Integrability of two independent random variables Suppose that $X$ and $Y$ are two independent random variables.
How does one show that if $X+Y\in L^1$, then both $X$ and $Y$ are in $L^1$?
I know that one can approach this problem using the fact that the joint law of $X+Y$ is the convolution of the laws of $X$ and $Y$, but I would like to know   more "elementary" ways of proving the claim.
That is, how does one establish that 
$$E[|X|: |X|>M] \to 0 $$ as $M \to \infty$?
 A: Use the following law
$$\mathbb{E}(|Z|) \leq \sum_{n=0}^{\infty} \mathbb{P}(|Z| > n) \leq \mathbb{E}(|Z|) + 1$$
which is known as layered representation of expectation. From assumption $X+Y$ is in $L^1$, we have
$$\sum_{n=0}^{\infty} \mathbb{P}(|X+Y| > n) \leq \mathbb{E}(|X+Y|) + 1 < \infty$$
Then note that
\begin{align*}
\mathbb{P}(|X+Y| > n) &\geq \mathbb{P}(|X| - |Y| > n)\\
&\geq \mathbb{P}(|X| > n + m \text{ and } |Y| < m)\\
&\geq \mathbb{P}(|X| > n + m) \cdot \mathbb{P}(|Y| < m) &\text{by independence}
\end{align*}
for any $m$. Choose $m$ sufficiently large so that $\mathbb{P}(|Y| < m) > 0$ (such $m$ must exists for $\lim_{m \rightarrow \infty} \mathbb{P}(|Y| < m) = 1$) so that we can bound the tail sum
$$\sum_{k \geq m} \mathbb{P}(|X| > k) = \sum_{n \geq 0} \mathbb{P}(|X| > n + m) \leq \frac{\sum_{n=0}^{\infty} \mathbb{P}(|X+Y| > n)}{\mathbb{P}(|Y| < m)} < \infty$$
This shows that
$$\sum_{k \geq 0} \mathbb{P}(|X| > k) < \infty$$
and so by the layered representation inequality,
$$\mathbb{E}(|X|) \leq \sum_{n=0}^{\infty} \mathbb{P}(|X| > n) < \infty$$
hence $X$ is in $L^1$. Likewise for $Y$.
A: Suppose that the positive part $X_+$ of $X$ is not integrable. So
$$
\sum 2^n P(2^n\le X_+<2^{n+1})=\infty  \quad\quad\quad\quad (1)
$$
Let's take $M>0$ so large that $P(|Y|\le M)\ge 1/2$. Let me write $A_n$ for the events from (1). Since $X,Y$ are independent, the probability that $A_n$ happens and at the same time $|Y|\le M$ is $\ge (1/2)P(A_n)$. This implies that $E(X+Y)_+=\infty$, by focusing on the contribution coming from those sets.
This contradicts our assumption that $X+Y\in L^1$.
