If $\mathbb E[|X_n-X|^2]\to 0$ can we say that $X_{n_k}(\omega )\to X(\omega )$ for a.e. $\omega $?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and let $X_n\to X$ in $L^2(\Omega )$ where $(X_n)$ is a sequence of random variable and $X$ is a random variable as well.

By a theorem of Lebesgue measure theory, we know that there is a subsequence such that $X_{n_k}(\omega )\underset{k\to \infty }{\to} X(\omega )$ a.e. Is it also true with random variable ? Because if yes, then the convergence in $L^2$ is very strong (and I don't get why $L^2$ convergence is called weak convergence). Indeed, we get that $$\mathbb P\{\lim_{k\to \infty }X_{n_k}=X\}=1,$$ which is (at my opinion) very strong.

Answer 1

Yes it's true. And it's indeed quite strong, but it's weaker than $$\mathbb P\left\{\lim_{n\to \infty }X_n=X\right\}=1,$$ But it's at least better (stronger) than convergence in distribution or convergence in probability, since we indeed have more information of the sequence $(X_n)$.

By the way, when I say that $L^2$ convergence than a.s. convergence it's in the sense that $L^2$ convergence implies $$\lim_{k\to \infty }X_{n_k}(\omega )=X(\omega ),\ \ a.s.,$$ which is not as well $$\lim_{n\to \infty }X_n(\omega )=X \ \ a.s.,$$ but notice that a.s. convergence doesn't implies $L^2$ convergence as $L^2$ convergence doesn't imply a.s. convergence. So the term "weaker" is not completely adapted for the situation.