Show that if $X$ is absolutely continuous and $g$ is absolutely continuous on bounded intervals, then $g(X)$ is absolutely continuous.

Question

I'm taking a course on probability, and the following question showed up in the reference textbook:

Show that if $X$ is absolutely continuous with pdf $f_X(\cdot)$ and $g$ is absolutely continuous on bounded intervals such that $g'(\cdot)>0 \ a.e \ (\lambda)$, then $Y = g(X)$ is also absolutely continuous with pdf $$ f_Y(y) = \frac{f_X(g^{-1}(y))}{g'(g^{-1}(y))} $$

I'm quite lost on how to solve this. It would be much help with someone could show how to do it.

Note: the $\lambda$ stands for the Lebesgue measure, $g:\mathbb R \rightarrow \mathbb R$ is Borel measurable, $X$ is a random variable with probability space $(\Omega, \mathcal F, P)$.

Answer 1

Maybe a hint: Since $g$ is invertible under these assumptions,

$ F_Y(t) = P( Y \le t) = P(X \le g^{-1}(t)). $

Calculate the derivative and in the chain rule apply the inverse function theorem of calculus.