In advanced probability we can just say:
\begin{align} & P(h(X) \in A, g(Y) \in B) \\[6pt] = {} & P(X \in h^{-1}(A), Y \in g^{-1}(B)) \\[6pt] = {} & P(X \in h^{-1}(A)) P(Y \in g^{-1}(B)) \\[6pt] = {}& P(h(X) \in A) P(g(Y) \in B) \end{align}
where $X,Y \in (\Omega, \mathscr F, \mathbb P)$ and are $(\mathbb R, \mathscr B(\mathbb R))$-valued, $A,B$ are Borel and $h,g$ are some kinds of functions (Borel? Bounded and Borel? $\mathscr F$-measurable?)
But what about in basic probability? Is there a double integral way to approach this?
The hint given is to show that
$$P(h(X) \in (-\infty, a], g(Y) \in (-\infty, b])$$
$$ = P(h(X) \in (-\infty, a]) P(g(Y) \in (-\infty, b])$$
I can sort of use the same argument as in advanced probability but then I would have to use
$$h^{-1}(-\infty, a]$$
Is that acceptable to use in basic probability?
I think
$$h^{-1}(-\infty, a] = \{x \mid h(x) \in (-\infty, a]\}$$
So $$X \in h^{-1}(-\infty, a] = \operatorname{Range}(X) \cap \{x \mid h(x) \in (-\infty, a]\} = \{\omega \in \Omega \mid X(\omega) \in \{x \mid h(x) \in (-\infty, a]\}\}$$
Regardless, I think $X$ and $Y$ are absolutely continuous continuous with pdfs.
I think there should be a double integral way to approach this.
Edit 4.5 years later: by the way, what's the condition for $h$ and $g$ to do this please? See here: Prove that for independent random variables $X_i$, we have $f_i(X_i)$ are independent.