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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 . For each A ( C ) show that there is a countable subcollection C A of C such that A ( C A ) . Solution Let D comprise all countable subcollections in C . Consider the union A of all ( D ) as D runs over D . Plainly, A contains 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 ( D n ) ; set D = n D n D (!) and note that each A n ( D ) . Thus, A is a algebra containing C and so ( C ) A (the reverse inclusion is obvious). Now, if A ( C ) then A A and we are done. Problem 1.2 Let ( , F ) and ( , F ) be measurable spaces and f : a function. (a) Prove that f 1 ( F ) is a algebra on . (b) Show that f ( F ) need not be a 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 but it need not be closed under complementation. Example: let = = [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 ( , F ) carrying finite measures and such that () = () but such that { A F : ( A ) = ( A ) } is not a algebra. 1 Solution At first sight, this seems false: plainly, the indicated class G contains and is closed under complementation; however, G need not be closed under (finite) union. Example: let = { p, q, r, s } with = p + q and = r + s on F = P () . Then A = { p, r } G and B = { q, r } G but A B = { p, q, r } / G ....
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 Fall '09
 Robinson

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