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 ?

  • $\begingroup$ Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory. $\endgroup$ – Stockfish Jan 11 at 12:40
  • $\begingroup$ Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ? $\endgroup$ – perturbation Jan 11 at 12:54
  • 1
    $\begingroup$ Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of $\{f_1 \neq f_2\}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity. $\endgroup$ – Mindlack Jan 11 at 12:54
  • $\begingroup$ @Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english). $\endgroup$ – perturbation Jan 11 at 13:25
  • 1
    $\begingroup$ By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=\int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’\circ g^{-1}|$, and $1/|g’\circ g^{-1}|$ can be unbounded. $\endgroup$ – Mindlack Jan 11 at 13:32

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.