Is There a Faster Way to Prove Part (b) Using Radon-Nykodim?

[Bacle2]

Hi, I'm trying to work this out. Just a small question:

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

 

[micromass]

It's very easy to verify it the usual way:
1) First verify it for step functions
2) Use monotone convergence to verify it for positive functions
3) Verify it for general functions using $f=f^+-f^-$.

 

[Bacle2]

Right; thanks, but I don't know to what extent this simplifies the necessary

calculations in Radon-Nykodim. R-N works, but it just seems long. I thought maybe

there was some corollary to help give it a short proof. Thanks, tho.