If $\mathbb E_{\mathbb P} \vert f(X,Y)\vert <\infty$, is also $\mathbb E_{\mathbb P_1\times \mathbb P_2} \vert f(X,Y)\vert <\infty$? Let $(E_1,\mathcal E_1)$ and $(E_2,\mathcal E_2)$ be two measurable spaces. Let $(E=E_1\times E_2,\mathcal E=\mathcal E_1\times \mathcal E_2)$. Let $\mathbb P$ a valid probability distribution on $(E,\mathcal E)$. Define the marginal distributions $\mathbb P_1(A)=\mathbb P (A\times E_2)$ and $\mathbb P_2(A)=\mathbb P (E_1\times A)$, and finally denote the product distribution by $\mathbb P_1\times \mathbb P_2$. Finally assume that we know
\begin{align*} \mathbb E_{\mathbb P} \vert f(X,Y)\vert <\infty.\end{align*}
My question is if it also holds that
\begin{align*} \mathbb E_{\mathbb P_1\times \mathbb P_2} \vert f(X,Y)\vert <\infty?\end{align*}
In particular, if it is false I hope to see a counterexample and I'd also be interested to know if it is true in the case $f(X,Y)=g(X)h(Y)$ for some functions $g$ and $h$.
 A: It is not true in general.
Let $E_1 = E_2 = \mathbb{R}$, and let $\mu$ be some probability measure on $\mathbb{R}$ with no first moment (e.g. Cauchy distribution).  Let $\mathbb{P}$ be the joint law of $(X, X)$ where $X \sim \mu$.  Finally let $f(x,y) = x-y$.  Under $\mathbb{P}$, we have $Y=X$ a.s., and therefore $f(X,Y)=0$ a.s., so $\mathbb{E}_{\mathbb{P}}|X-Y| = 0$.   But under $\mathbb{P}_1 \times \mathbb{P}_2$, we have $X,Y$ are iid $\mu$, and it follows that $\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2} |X-Y| = \infty$.  To see this, use Tonelli's theorem to write
$$\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2} = \int \mathbb{E}|X-y| \,\mathbb{P}_1(dy)$$ where the integrand is identically $+\infty$, using $|X-y| \ge |X|-|y|$.
It is not even true for $f(x,y) = g(x) h(y)$.  Let $W$ be a random variable with infinite first moment, and $Z$ a Bernoulli(1/2) random variable independent from $W$.  Let $\mathbb{P}$ be the joint law of $(WZ, W(1-Z))$, and set $f(x,y) = xy$, which is of the form $g(x) h(y)$.  Then
$$\mathbb{E}_{\mathbb{P}}|XY| = \mathbb{E} |W^2 Z (1-Z)| = 0$$ because $Z(1-Z)=0$ a.s.  But $WZ, W(1-Z)$ are identically distributed, so we have
$$\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2}|XY| = (\mathbb{E}|WZ|)^2 = (\mathbb{E}|W| \cdot \mathbb{E}|Z|)^2 = \frac{1}{4} (\mathbb{E}|W|)^2 = \infty.$$
