Suppose $X$ and $Y$ are independent random variables with $E(X)=0$, $E(Y)=0$. Does this already imply that $E(XY)=0$? Often when we talk about independence implies that the covariance is zero, we assume that $E(X^2)<\infty$ and $E(Y^2) < \infty$. In this case it is clear that $E(|XY|)$ is well defined, and then an elementary calculation gives $E(XY)=0$. It seems though that when the variables are independent, we do not need second order moments to make this conclusion. The way I see this is as follows:
Proof: By Tonnelli's Theorem,
$E(XY) = E( [X^+ - X^-][Y^+-Y^-])=E(X^+Y^+)-E(X^+Y^-)-E(X^-Y^+)+E(X^-Y^-)=E(X^+)E(Y^+)-E(X^+)E(Y^-)-E(X^-)E(Y^+)+E(X^-)E(Y^-) = E(X)E(Y). $
Is this overkill?
First, note that $\mathbb E[XY]$ is not defined, as long as you do not know that $XY\in L^1$. That means, that you first need to prove that $\mathbb E[\vert XY\vert]$ is finite. (Here, the expectation is defined, because $\vert XY\vert$ is a.s. positive.) You can check that $\vert X\vert$ and $\vert Y\vert$ are still independent! In particular, that means that $$ \mathbb E[ g(\vert X\vert)h(\vert Y\vert)] = \mathbb E[ g(\vert X\vert)]\cdot \mathbb E[h(\vert Y\vert)] $$ for all continuous bounded functions. Take $g_n(x) := h_n(x) := \max(0, \min(x, n))$. Then, $g_n$ is bounded continuous and we may use the above. Furthermore, $g_n(\vert X\vert)$ is monotonously increasing to $\vert X\vert$ and similarly for $g_n(\vert Y\vert)$ and $g_n(\vert X\vert)g_n(\vert Y\vert)$. Hence, monotone convergence implies that $$ \mathbb E[\vert XY\vert] = \lim_n \mathbb E[g_n(\vert X\vert)g_n(\vert Y\vert)] = \lim_n \mathbb E[g_n(\vert X\vert)]\mathbb E[g_n(\vert Y\vert)] = \mathbb E[\vert X\vert]\mathbb E[\vert Y\vert]. $$
Now we know that $XY\in L^1$. In particular, we can now use the same functions $g_n$ and Dominated Convergence to finally get $$ \mathbb E[XY] = \mathbb E[X]\mathbb E[Y]. $$
There may be an easier way...