If $X$, $Y$ are independent random variables and $E[X+Y]$ exists, then $E[X]$ and $E[Y]$ exist. I've been trying to show E$|X+Y|$ < $\infty$ $\Rightarrow$ E$|X|$ < $\infty$ by showing E$|X|$ $\leq$ E$|X+Y|$, but I'm stuck and cannot proceed from here.
Someone can help me, please?
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$$E|X+Y| = \int\int|x+y|f_X f_Y\mathsf dx\mathsf dy = \int E|X+y|f_Y\mathsf dy < \infty.$$
So, can I say that $E|X+y| < \infty$ for almost every $y$ including $y=0$?
 A: Your second approach works. If $X$ and $Y$ are independent and $|X+Y|$ is integrable, then
$$
E|X+Y| = \int|x+y|dP_{(X,Y)}(x,y)=\int\left[\int|x+y|dP_X(x)\right]dP_Y(y)
$$
by Fubini's theorem, and moreover the function $$y\mapsto\int|x+y|dP_X(x)=E|X+y|$$ is integrable with respect to $P_Y$, hence finite almost surely. Pick any  $y$ for which $E|X+y|$ is finite and use
$$
|X|=|X+y-y|\le|X+y|+|y|$$
to conclude that  $X$ is integrable.
A: Let $X$ and $Y$ independent random variables such that their cumulative distribution functions are $F$ and $G$. Then the CDF of their sum is $$H(t)=\int_{-\infty}^\infty F(t-x)\,dG(x) $$
If $X+Y$ is integrable then it's absolute integrable and $E|X+Y|\geq \int\limits_{0}^\infty 1-H(t)\,dt $. By using Fubini's theorem we get
$$\int\limits_{0}^\infty 1-H(t)\,dt= \int\limits_{-\infty}^\infty\int\limits_{0}^\infty 1-F(t-x)\,dt\, dG(x)$$
where $$\int\limits_{0}^\infty 1-F(t-x)\,dt=\left[t(1-F(t-x))\right]_{t=0}^\infty-\int\limits_{0}^\infty t\,d[1-F(t-x)]=E(X\cdot I[X>-x]) +x$$ is finite for almost every $x$, thus $E(X^+)$ is finite. Similarly we can show, that $E(X^-)$ is finite, hence $EX=EX^+-EX^-$ is finite.
