Hi, I'm trying to work this out. Just a small question:(adsbygoogle = window.adsbygoogle || []).push({});

Let (X, m, μ) be a measure-triple (X a set; m a sigma-algebra, μ

a measure) , and let,

f:X-->ℝ be measurable. For B a Borel set, define :

v(B):=μ(f-1(B)). Then show:

a)v is a measure in the Borel subsets of R , and

b)If g: ℝ-->[0 , oo) is any Borel-measurable function on R,

∫ℝ (gdv) = ∫X (gof)dμ

First one is easy, but my method for b seems too long b). I

am trying to work this with Radon-Nykodim theorem. I'm pretty sure we can pull back

the measure on R to have the absolute continuity condition satisfied, and same thing

for the sigma-finiteness, i.e., by pulling back the Borel measure on R.

I'm just curious as to whether there is a faster way of doing this problem.

TIA

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# Simplifying Radon-Nykodim App?

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