Finding Radon-Nikodym derivative $d\mu/dm$ where $m$ is the Lebesgue measure on $[0,1]$, $f(x)=x^2$, and $\mu(E)=m(f(E))$

Question

Consider the function $f:[0,1]\to \Bbb R$, $f(x)=x^2$. Let $m$ denote the Lebesgue measure on $[0,1]$ and define $\mu(E)=m(f(E))$. Since $f$ is absolutely continuous and nondecreasing, $f$ maps measure zero sets to measure zero sets, so $\mu$ is absolutely continuous with respect to $m$, so we can consider the Radon-Nikodym derivative $d\mu/dm$. Can we explicitly find $d\mu/dm$?

Answer 1

For interval $[a,b]\subseteq[0,1]$ we find:

$$\mu\left(\left[a,b\right]\right)=m\left(f\left(\left[a,b\right]\right)\right)=m\left(\left[a^{2},b^{2}\right]\right)=b^{2}-a^{2}=\left[x^{2}\right]_{a}^{b}=\int_{a}^{b}2xdx$$

Indicating that the derivative is $2x$.