Let $X$ be a square-integrable random variable defined on a probability space and $(\Omega, \mathcal F, \mathbb P)$. Let $(\mathcal F_t, t \ge 0)$ be a filtration and $T>0$. Let $Y_t := \mathbb E [X^2 |\mathcal F_t] \ge 0$ for all $t\ge 0$. Then $Y_t$ is integrable and $\mathcal F_t$-measurable.
I'm trying to prove a claim mentioned in this answer, i.e.,
Theorem $(Y_t, t\ge 0)$ is uniformly integrable.
Could you verify if my below attempt is fine?
Proof Fix $\varepsilon>0$. By absolute continuity of integral, there is $\delta>0$ such that if $B \in \mathcal F$ and $\mathbb P [B] < \delta$ then $\mathbb E [ X^2 1_B] < \varepsilon$.
By Chebyshev's inequality, $$ \mathbb P [Y_t \ge K] \le \frac{\mathbb E [Y_t]}{K} = \frac{\mathbb E [X^2]}{K} \quad \forall K>0, \forall t \ge 0. $$
There is $K>0$ such that $\frac{\mathbb E [X^2]}{K} < \delta$. This implies $\mathbb P [Y_t \ge K] < \delta$ for all $t\ge 0$. Notice that $\{Y_t \ge K\} \in \mathcal F_t$. Then $$ \mathbb E [ Y_t 1_{\{Y_t \ge K\}}] = \mathbb E [ X^2 1_{\{Y_t \ge K\}}] < \varepsilon \quad \forall t \ge 0. $$
This completes the proof.
