What's a condition that's weaker than uniform integrability and implies convergence of an integral to $0$? Let $(\Omega, \mathcal{F}, P)$ be a probability space, and let $(X_n)$, $(Y_n)$ be two sequences of nonnegative, integrable random variables. Suppose $X_n \leq 1$, $X_n \to 0$, and $Y_n \to Y$ almost surely. Then, $X_nY_n \to 0$ almost surely.
I am wondering when $\int X_nY_n dP \to 0$.
If $(Y_n)$ is uniformly integrable, then so is $(X_nY_n)$, in which case $\int X_nY_n dP \to 0$. So my question becomes

What's a condition on $(Y_n)$ that's weaker than the uniform integrability and guarantees $\int X_nY_n dP \to 0$?

As Davide Giraudo points out below, if $(X_nY_n)$ is uniformly integrable then $\int X_nY_ndP \to 0$, so yet another way to ask my question is

What's a condition on $(Y_n)$ that's weaker than uniform integrability and implies that $(X_nY_n)$ is uniformly integrable?

 A: Let $\left(Z_n\right)_{n\geqslant 1}$ be a sequence of random variables such that $Z_n\to 0$ almost surely. Then uniform integrability of $\left(Z_n\right)_{n\geqslant 1}$ is equivalent to convergence of $\left(Z_n\right)_{n\geqslant 1}$ to $0$ in $\mathbb L^1$. Therefore, uniform integrability of $\left(X_nY_n\right)_{n\geqslant 1}$ is enough.
One of the most reasonable attempt could be to impose boundedness in $\mathbb L^1$ of $\left(Y_n\right)_{n\geqslant 1}$. But it does not work. If $\left(Y_n\right)_{n\geqslant 1}$ is bounded in $\mathbb L^1$ but not uniformly integrable, then there exists a positive $\varepsilon_0$ and $I\subset\mathbb N$ infinite  such that for all $\delta$, there exists a measurable set $B_\delta$of measure smaller than $\delta$ for which $\mathbb E\left[Y_n\mathbf 1_{B_\delta}\right]\gt \varepsilon_0$ foreach $n\in I$. Let $A_n:=B_{2^{-n}}$ and let $X_n$ be the indicator of $A_n$. Then $0\leqslant X_n\leqslant 1$, $X_n\to 0$ almost surely,  but $\mathbb E\left[X_nY_n\right]\geqslant \varepsilon_0$ for all $n\in I$.
