We often encounter absolutely continuous RV's with PDF and are interested in the PDF of a function of the RV. At this point, textbooks usually provide the distribution function technique, which is all fine. But given an absolutely continuous RV $X$ what conditions in general must be obeyed by a Borel measurable $g:\mathbb R\to \mathbb R$, so that $g\circ X$ is absolutely continuous and the method can be applied in the first place?
I am aware of the conditions given in this question, but I will prefer general necessary and sufficient conditions and a reference to the proof (or proof sketch).
An RV is defined to be absolutely continuous if it's distribution function is absolutely continuous.
Best Answer
I will assume that a $\mathbb{R}^n$-valued random variable $Y$ being absolutely continuous means that the measure $P_Y$ on $\mathcal{B}(\mathbb{R}^n)$ defined by $P_Y(A) = P(Y \in A)$ is absolutely continuous wrt Lebesgue measure, i.e. has a density $f_Y : \mathbb{R}^n \to [0, \infty)$. I think this is equivalent to your definition when $n = 1$.
Assume $X$ is $\mathbb{R}^n$-valued and absolutely continuous and $g : \mathbb{R}^n \to \mathbb{R}^n$ is Borel measurable. One strategy to see when $g(X)$ is absolutley continuous is to appeal to the Radon-Nikodym theorem. Suppose $N \subset \mathbb{R}$ and $m(N) = 0$. Then $P(g(X) \in N) = P(X \in g^{-1}(N))$. Since $X$ is absolutely continuous, this will be $0$ provided that $m(g^{-1}(N)) = 0$. So if $g$ preserves null sets, then we can conclude that $g(X)$ is absolutely continuous. This is a pretty general criterion.
A less general criterion, but one that actually gives you a formula for the density of $Y = g(X)$ is to use the change of variables theorem. Assume that $U \subset \mathbb{R}^n$ is open and $P(X \in U) = 1$. Also assume only that $g : U \to \mathbb{R}^n$ and that $g$ is injective. We have $P(g(X) \in g(U)) = 1$. For the moment, let us assume that $g(U)$ is open and that $g : U \to g(U)$ is a $C^1$ diffeomorphism. For $A \in \mathcal{B}(g(U))$, $$P(g(X) \in A) = P(X \in g^{-1}(A)) = \int_{g^{-1}(A)}f_X(x)\,dx.$$ Use the change of variables theorem to make the substitution $$x = g^{-1}(y),$$ $$dx = J(y)\,dy,$$ where $J(y) = |\det Dg^{-1}(y)|$. We get $$\int_{g^{-1}(A)}f_X(x)\,dx = \int_{A}f_X(g^{-1}(y))J(y)\,dy.$$ Hence for $y \in g(U)$, $$f_{Y}(y) = f_X(x)J(y),\hspace{20pt}x = g^{-1}(y).$$ Now the question is how much can we weaken the regularity of $g$ and of $U$. We need $g^{-1}$ to be Lipschitz and $g(U) \in \mathcal{B}(\mathbb{R}^n)$ in order satisfy the hypotheses of the change of variables formula. Hence it is enough that $g^{-1}$ is Lipschitz and that $g(U) \in \mathcal{B}(\mathbb{R}^n)$.
In practice though, $g$ is usually not injective, but rather $U$ can be partitioned into disjoint Borel sets $U_1, \dots, U_N$ such that $g : U_i \to g(U)$ is bijective with Lipschitz inverse. The change of variables theorem can be used to get the formula for $f_Y$ in this case using a similar analysis as before. We get that for $y \in g(U)$, $$f_Y(y) = \sum_{i = 1}^{N}f_X(x_i)J_i(y),\hspace{20pt}x_i = g|_{U_i}^{-1}(y), J_i = |\det D(g|_{U_i}^{-1})|.$$
This is the situation when $g(x) = x^2$, with $U_1 = (-\infty, 0)$, $U_2 = (0, \infty)$. It is also the situation when $g(x_1, \dots, x_n) = (x_{(1)}, \dots, x_{(n)})$ is the order statistics function, with $n!$ different $U_i$s.
References: For the change of variable formula for Lipschitz functions, see THEOREM 3.8 (Area Formula) of "Measure Theory and Fine Properties of Functions" by Evans and Gariepy. For the (much simpler) proof of the version for $C^1$ diffeomorphisms, see Folland's book. For a general reference on probability theory and measure theory, see "Probability Theory: A Comprehensive Course" by Klenke.