No2(2004)solutions

No2(2004)solutions - MATH 587 Assignment 2 Solutions...

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Unformatted text preview: MATH 587 Assignment 2 Solutions November 1, 2004 (1) If B B ( R ), then h 1 ( B ) = [ h 1 ( B ) A ] [ h 1 ( B ) A c ] = [ f 1 ( B ) A ] [ g 1 ( B ) A c ] F , so h is measurable. (2) Let B = { B R n | x + B B ( R n ) } . Since x + = and x + R n = R n , then , R n B . If B B , then x + B c = ( x + B ) c B ( R n ) so B c B . If B 1 , B 2 , . . . B , then x + i =1 B i = i =1 ( x + B i ) B ( R n ), so i =1 B i B . Hence B is a -algebra of subsets of R n . Since x + ( c, d ) = ( x + c, x + d ) B ( R n ), then ( c, d ) B for all intervals ( c, d ) in R n . Hence B B ( R n ). Next, let B = { B R n : cB B ( R n ) } . As in the previous paragraph, B B ( R n ). (3) We shall use the following fact, which is easily deduced from a result in class: if and are two measures on ( F ) which agree on the algebra F and if is -finite with respect to F , then and agree on ( F )....
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No2(2004)solutions - MATH 587 Assignment 2 Solutions...

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