# The composition of a L^2 function and a diffeomorphism is well-defined?

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 ?

• 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. – Stockfish Jan 11 at 12:40
• 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 ? – perturbation Jan 11 at 12:54
• 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. – Mindlack Jan 11 at 12:54
• @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). – perturbation Jan 11 at 13:25
• 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. – Mindlack Jan 11 at 13:32