Transformation of lebesgue integral

  • #1
define F(x)=x, then uF is the lebesgue stiljes measure
duF=dx
Let y=x^2
we all know that how to transform [tex]\int f(y)dy [/tex] into [tex]\int f(x^2)2xdx [/tex] (***)
But how exactly would one use the transformation theorem ?
Ie. T be a measurable transformation from X to Y, u is a measure on X
[tex]\int_{Y}fduT^{-1}=\int_{X}fTdu[/tex]

I want to see the transformation theorem in action otherwise it's too abstract for me to understand. My question is, how would one use the transformation theorem to obtain the same result as equation (***)?
 

Answers and Replies

  • #2
morphism
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You should ask yourself the following question: what are X, Y and T in this case?
 
  • #3
X is R
Y is R
T is sqrt(y)
T-1 is y^2

then?
[tex]\int_{Y}fduT^{-1}=\int_{X}fdu y^2 ? [/tex] doesn't make a lot of sense

what is the explicit expression of duT^-1 in this example?
 
  • #4
morphism
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You need a Radon-Nikodym derivative in there. We have,

[tex]\int g(T(x)) d\mu(x) = \int g(y) \frac{d(\mu T^{-1})}{d\mu}(y) \: d\mu(y).[/tex]
 
  • #5
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You need a Radon-Nikodym derivative in there. We have,

[tex]\int g(T(x)) d\mu(x) = \int g(y) \frac{d(\mu T^{-1})}{d\mu}(y) \: d\mu(y).[/tex]

I wonder whether someone could write the next step on the solution of this problem. I must confess that I didn't figure out how to use Radon-Nikodym derivative in this case.

Many thanks
 

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