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Unformatted text preview: Measure and Integration Solutions 1 Problem 1.1 Let C be a collection of subsets of the set Ω . For each A ∈ σ ( C ) show that there is a countable subcollection C A of C such that A ∈ σ ( C A ) . Solution Let D comprise all countable subcollections in C . Consider the union A of all σ ( D ) as D runs over D . Plainly, A contains Ω and is closed under complementation. Moreover, A is closed under countable unions: if A n ∈ A for n ∈ N then choose D n such that A n ∈ σ ( D n ) ; set D = ∪ n D n ∈ D (!) and note that each A n ∈ σ ( D ) . Thus, A is a σalgebra containing C and so σ ( C ) ⊂ A (the reverse inclusion is obvious). Now, if A ∈ σ ( C ) then A ∈ A and we are done. Problem 1.2 Let (Ω , F ) and (Ω , F ) be measurable spaces and f : Ω → Ω a function. (a) Prove that f 1 ( F ) is a σalgebra on Ω . (b) Show that f ( F ) need not be a σalgebra even if f is surjective. Solution Part (a) is entirely straightforward. Part (b): f ( F ) will be closed under countable unions and (when f is surjective) contain Ω but it need not be closed under complementation. Example: let Ω = Ω = [0 , 2] and let F = {∅ , [0 , 1) , [1 , 2] , [0 , 2] } ; let f : [0 , 2] → [0 , 2] map t ∈ [0 , 1) to 2 t and t ∈ [1 , 2] to 2 t 2 (sketch). In this case, f ( F ) = {∅ , [0 , 2) , [0 , 2] } contains [0 , 2) but not its complement { 2 } . 2 Problem 2.1 Find a measurable space (Ω , F ) carrying finite measures μ and ν such that μ (Ω) = ν (Ω) but such that { A ∈ F : μ ( A ) = ν ( A ) } is not a σalgebra. 1 Solution At first sight, this seems false: plainly, the indicated class G contains Ω and is closed under complementation; however, G need not be closed under (finite) union. Example: let Ω = { p, q, r, s } with μ = δ p + δ q and ν = δ r + δ s on F = P (Ω) . Then A = { p, r } ∈ G and B = { q, r } ∈ G but A ∪ B = { p, q, r } / ∈ G ....
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 Fall '09
 Robinson
 µ, Topological space, measure, Lebesgue integration, measurable space

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