If the sum of two independent random variables is in $L^2$, is it true that both of them are in $L^1$? Let $X$ and $Y$ be two independent random variables. If $\mathbb E(X+Y)^2 < \infty$, do we have $\mathbb E |X| < \infty$ and $\mathbb E |Y| < \infty$?
What I actually want is that $X$ and $Y$ are both in $L^2$, i.e., $\mathbb E X^2 < \infty$ and $\mathbb E Y^2 < \infty$. But this can be reduced to $\mathbb E |X|\mathbb E |X| < \infty$. Thus it suffices to show that $X$ and $Y$ are in $L^1$.
It might be helpful to see that since $|X| < |Y| + |X+Y|$, (and by symmetry), either both of $X$ and $Y$ or none of them are in $L^1$. So we may assume $\mathbb E |X| = \infty$ and $\mathbb E |Y| = \infty$ and try to find a contradiction. But here is where I got stuck.

If they have the finite mean x=$E X < \infty$  and $y= E Y <\infty$ then yes 
$E(X+Y)^2 = E(X^2)+E(Y^2)+2 E(X)*E(Y) = E(X^2)+E(Y^2)+2xy < \infty$. If no, I am affraid one can found a pathological case  when this is not true. 
Let me update myself. I think the previous answer given by David Giraudo is correct. Let me just a little expand it. $E(X+y)^2<\infty$ for some $y$ means 
that $E(X+y)=E(X)+y<\infty$ for this $y$. This means that $E(X)<\infty$. This means that   $E(X+y)^2=E(X^2)+2E(X)y+y^2 <\infty$. So $E(X^2)<\infty$.
