Let $(X_t,Y_t)_{t\in\mathbb N}$ be a bivariate real stationary process, and let $\mathcal F_t:=\sigma(Y_s :s\leq t)$ be the filtration generated by $Y_t$. Assuming the following convergence result
$$\frac{1}{T}\sum_{t=1}^T X^2_tY_t^2 \overset{p}{\to} E[X_t^2Y_t^2]<\infty \text{ as } T\to \infty,$$
holds, can I show then that
$$\frac{1}{T}\sum_{t=1}^T E[X^2_t|\mathcal F_t]Y^2_t \overset{p}{\to} E[X_t^2Y_t^2] \text{ as } T\to \infty $$? (Assume $E[X^2_t]<\infty $ and $Y_t$ bounded).
EDIT: This lemma can be found in Bauer's book measure and integration.
The lemma implies that we have $\frac{1}{T}\sum_{t=1}^T X^2_tY_t^2 \overset{L^1}{\to} E[X_t^2Y_t^2]\text{ as } T\to \infty$. We can then almost prove the claim: if we replace $\mathcal F_t$ by $\mathcal F_T$ then for any $\epsilon>0$ we have
$$P\bigg[\bigg|\frac{1}{T}\sum_{t=1}^T E[X^2_t|\mathcal F_T]Y_t^2 - E[X_t^2Y_t^2]\bigg|\geq \epsilon\bigg ]$$
$$=P\bigg[\bigg|E\bigg[\frac{1}{T}\sum_{t=1}^T X_t^2Y_t^2 - E[X_t^2Y_t^2] \big|\mathcal F_T\bigg]\bigg|\geq \epsilon\bigg ]$$
$$\leq P\bigg[E\bigg[ \bigg|\frac{1}{T}\sum_{t=1}^T X_t^2Y_t^2 - E[X_t^2Y_t^2] \bigg| \big|\mathcal F_T\bigg]\geq \epsilon\bigg ]$$
$$\leq \frac{1}{\epsilon} E\bigg[ \bigg|\frac{1}{T}\sum_{t=1}^T X_t^2Y_t^2 - E[X_t^2Y_t^2] \bigg|\bigg ] \to 0 \text{ as } T\to \infty,$$
where the first inequality is Jensen's inequality for conditional expectations and the second inequality is because $\epsilon1_A\leq 1_AE\bigg[ \bigg|\frac{1}{T}\sum_{t=1}^T X_t^2Y_t^2 - E[X_t^2Y_t^2] \bigg| \big|\mathcal F_T\bigg]$ with $A=\bigg\{E\bigg[ \bigg|\frac{1}{T}\sum_{t=1}^T X_t^2Y_t^2 - E[X_t^2Y_t^2] \bigg| \big|\mathcal F_T\bigg]\geq \epsilon\bigg\}$.