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Suppose $(X,\mathcal A, \mu)$ is a measure space and $g:X \to \mathbb R$ a measurable function. Then $\mu\circ g^{-1}$ is a measure on $\mathbb R$. Let $f:\mathbb R \to \mathbb R$ be a measurable function.

I would like some help with the following

Question: Can I write $$\int_\mathbb{R} f d(\mu\circ g^{-1})=\int_{X} f\circ g \ d\mu \text{?}$$

If so, do I need any other conditions on $g$ besides measurability? This identity falls under which theorem? Is this an example of change of variables or stands alone as a separate fact?

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$f\circ g$ is not defined for $f,g:X\to\Bbb R$ – Ilya Mar 21 '13 at 16:34
Thanks Ilya, now it should make sense. – Cantor Mar 21 '13 at 16:42
Well, it certainly holds e.g. for all positive $f$ and all positive measurable $g$ (to assure that $\mu\circ g^{-1}$ is a positive measure). You can show this by approximating $f$ with simple functions. Are these droids you are looking for? – Ilya Mar 21 '13 at 16:44
@Ilya: $g$ doesn't need to be positive in order for $\mu\circ g^{-1}$ to be a measure. – Stefan Hansen Mar 21 '13 at 16:45
@Cantor This is the change of variables theorem. – Byron Schmuland Mar 21 '13 at 17:00
up vote 1 down vote accepted

You just need $g$ to be measurable, since this will ensure that $\mu\circ g^{-1}$ is a well-defined measure. In order to conclude the equality you need $f$ to be a measurable function such that both integrals exist. A dicussion of when they exist is given e.g. here.

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I was looking for the linked answer of yours, btw, to refer the OP, and didn't find it. Not that I remember it was yours – Ilya Mar 21 '13 at 16:57

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