Fubini's theorem and uncorrelated/independent random variables Suppose $X$ and $Y$ are two real-valued random variables, and $f:\mathbb{R}^2\to \mathbb{R}$ is Borel measurable.
I was wondering if $X$ and $Y$ being uncorrelated or independent implies that
$$
\mathrm{E}_{X,Y} f(X,Y) = \mathrm{E}_X [\mathrm{E}_Y f(X,Y)]  = \mathrm{E}_Y [\mathrm{E}_X f(X,Y)]?
$$
Thanks and regards!
 A: Independence is equivalent to the measure on the product space being the product measure.  This is actually just a restatement of the definitions:
$\mu_{X_1,...,X_n}(E_1,..,E_n) = P((X_1,...,X_n) \in (E_1,...,E_n) = \prod P(X_i \in E_i) = \prod \mu_i (E_i) = \mu_1 \times ... \times \mu_n (E_1 \times ... \times E_n)$
This is sufficient for Fubini's Theorem (modulo existence of integrals).
edit: Looking back at this, I suppose I should have been clearer since I may have misinterpreted your question.  This is unambiguously true for product measures and could be true depending on what you mean by $E_X$ and $E_Y$ more generally.  Let's just move down to the absolutely continuous case to make this a bit clearer:  In general, we have $f_{X,Y}(x,y) = f_{X|Y}(x|y) f_Y (y)$ (this does generalize beyond the absolutely continuous case, but the notation is cumbersome).  If your question does not involve conditioning then find any random variable for which the conditional probability $f_{X|Y} (x|y)$ does not equal the marginal $f_X (x)$.  This is how I interpreted your original question, but that is probably not what you meant.
In light of your comment I will give a standard counterexample.  Let $B = 1$ or $0$ each with probability $\frac{1}{2}$ and $D = 1$ or $-1$ each with probability $\frac{1}{2}$.  Then $A = BD$ is uncorrelated with $B$, but $E A^2 B$ = $\frac{1}{2} \neq E_A E_B A^2 B = E_A A^2 E_B B = \frac{1}{4}$ where I am interpreting $E_B$ to mean the integral with respect to the marginal distribution of $B$.
