# $\sum_{n=1}^{\infty}P\left( |X_n-X|>\epsilon\right)<\infty$ for each $\epsilon>0$ $\implies$ $X_n\to X$ a.s.

Problem. Let $\left( \Omega, \mathcal A, P \right)$ be a probability space and $X,X_1,X_2,...$ random variables on $\Omega$. If $\sum_{n=1}^{\infty}P\left( |X_n-X|>\epsilon\right)<\infty$ for each $\epsilon>0$, then $X_n\to X$ almost surely as $n\to \infty$.

My work. Since $$\sum_{n=1}^{\infty}P\left( |X_n-X|>\epsilon\right)<\infty$$ Borel-Cantelli implies that $P(\limsup_{n\to \infty} \{|X_n-X|>\epsilon\})=0$ for each $\epsilon>0$.

Now let $A_k=\limsup_{n\to \infty} \{|X_n-X|>1/k\}$. Then $A_k$ has probability $0$ and hence $A_k^c=\liminf_{n\to \infty} \{|X_n-X|\leq\ 1/k\}$ has probability $1$. Then the intersection $A=\cap_{k\in \Bbb{N}}A_k^c$ also has probability $1$ and this is precisely the set of all $\omega\in \Omega$ that satisfy $\lim_{n\to\infty}X_n(\omega)=X(\omega)$. So $A$ is a set of probability $1$ on which we have pointwise convergence and so $X_n\to X$ almost surely.

Does that solution work? It is true that the intersection of countably many sets of probability $1$ again has probability $1$, right?