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 σ-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 µ and µ are two measures on σ ( F ) which agree on the algebra F and if µ is σ-finite with respect to F , then µ and µ agree on σ ( F )....
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## This note was uploaded on 01/25/2011 for the course STAT 235a at Stanford.

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

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