- #1
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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
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