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Unformatted text preview: Stat 310A/Math 230A Theory of Probability Homework 2 Solutions Andrea Montanari Due on 10/7/2010 Exercises on measurable functions and Lebesgue integration Exercise [1.2.14] The same method works for all four parts. 1. Since B = σ ( { (∞ ,α ] : α ∈ R } ), it follows from Theorem 1.2.11 that X is measurable with respect to the right hand side (RHS), which hence also contains the left hand side (LHS). But the RHS is generated by elements of the σalgebra on the LHS, so the LHS contains the RHS as well. 2. For 1 ≤ i ≤ n , each X i is by Theorem 1.2.11 measurable with respect to the RHS. Therefore, the RHS contains the LHS. Again, the RHS is generated by sets from the LHS, so the latter contains the former. 3. Exactly the same method applies. 4. Since each X k is measurable with respect to the RHS, the latter contains the LHS. By definition σ ( X k ,k ≤ n ) is contained in the LHS for each n , hence so is the union of these collections, implying that the LHS contains the RHS as well. Exercise [1.2.20] 1. If g is l.s.c. and x n is a sequence of points that converge to x and such that g ( x n ) ≤ a for all n then necessarily also g ( x ) ≤ liminf n g ( x n ) ≤ a . So, we see that { x : g ( x ) ≤ a } is closed. 2. From part (a) we see that g 1 ((∞ ,a ]) is a closed set for l.s.c. g and g 1 ((∞ ,a )) is an open set for u.s.c. g , hence both are in B R n . Since (∞ ,a ], a ∈ R generate B (see Exercise 1.1.17), as do (∞ ,a ), a ∈ R , it follows (from Theorem 1.2.11) that any such g is a Borel function. 3. Since continuous functions are also l.s.c. we use part (b) here. Exercise [1.2.30] Since { ω : limsup n →∞ X n ( ω ) ≤ X ∞ ( ω ) } = { ω : limsup n →∞ ( X n ( ω ) X ∞ ( ω )) ≤ } , it suffices to consider the case of X ∞ = 0. By the definition of limsup, { ω : limsup n →∞ X n ( ω...
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 '11
 MONTANARI,A.
 Probability, LHS, RHS, Lebesgue integration, lim sup xn

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