}) = 0. (b) We have (fn − f )2 → 0 in measure. Using (a) we have (fn − f )2 ≤ (2g)2 = 4g 2 , and 2 4g ∈ L1 , so the result follows from Exercise 34b in Chapter 2. Pn |F (xj ) − F (xj−1 )| ≥ TF (x) − . Then (3) Choose −∞ P < x0 < ... < xn = x with 1P TF (x) − ≤ limk n1 |Fk (xj ) − Fk (xj−1 )| = lim inf k n1 |Fk (xj ) − Fk (xj−1 )| ≤ lim inf k TFk (x). Since is arbitrary, this shows TF (c) ≤ lim inf TFk (x). (4), (5) See Ch. 3 #12, and (II), in Assignment 8 solutions. (6)(a) There are 3 things to establish, as follows. (i) χφ = 0 so ν(φ) = φ(0) = 0 by linearity. (ii) |ν(E)| ≤ ||φ|| ||χE ||1 = ||φ||µ(E)P< ∞, forPall E. P∞ P∞ (iii) For E1 , E2 , ... disjoint we have || n1 χEj − ∞ n+1 µ(Ej ) → n+1 χEj ||1 = P 1 χEj ||1 = || 0 as n P → ∞, since µP is finite. Hence since φ is continuous, ν(∪j Ej ) = φ( ∞ 1 χEj ) = P ν(E ). limn φ( n1 χEj ) = limn n1 ν(Ej ) = ∞ j 1 Together, (i) – (iii) show ν is a signed measure. (b) If E is µ-null then χE = 0 as an element of L1 , so ν(E) = φ(χE ) = φ(0) = 0. (c) Proof of hint: ⇒ is clear. Conversely, if dν/dµ > M on a set E of positive µ-measure then the second inequality R R dν in the Rhint is false for that E. This proves the hint. Now for every E, E dµ dµ = E dν = ν(E) ≤ ||φ|| ||χE ||1 = ||φ||µ(E) = E ||φ|| dµ. By the hint this means dν/dµ ≤ ||φ|| a.e.(µ). ≥ −||φ|| a.e.(µ). R dνSimilarlyR dν/dµ dν (d) For all E, φ(χE ) = ν(E) = E dµ dµ = χE dµ dµ so the statement is true for characteristic functions. By linearity it is true for simple functions. For general f take 1 simple fn → f pointwise Rwith |fn | % |f |. By Dominated Convergence, (µ) so R dν R dν Rfn →dνf in L dν dν φ(f ) = limn φ(fn ) = limn fn dµ dµ (*). Now | fn dµ dµ − f dµ dµ| ≤ |fn dµ − f dµ | dµ ≤ R R dν ||φ|| |fn − f | dµ → 0 so the limit (*) is f dµ dµ.