# Convergence of conditional expectation of multiplication of two sequences

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 $$\begin{eqnarray} \mathbb{E}[X_nY_n \mid \mathcal{F}]\to 0 ~~\text{a.s?} \end{eqnarray}$$

• If $X_n$ is $\mathcal{F}$-measurable, then this is correct; otherwise it is in general wrong.
– saz
Nov 1 '18 at 17:52
• Thank you saz. In my case, $X_n$ is not $\mathcal{F}$-measurable. Could you give me a counter-example? Nov 1 '18 at 17:56

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.