If $X\sim f$, what is the variable $f(X)$? For some real random variables $X$ with density $f\in C^\infty(\mathbb R)$, I am looking to find the distribution of the random variable $f(X)$.
I found that
$$\mathbb P(t< f(X)\le t+h)=\mathbb E[1_{(t,t+h]}(f(X))]=\int_{-\infty}^\infty  1_{(t,t+h]}(f(x))f(x)\ dx$$
$$\approx f(t) \int_{-\infty}^\infty  1_{(t,t+h]}(f(x))\ dx$$
But now I don't know how to proceed.
Is there a way the distribution of $f(X)$ in a simple form?
Is there someone who has already studied this type of variables?
 A: From what you have already found, the probability density would then be $$\lim_{h \to 0^+}\mathbb{P}\left( t < f(X) \le t+h \right) = \lim_{h \to 0^+}\frac{1}{h}\int_{-\infty}^\infty  1_{(t,t+h]}(f(x))f(x)\ dx$$
Now because $h$ is approaching $0$, consider it to be small enough that each region such that $t < f(x) \le t+h$ is disjoint and that $f(x)$ is monotone (the second condition will be true for all $x$ where $f(x)$ isn't a local extrema or constant). Let $x_n$ be the $n$th value of $x$ such that $f(x) = t$ and $f^{-1}_n(x)$ be the inverse of $f(x)$ near $x_n$. Then the limit of the integral is $$\lim_{h \to 0^+}\frac{1}{h}\left( \sum_{n} \left| \int_{f_n^{-1}(t)}^{f_n^{-1}(t+h)}f(x)dx \right| \right) = \sum_{n}\lim_{h \to 0^+}\frac{1}{h}\left(  \left| \int_{f_n^{-1}(t)}^{f_n^{-1}(t+h)}f(x)dx \right| \right)$$
This evaluates to $$\sum_n \lim_{h \to 0^+} \frac{\left| F\left(f_n^{-1}(t+h)\right)-F\left(f_n^{-1}(t)\right) \right|}{h}$$
Then using L'Hopital's, this becomes $$\sum_n \lim_{h \to 0^+} \left|f(f_n^{-1}(t+h))f_n^{-1}{'}(t+h)\right|$$
This simplifies to $$\sum_n \left|\frac{t}{f'(x_n)}\right| = |t|\sum_n \left|\frac{1}{f'(x_n)}\right|$$
