Let $ X_n \geq 0$ be a sequence of integrable random variables defined on a probability space $(\Omega, \mathcal{F}, P)$ and given a sub-$\sigma$-algebra $\mathcal{G}\subset\mathcal{F}$, we assume that
$$
E\big[ X_n \vert \mathcal{G} \big]
$$
is uniformly bounded in $L^2(\Omega)$ over $n$. Suppose also that $X_n $
converges to zero almost surely. Is that true
$$
E\big[ X_n \vert \mathcal{G} \big] \to 0 \quad\text{almost surely?}
$$
I guess not, but I hope to see a counterexample. Or a proof of necessity is also greatly appreciated.
An example that $X_n \to 0$ almost surely but $\mathbb{E}(X_n | \mathcal{G} ) \not\to 0$.
probability theory
Best Answer
No, this does not need to be true. Let $(Z_n)$ be a sequence of i.i.d. Bernoulli random variables with $P(Z_n = 2) = P(Z_n = 0) = \frac 12$, and $X_n := \prod_{k=1}^n Z_k$. Then $X_n \ge 0$, and $X$ is a martingale so $\mathbb{E}[X_n] = 1$, but $(X_n) \rightarrow 0$ almost surely. Taking $\mathcal G$ to be the trivial $\sigma$-algebra gives the counterexample. More generally, if we let $(\mathcal F_n)$ be the natural filtration of $(X_n)$ then we have for $\mathcal G = \mathcal F_m$ where $m$ is some fixed integer that $\mathbb{E}[X_n|\mathcal G] = X_{n \wedge m}$ so $$\mathbb{E}[\mathbb{E}[X_n|\mathcal G]^2] \le \mathbb{E}[X_m^2] \le 4^m < \infty,$$ i.e. $\mathbb{E}[X_n|\mathcal G]$ is bounded in $\mathcal L^2$, but $\lim_{n \rightarrow \infty} \mathbb{E}[X_n|\mathcal G] = X_m \ne 0$.