MI_Solutions - Measure and Integration Solutions 1 Problem...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 σ-algebra 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 σ-algebra on Ω . (b) Show that f ( F ) need not be a σ-algebra 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 σ-algebra. 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 ....
View Full Document

This note was uploaded on 11/09/2011 for the course MAA 5228 taught by Professor Robinson during the Fall '09 term at University of Florida.

Page1 / 5

MI_Solutions - Measure and Integration Solutions 1 Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online