If $X$ and $Y$ are independent and for some $p>0$, $\mathscr{E}|X+Y|^p<\infty$, then $\mathscr{E}|X|^p<\infty$ and $\mathscr{E}|Y|^p<\infty$.
$\bullet$ Proof.
Since $X$ and $Y$ are independent, by Fubini's theorem $$\mathscr{E}|X+Y|^p=\int\int|x+y|^p\,\mu^2(dx,dy)=\int\left[\int|x+y|^p\,\mu_x(dx)\right]\mu_y(dy)<\infty,$$which implies $$\int|x+y|^p\,\mu_x(dx)<\infty$$for $\mu_y$ almost everywhere. Thus there exists $Y=y_0$ as a constant such that $$\int|x+y_0|^p\,\mu_x(dx)<\infty.$$Then we use the following inequality.
[Inequality]
If $X\geq0$ and $Y\geq0$, $p\geq0$, then $$\mathscr{E}\{(X+Y)^p\}\leq2^p\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}.$$If $p>1$, the factor $2^p$ may be replaced by $2^{p-1}$. If $0\leq p\leq1$, it may be replaced by $1$.
For $p>0$, $$\begin{array}{rl}\mathscr{E}|X|^p&=\mathscr{E}|X+y_0-y_0|^p\\ &\leq2^p\left\{\mathscr{E}|X+y_0|^p+\mathscr{E}|y_0|^p\right\}\\ &=2^p\left\{\int|x+y_0|^p\,\mu_x(dx)+\int|y_0|^p\,\mu_x(dx)\right\}\\ &=2^p\left\{\int|x+y_0|^p\,\mu_x(dx)+|y_0|^p\right\}<\infty.\end{array}$$Similarly, by fixing $X=x_0$ as a constant, we have $\mathscr{E}|Y|^p<\infty.$
$\Box$
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