I really could use a hand on this question Let f : <mrow class="MJX-TeXAtom-ORD"> <

Sylvia Byrd

Sylvia Byrd

Answered question

2022-07-07

I really could use a hand on this question
Let f : R f ( R ) := E, a continuous function. Let μ e be a probability on ( E , B ( E ) ).
Show that there exists a measure ν on R such that :
B B ( E ) , μ e ( B ) = ν ( f 1 ( B ) )
I really can't think of a single way to approache this. The obvious choice would seem to be ν ( B ) = μ e ( f ( B ) ), but this doesn't seem to work. Another idea would be to construct an outer measure with one of the methods available but I don't know how I would go about it
I thank you all in advance

Answer & Explanation

persstemc1

persstemc1

Beginner2022-07-08Added 18 answers

Hint:
E is an interval (possibly a single point) since f is continuous. Let A = { f 1 ( B ) : B B ( E ) }. This is a σ-algebra, define ν ( f 1 ( B ) ) = μ ( B ).
ν is well defined since f : R E is surjective; thus, if f 1 ( B ) = f 1 ( B ), then B = f ( f 1 ( B ) ) = f ( f 1 ( B ) ) = B .
If { f 1 ( B n ) : n N } are disjoint, then { B n : n N } are disjoint since f is surjective. Thus, if the { B n : n N } B ( E ),
ν ( n f 1 ( B n ) ) = ν ( f 1 ( n B n ) ) = μ ( n B n ) = n μ ( B n ) = n ν ( f 1 ( B n ) )

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