Let $X,Y$ be independent random variables. Suppose $E|X+Y|$ is finite. Does it follow $E|X|<\infty$?
I believe the answer is no and tried to come up with a counter-example. Below is my attempt where I couldn't complete. If it follows that $E|X|$ is finite, please ignore my attempt.
Attempt (Constants are ignored for readability.)
Let $X$ be defined on $\mathbb{N}$ with distribution $P(X=k) = 1/k^2$. Note $EX = \sum_k k/k^2=\infty$.
Let $Y$ be identical to $X$. Distribution of $X+Y$ is given by;
\begin{equation}
\begin{aligned}
P(X+Y = z) &= \sum_{k=1}^{z-1} P(X=z-k)P(Y=k)\\
&=\sum_{k=1}^{z-1} \frac{1}{(z-k)^2k^2}
\end{aligned}
\end{equation}
Hence expected is given by;
\begin{equation}
E(X+Y) = \sum_{z=2}^{\infty}\sum_{k=1}^{z-1} \frac{z}{(z-k)^2k^2}
\end{equation}
I couldn't argue that the sum is finite. Is above sum convergent?
Since $|X| \le |X+Y| + |Y|$, $E[|X|] = E[ |X| \mid Y] \le E[|X+Y| \mid Y] + E[|Y| \mid Y]$ and $E[|X|] \le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < \infty$. Now $|X+Y| \ge |X| - |Y|$, so $$R = E[|X+Y|] \ge E[|X|-|Y|] = E[E[|X|-|Y| \mid Y]] = E[E[|X|] - |Y|]$$ But that implies $\text{Prob}(E[|X|]-|Y| \le R) > 0$ and thus $E[|X|] \le R + s < \infty$ for some $s$ where $\text{Prob}(|Y|<s) > 0$.