Is this true: $\mathbb{E}(f(X)g(Y)) = \mathbb{E}(f(X))\mathbb{E}(g(Y))$ if $X,Y$ are independent? 
Is this true: $\mathbb{E}(f(X)g(Y)) = \mathbb{E}(f(X))\mathbb{E}(g(Y))$ if $X,Y$ are independent?

I ask because I noticed that
$$\mathbb{E}(X^2 Y) = \iint_{\mathbb{R}^2} x^2 y f_{X,Y}(x,y)dydx = \int_{\mathbb{R}} x^2 f_{X}(x) dx \int_{\mathbb{R}}yf_{Y}(y)dy$$
by Fubini's Theorem and independence of $X,Y$, which equals
$$\mathbb{E}(X^2)\mathbb{E}(Y).$$
 A: Yes. Provided that $f(X)g(Y)$, $f(X)$, and $g(Y)$ are integrable. It is because $f(X)$ and $g(Y)$ are independent.
A: Yes, and basically for the reason you have shown.   But let's use $h$ and $g$ to avoid confusion with the probability density functions.
So because independence of $X,Y$ means $f_{X,Y}(x,y)=f_X(x)~f_Y(y)$ , therefore:
$${\mathsf E(h(X)\,g(Y)) ~{= \iint_{\Bbb R^2} h(x)\,g(x)\,f_{X}(x)\,f_Y(y)~\mathrm d x~\mathrm d y \\ = \int_\Bbb R h(x)\,f_X(x)\mathrm d x\cdot\int_\Bbb R g(y)\,f_Y(y)\mathrm d y \\ = \mathsf E(h(X))\;\mathsf E(g(Y))}\\\blacksquare}$$
Further, it can be shown that as long as the expectations have finite existance, this is true for any independent random variales (not just continuus ones).  $~\mathsf E(h(X)~g(Y))=\mathsf E(h(X))~\mathsf E(g(Y))$.

Intuitively this is sensible, because when knowing anything about the value of $Y$ tells us nothing about $X$, then likewise knowing anything about a function of $Y$ will tell us nothing about a function of $X$.  Ie $h(X)$ and $g(Y)$ shall be independent whenever $X$ and $Y$ are.
