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Define the transformation of a measure $v(X)$ for $X \subseteq {\mathbb R}^d$ by function $A: {\mathbb R}^d \rightarrow {\mathbb R}^d$ as $$ A(v)(X) = v(A^{-1}(X)) $$

Then the density function $v(x)$ for $x \in {\mathbb R}^d$ is not transformed in the same way (except in certain special cases).

Now I would guess that if A is an affine function, that we can say

$$ A(v)(x) = \frac{1}{\mbox{det}|A|} v(A^{-1}(x)) $$

Unfortunately I lack the knowledge of measure theory (if that is what is required, maybe it's just calculus?) to work out a proof, and Google and Wikipedia have not been very helpful.

My main question is whether this is right, and if not, what the correct equation is. Secondly, what would the proof look like, and would there be some book on some subject that actually discusses it in this form?

PS. I'm no expert in measure theory. If it helps, the measures can be simplified to whatever category of well behaved measures are available. For my purposes they only need to be probability measures on ${\mathbb R}^d$. Perhaps it is required to approximate a uniform measure for sufficiently small neighbourhoods?

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up vote 1 down vote accepted

By definition, the image measure $A(v)$ satisfies, for bounded measurable $g$: $$\int g(x) A(v)(dx)=\int g(Ax) v(dx)=\int g(Ax) v(x)\,dx.\tag1$$ Here I assume that the measure $v$ admits a density (called $v(x)$) with respect to Lebesgue measure.

Now perform the change of variables $w=A(x)$. The inverse map is $x=A^{-1}(w)$ and $dx=|A^{-1}| dw$. Substituting into (1), we get $$\int g(x) A(v)(dx)=\int g(w) v(A^{-1}w) |A^{-1}|\,dw.$$ Thus the density of the measure $A(v)$ with respect to Lebesgue measure is $v(A^{-1}w) |A^{-1}|$, as you expected.

We assumed that $A$ is invertible to make this work.

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