Showing that the product of two integrable complex valued random variables is integrable

Question

Let $S_1 = A_1 + iB_1, S_2 = C_2 + iD_2$ be two integrable complex valued random variables on the probability space $(\Omega, F, P)$, where $A_1, B_1, C_2, D_2$ are real valued r.v.s on the same space. Then, in order to show that the product $S_1S_2$ is integrable, I have to show that the products of the real and imaginary parts of $S_1, S_2$ are integrable as well, i.e. $\mathbb{E}(A_1B_1) < +\infty$ etc. I do know that the product of two independent integrable r.v.s is integrable as well, but evidently we do not have that luxury here. Therefore, I was wondering whether the following applicaiton of the Fubini's theorem would help: Define $T_1, T_2 \in \{A_1, B_1, C_2, D_2\}, T_1 \neq T_2$. Then clearly $\left|T_1\cdot T_2\right|$ is measurable w.r.t. the product sigma-algebra $F \otimes F$ (isn't it true that this would hold event if $S_1$ and $S_2$ would live in different probability spaces?). So now the expected value of the absolute value of the component product is: $\mathbb{E}(|T_1\cdot T_2|) = \int_{\Omega^2}|T_1\cdot T_2|d(P\otimes P) = \int_{\Omega_1}\left(\int_{\Omega_2}|T_1(\omega_1)\cdot T_2(\omega_2)|dP(\omega_2)\right)dP(\omega_1)$ from which we obtain that $\int_{\Omega_1}|T_1(\omega_1)|\left(\int_{\Omega_2}|T_2(\omega_2)|dP(\omega_2)\right)dP(\omega_1) = \int_{\Omega_1}|T_1(\omega_1)|\mathbb{E}(T_2)dP(\omega_1) = \mathbb{E}(T_1)\mathbb{E}(T_2) < +\infty$ by assumption, as $S_1, S_2 \in \mathcal{L}^1(P)$.

If the above reasoning was valid, then this would show that the product of $S_1$ and $S_2$ is integrable as a linear combination of integrable functions.

Answer 1

Note that by definition, $X \in L^1$ if and only if $|X| \in L^1$. Since $|S_1S_2| = |S_1||S_2|$, this means that you can WLOG assume that $S_1, S_2 \geq 0$. Still, this reduction doesn't help since the proposition is false. The product of two $L^1$ variables is in general not $L^1$. For example, pick $p \in \mathbb{R}$ such that $x^p \in L^1([0, 1])$ but $x^p \cdot x^p = x^{2p} \notin L^1([0, 1])$.