If $X \perp Y$ and $X+Y \in L^1$ then $X,Y \in L^1$

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.

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

$$ \Bbb{E}[X+Y] = \int_{\Omega}\int_{\Omega} X(\omega)+Y(\eta) \, \Bbb{P}(d\omega)\Bbb{P}(d\eta). $$