by perturbation
Last Updated January 11, 2019 12:20 PM - source

Let $f \in L^2((0,1);\mathbb{R})$ with respect to the Lebesgue measure. Let $g$ be a $C^1$-diffeomorphism from $(0,2)$ into its image $g((0,2)) \subset (0,1)$. Can I define the composition function $h:(0,2) \rightarrow \mathbb{R}$ by $$h(x)=f(g(x)),$$ for almost every $x \in (0,2)$ ? Is it enough to say that the measure of $g((0,2))$ is not zero ?

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