If $X \perp Y$ and $X+Y \in L^1$ then $X,Y \in L^1$ Suppose $X,Y$ are independent random variables and that $X+Y \in L^1$. How can I show that $X,Y \in L^1$.
The theorem obviously fails if independence is left off, but I don't see how to use it. Intuitively, if $X$ failed to be in $L^1$, then it is big somewhere, and the independence of $Y$ should says that $Y$ can't selectively cancel out $X$ where $X$ is big.
 A: Since $X$ and $Y$ are independent, $E[|X+Y|]=E[u(X)]$ where $u(x)=E[|x+Y|]$. If $X+Y$ is integrable, then $u(X)$ is integrable, in particular, $u(X)$ is almost surely finite, thus there exists at least one value $x$ such that $u(x)$ is finite. Then $x+Y$ is integrable, hence $Y$ is integrable. Finally, $X=(X+Y)+(-Y)$ is integrable as a sum of integrable random variables.
A: My idea is to notice that the measure $d\Bbb{P}(X+Y \leq z)$ is the convolution of two measures
$$d\Bbb{P}(X \leq x) \quad \text{and} \quad d\Bbb{P}(Y \leq y).$$
Thus by Tonelli's theorem it follows that
\begin{align*}
\int_{\Omega} |X+Y| \, d\Bbb{P}
&= \int_{\Bbb{R}} |z| \, d\Bbb{P}(X+Y \leq z) \\
&= \int_{\Bbb{R}}\int_{\Bbb{R}} |x+y| \, d\Bbb{P}(X \leq x) \, d\Bbb{P}(Y \leq y) \\
&= \int_\Omega \int_\Omega |X(\omega)+Y(\eta)| \, \Bbb{P}(d\omega) \Bbb{P}(d\eta),
\end{align*}
which, in particular, is finite.
Thus applying Fubini's theorem we have
$$ \int_{\Omega} (X+Y) \, d\Bbb{P}
= \int_{\Omega} \int_{\Omega} \{ X(\omega)+Y(\eta) \} \, \Bbb{P}(d\omega)\Bbb{P}(d\eta). \tag{1} $$
In fact, Fubini's theorem says more than the equality:
$$ \int_{\Omega} \{ X(\omega)+Y(\eta) \} \, \Bbb{P}(d\omega) $$
is integrable for a.s. $\eta$ and likewise for the mapping with the role of $\omega$ and $\eta$ interchanged. In particular, this implies that $X$ and $Y$ are also in $L^{1}$.

Proof of $(1)$ directly: Let $F_{X}(x) = \Bbb{P}(X \leq x)$ be the CDF of $X$ and likewise for $Y$ and $X+Y$. Then it holds that
\begin{align*}
F_{X+Y}(z) &= \int_{\Bbb{R}} F_{Y}(z - x) \, dF_X(x). \tag{2}
\end{align*}
(Actually I wrote down a proof of $(2)$ but soon deleted it because it was too lengthy while not relevant to the rest of the argument.) Applying this, for any bounded $C^{1}$ function $f$ we have
\begin{align*}
\int_{(a, b]} f'(z) F(z) \, dz
&= \int_{(a, b]} \int_{\Bbb{R}} f'(z) F_{Y}(z - x) \, dF_{X}(x) dz \\
&= \int_{\Bbb{R}} \int_{(a, b]} f'(z) F_{Y}(z - x) \, dz \, dF_X(x) \\
&= \int_{\Bbb{R}} \left\{ f(b)F_{Y}(b-x) - f(a)F_{Y}(a-x) - \int_{(a, b]} f(y) \, dF_{Y}(y-x) \right\} \, dF_X(x) \\
&= f(b)F_{X+Y}(b) - f(a)F_{X+Y}(a) - \int_{\Bbb{R}} \int_{(a-x, b-x]} f(y+x) \, dF_{Y}(y) \, dF_X(x).
\end{align*}
Thus it follows that
\begin{align*}
\int_{(a, b]} f(z) \, dF_{X+Y}(z)
&= f(b)F_{X+Y}(b) - f(a)F_{X+Y}(a) - \int_{(a, b]} f'(z) F_{X+Y}(z) \, dz \\
&= \int_{\Bbb{R}} \int_{(a-x, b-x]} f(y+x) \, dF_{Y}(y) \, dF_{X}(x) \\
&= \int_{\Bbb{R}} \int_{\Bbb{R}} f(y+x) \mathbf{1}_{(a-x, b-x]}(y) \, dF_{Y}(y) \, dF_{X}(x).
\end{align*}
Since both the integrand is bounded, taking $(a, b) \to (-\infty, \infty)$ gives
\begin{align*}
\Bbb{E}[f(X+Y)]
&= \int_{-\infty}^{\infty} f(z) \, dF_{X+Y}(z) \\
&= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x+y) \, dF_{X}(z) \, dF_{Y}(z) \\
&= \int_{\Omega}\int_{\Omega} f(X(\omega)+Y(\eta)) \, \Bbb{P}(d\omega)\Bbb{P}(d\eta).
\end{align*}
Using the condition that $\Bbb{E}[|X+Y|] < \infty$, approximating $x \mapsto |x|$ by bounded $C^{1}$ functions in an appropriate way, it follows that
$$ \Bbb{E}[|X+Y|] = \int_{\Omega}\int_{\Omega} |X(\omega)+Y(\eta)| \, \Bbb{P}(d\omega)\Bbb{P}(d\eta). $$
Then approximating $x \mapsto x$ by bounded $C^{1}$ functions dominated by $|x|$, dominated convergence theorem gives
$$ \Bbb{E}[X+Y] = \int_{\Omega}\int_{\Omega} X(\omega)+Y(\eta) \, \Bbb{P}(d\omega)\Bbb{P}(d\eta). $$
