Convergence of conditional expectation of multiplication of two sequences

Question

Let $\mathcal{F}$ be a sub-sigma field and $(X_n)_{n=1}^\infty, ~ (Y_n)_{n=1}^\infty$ be two sequences of non-negative random variables such that $|X_n| \leq 1$ and \begin{eqnarray} &X_n\to 0 ~~\text{a.s,}\\ &\mathbb{E}[Y_n \mid \mathcal{F}]\to Y ~~\text{a.s,} \end{eqnarray} where $Y$ is an $\mathcal{F}$-measurable random variable. Then can I have \begin{eqnarray} \mathbb{E}[X_nY_n \mid \mathcal{F}]\to 0 ~~\text{a.s?} \end{eqnarray}

Answer 1

This is, in general, wrong.

Consider for instance $\Omega=(0,1)$ endowed with the Lebesgue measure (restricted to $(0,1)$). Set $$X_n(\omega) := 1_{(0,1/n)}(\omega) \quad \text{and} \quad Y_n(\omega) := n 1_{(0,1/n)}(\omega) = n X_n(\omega).$$ Clearly, $X_n \to 0$ almost surely and $0 \leq X_n \leq 1$. For $\mathcal{F} := \{\emptyset,\Omega\}$ we have $$\mathbb{E}(Y_n \mid \mathcal{F}) = \mathbb{E}(Y_n)=1 \to 1=:Y \quad \text{a.s.}$$ On the other hand$$\mathbb{E}(X_n Y_n \mid \mathcal{F}) = n \mathbb{E}(X_n^2) = n \mathbb{E}(X_n) = 1$$ for all $n \in \mathbb{N}$, i.e. $\mathbb{E}(X_n Y_n \mid \mathcal{F}) \to 0$ does not hold true.