It is known that if a sequence of random variables converges in norm then there exists a subsequence which converges almost surely. That is: let $\left(X_{n}\right)_{n\in\mathbb{N}}\subseteq L^{2}\left(\Omega,\mathcal{F},P\right)$ be a sequence of square integrable random variables on the probability space $\left(\Omega,\mathcal{F},P\right)$ . If $X\in L^{2}\left(\Omega,\mathcal{F},P\right)$ is such that $\mathrm{E}_{P}\left(\left(X_{n}-X\right)^{2}\right)\longrightarrow0$ as $n\longrightarrow\infty$ , then there exists a subsequence $\left(X_{n_{k}}\right)_{k\in\mathbb{N}}\subseteq\left(X_{n}\right)_{n\in\mathbb{N}}$ such that $X_{n_{k}}\longrightarrow X$ almost surely as $k\longrightarrow\infty$ . My question is the following: does this holds for an infinite direct sum of $L^{2}$ spaces? To be precise: consider a filtered probability space $\left(\Omega,\left(\mathcal{F}_{t}\right)_{t\in\mathbb{N}},\mathcal{F},P\right)$ . Define: $$\oplus_{t=1}^{\infty}L^{2}\left(\Omega,\mathcal{F}_{t},P\right)\equiv\left\{ \left(X\left(t\right)\right)_{t=1}^{\infty}:\, X\left(t\right)\in L^{2}\left(\Omega,\mathcal{F}_{t},P\right)\forall t,\,\sum_{t=1}^{\infty}\mathrm{E}_{P}\left(X\left(t\right)^{2}\right)<\infty\right\}.$$ Let $\left(\left(X_{n}\left(t\right)\right)_{t=1}^{\infty}\right)_{n\in\mathbb{N}}\subseteq\oplus_{t=1}^{\infty}L^{2}\left(\Omega,\mathcal{F}_{t},P\right)$ be such that $\sum_{t=1}^{\infty}\mathrm{E}_{P}\left(\left(X_{n}\left(t\right)-X\left(t\right)\right)^{2}\right)\longrightarrow0$ as $n\longrightarrow\infty$ for some $\left(X\left(t\right)\right)_{t=1}^{\infty}\in\oplus_{t=1}^{\infty}L^{2}\left(\Omega,\mathcal{F}_{t},P\right).$ Does there exists a subsequence $\left(\left(X_{n_{k}}\left(t\right)\right)_{t=1}^{\infty}\right)_{k\in\mathbb{N}}\subseteq\left(\left(X_{n}\left(t\right)\right)_{t=1}^{\infty}\right)_{n\in\mathbb{N}}$ such that, for every $t\geq1$ , $X_{n_{k}}\left(t\right)\longrightarrow X\left(t\right)$ almost surely? I worked out a proof that seems to be sound, but I'm not sure of a couple of passage I did. The proof is as follows. I indicate inside parenthesis the passage I am not sure about. First I claim that $$\sum_{t=1}^{\infty}E\left(\left(X_{n}\left(t\right)-X\left(t\right)\right)^{2}\right)\rightarrow0\Rightarrow P\left[\sup_{t}\left|X_{n}\left(t\right)-X\left(t\right)\right|>\epsilon\right]\rightarrow0 $$ for every $\epsilon>0 $.
Proof of the claim. I prove the counterpositive. Suppose that there exist $\epsilon,\delta>0$ such that, for every $\bar{n}$, there exists an $n\left(\bar{n}\right)\geq\bar{n}$ such that $$P\left[\sup_{t}\left|X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right|>\epsilon\right]\geq\delta.$$ Then it must be true that (this the passage I am not sure about) $$P\left[\sum_{t=1}^{\infty}\left(X_{n\left(\bar{n}\right)}\left(t\right)-X_{n\left(\bar{n}\right)}\left(t\right)\right)^{2}\geq\epsilon^{2}\right]\geq\delta.$$ Call $E=\left[\sum_{t=1}^{\infty}\left(X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right)^{2}\geq\epsilon^{2}\right]\in\mathcal{F}$. We have:
$$\sum_{t=1}^{\infty}E\left(\left(X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right)^{2}\right) = E\left(\sum_{t=1}^{\infty}\left(X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right)^{2}\right) \geq\int_{E}\sum_{t=1}^{\infty}\left(X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right)^{2}\mathrm{d}P \geq \epsilon^{2}\delta.$$
Then, for every $\bar{n}$, there exists an $n\left(\bar{n}\right)\geq\bar{n}$ such that $$\sum_{t=1}^{\infty}E\left(\left(X_{n\left(\bar{n}\right)}\left(t\right)-X\left(t\right)\right)^{2}\right)\geq\epsilon^{2}\delta$$ and this completes the proof of the claim. $\square$
Now I show that if $P\left[\sup_{t}\left|X_{n}\left(t\right)-X\left(t\right)\right|>\epsilon\right]\rightarrow0$ then there exists a subsequence $\left(\left(X_{n_{k}}\left(t\right)\right)_{t\in\mathbb{N}}\right)_{k\in\mathbb{N}}\subseteq\left(\left(X_{n}\left(t\right)\right)_{t\in\mathbb{N}}\right)_{n\in\mathbb{N}}$ such that $$\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|\longrightarrow0\qquad\mbox{a.s.}$$ First observe that, for every $k>0$ , there exists an integer $n_{k}$ such that$$P\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\frac{1}{k}\right]\leq\frac{1}{2^{k}}.$$ This implies that$$\sum_{k=1}^{\infty}P\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\frac{1}{k}\right]<\infty.$$ Finally, observe that, for every $\epsilon>0$ , $$\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\epsilon\,\mbox{i.o.}\right]\subseteq\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\frac{1}{k}\,\mbox{i.o.}\right],$$ so that$$\sum_{k=1}^{\infty}P\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\frac{1}{k}\right]<\infty \Rightarrow P\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\frac{1}{k}\,\mbox{i.o.}\right]=0 \Rightarrow P\left[\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|>\epsilon\,\mbox{i.o.}\right]=0 \forall\epsilon>0 \Rightarrow\sup_{t}\left|X_{n_{k}}\left(t\right)-X\left(t\right)\right|\rightarrow0 \mbox{a.s.}$$ and this completes the proof $\square\square$