$L^1$ norm of product of independent random variables I am trying to show that $\|XY\|_1 = \|X\|_1\|Y\|_1$ for $X,Y$ independent random variables, where $\|X\|_1 = \int{|X| d\mathbb{P}}$.
I have a feeling that this result is intuitive, but could anyone explain from a measure theoretic perspective why this is true? Thanks
 A: To say that $X$ and $Y$ are independent means exactly that the joint distribution $P_{X, Y}$ factors as the product measure of $P_X$ and $P_Y$. By the change of variable formula, we have 
$$\int_{\Omega} |XY|\, d P=\int_{\mathbb{R}^2} |xy|\, P_{X, Y}(dxdy)=\int_{\mathbb{R}^2}|xy|\, P_X(dx)P_Y(dy),$$
and clearly the last integral factors as $\int_{\mathbb{R}}|x|P_X(dx)\int_{\mathbb{R}}|y|P_Y(dy)$. Applying again the change of variable formula we see that this is exactly $\int_{\Omega}|X|\, dP\int_{\Omega}|Y|\, dP$.
A: This can be proved only using the definition of independence $\mathbb P(AB)=\mathbb P(A)\cdot \mathbb P(B)$ and Lebesgue integral.
We assume that $X$ and $Y$ are non-negative (otherwise we consider their absolute values which still are independent). 
Notice that $X$ is the pointwise non-decreasing limit of characteristic functions of measurable sets, and these sets belong to $\sigma(X)$. We do the same for $Y$. 
By monotone convergence, it's enough to do it when $X$ and $Y$ are simple and in this case the relationship can be checked easily. 
