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

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.

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.