If $X_n \to X$ in $L^1$ and $Y_n \to Y $ in $L^1$, I am trying to analyze the behavior of convergence of $X_nY_n$ and $\frac{Y_n}{X_n}$. I have an intuition that they both don't converge to $XY$ and $\frac{Y}{X}$ respectively. but I am not getting any counterexample. Am I thinking correct or is there something I am missing?
Note: In the second case we can assume that $P(X=0)=0$.
It is not even clear that $X_nY_n, Y_n/ X_n, XY$ or $Y/X$ are in $L^1(\Omega)$ anymore. Considering $\Omega = [0,1]$ with the Lebesgue measure, define $$X_n(\omega) = Y_n(\omega) = \frac{1}{\omega^{1/2 + 1/n}},\,\,\,\, n =1,2,3,\ldots, \,\,\,\, \omega \in [0,1].$$ Then each $X_n$ and each $Y_n$ is in $L^1([0,1])$ but none of the products $X_nY_n$ are integrable. We do however have a sort similar result.
Theorem. Suppose $\Omega$ is a probability space (with measure $\mu$) and that $X_n \to X$ and $Y_n \to Y$ in $L^2(\Omega)$. Then $X_nY_n \to XY$ in $L^1(\Omega)$.
Proof. Notice \begin{align*} \lvert X_nY_n - XY \rvert &= \lvert X_nY_n - X_n Y + X_n Y - XY\rvert \\ &\le \lvert X_n \rvert \cdot \lvert Y_n - Y \rvert + \lvert Y \rvert \cdot \lvert Y_n - Y \rvert. \end{align*} Integrating on both sides and using Holder's inequality, we see \begin{align*} \int_\Omega \lvert X_nY_n - XY \rvert d\mu &\le \int_\Omega \lvert X_n \rvert \cdot \lvert Y_n - Y \rvert d\mu + \int_\Omega \lvert Y \rvert \cdot \lvert Y_n - Y \rvert d\mu \\ &\le \| X_n \|_{L^2(\Omega)} \| Y_n - Y \|_{L^2(\Omega)} + \| Y \|_{L^2(\Omega)} \| X_n - X \|_{L^2(\Omega)}. \end{align*} The latter goes to zero as $n\to \infty$ since both $ \| X_n - X \|_{L^2(\Omega)}$ and $ \| Y_n - Y \|_{L^2(\Omega)}$ go to zero and because $ \| X_n \|_{L^2(\Omega)}$ remains bounded. This shows that $\int_\Omega \lvert X_nY_n - XY \rvert d\mu \to 0$ so $X_nY_n \to XY$ in $L^1(\Omega)$.