# 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?

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 $n\in I$.
• I see, thanks. So if $(Y_n)$ is not UI, then there is some $(X_n)$ such that $E(X_nY_n) \not \to 0$. But this doesn't rule out the possibility that for some $(X_n)$, $E(X_nY_n) \to 0$ even when $(Y_n)$ is not UI. Is there a straightforward way of describing those $(X_n)$ for which $E(X_nY_n) \to 0$ holds without the UI of $(Y_n)$? – grndl Jul 9 '18 at 19:10
• So, after thinking about it some more, I think what I'm trying to do is formalize the idea that $(Y_n)$ may not be uniformly integrable, but $(X_n)$ goes to $0$ "fast enough" to make the product sequence $(X_nY_n)$ uniformly integrable. Any idea how to make that precise? – grndl Jul 9 '18 at 23:42