Let $X, Y$ be two independent, integrable random variables. Starting just from $||XY||_1=||X||_1||Y||_1$, is it possible to deduce that $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$?
I'm interested in a measure theoretic point of view.
My thoughts: Since $X, Y$ are integrable, so are $|X|, |Y|$ and from $||XY||_1=||X||_1||Y||_1$, $|XY|$ is also integrable. So we have $\mathbb{E}(|XY|)=\mathbb{E}(|X|)\mathbb{E}(|Y|)$. But how does this imply $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$?
If $X=Y$ and $X,Y$ take values $\pm 1$ with probability $\frac 1 2$ each then $\|XY\|_1=1=\|X\|_1\|Y\|_1$ but $EXY \neq EX EY$. So deducing $EXY = EX EY$ just from $\|XY\|_1=1=\|X\|_1\|Y\|_1$ (without using independence) is not possible.