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In the above theorem, doesn't $\phi$ need to be injective too? The inverse function theorem merely implies that $\phi$ is locally injective -- is this sufficient? I ask because Marsden, in his Elementary Classical Analysis, actually does stipulate the condition that $\phi$ be (globally) injective.

p.s. In the above, "a set has volume" is equivalent to "the boundary of the set has measure zero."

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Yes, the theorem you quoted is missing an injectivity assumption. It is possible to change variables by a non-injective map, but then one must account for multiple preimages in the integral on the right (multiply $f(x)$ by the number of preimages of $x$ under $\phi$).

As is, the following is a counterexample: let $n=2$, let $A$ be the annulus described in polar coordinates $(r,\theta)$ as $1<r<2$, and let $\phi(r,\theta)=(r,2\theta)$ be the "winding" map. Then $\phi(A)=A$, and $J\phi\equiv 2$. We can set $f\equiv 1$ and obtain a contradiction: $$ \int_A 2 = \int_A 1 $$

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