Composition of measureable function with continuou function in $L^2[0,1]$

Question

I was wondering, for general knowledge, if this claim is correct.

Let $X=L^2[0,1]$, $f\in X$, $g:[0,1]\to[0,1]$ invertible. Particularly, it's image is $[0,1]$ so everything is well defined. Specifically I'm interested in $\sqrt x$ so if you could also answer about this one specifically, but it is also interesting for general functions.

Is $f\circ g \in L^2[0,1]$?

Thanks so much

Answer 1

Consider $f(x)=\frac{1}{x^{\frac{1}{4}}}\in L^2$ and $g(x)=x^2\in C^0$. Then $g(f(x))\not\in L^2(0,1)$.

Let $g(x)=\sqrt{x}$, then $f(g(x))=f(\sqrt{x})$. We have

$$\int_0^1 f^2(\sqrt{x})dx=4\int_0^1 f^2(y)y^2dy.$$

Since $y^2\leq 1$ we have $\int_0^1 f^2(\sqrt{x})dx\leq4\int_0^1 f^2(y)dy<\infty$, because $f\in L^2$.