# Criteria for almost sure convergence

Given a sequence of random variables $$\{X_n \}_n$$ I have obtained that: $$X_n \overset{\mbox{a.s.}}{\longrightarrow} X \; \;\mbox{iff}\;\; \forall \epsilon>0 \; \lim_{n \to \infty} P(|X_n-X|< \epsilon)=1$$ Please, could anyboby check my proof and tell me where I did a mistake?

For $$\epsilon>0$$ and $$n \in \mathbb{N}$$ I defined: $$A_n^{\epsilon} \overset{def}{=} \{\omega: \; |X_n(\omega) - X(\omega)| < \epsilon\}$$ By resorting to the definition of $$\underset{n \to \infty}{\text{lim inf } A_n^{\epsilon}}$$ I got: $$P(\{\omega:\; \forall \epsilon>0\; \exists N\;\forall n\ge N |X_n(\omega) - X(\omega)| < \epsilon \}) = 1 \; \mbox{iff} \; \; \forall \epsilon > 0 \; P(\underset{n \to \infty}{\text{lim inf } A_n^{\epsilon}})=1$$ Being $$\big\{ \bigcap_{n=1}^{\infty}A_n^\epsilon \big\}_n$$ an increasing sequence, $$\big\{ \bigcap_{n=i}^{m}A_n^\epsilon \big\}_m$$ a decreasing sequence and knowing that $$\bigcap_{n=i}^{\infty}A_n^\epsilon = \bigcap_{m=1}^{\infty}\bigcap_{n=i}^{m}A_n^\epsilon$$, from continuity of measures from above and from below I obtained: $$P(\underset{n \to \infty}{\text{lim inf } A_n^{\epsilon}}) = P\big(\cup_{n=1}^{\infty}\cap_{i=n}^{\infty}A_{i}^{\epsilon} \big) = \lim_{n \to \infty}P(\cap{}_{i=n}^{\infty} A_i^\epsilon)=\lim_{n \to \infty}P(A_n^\epsilon)=1$$

• The RHS of your iff statement depends on $n$ and the left doesn't. If you mean to have a limit in front of the $P$ on the RHS, then the RHS is convergence in probability, which is strictly weaker than almost sure convergence. As for the proof, it is correct that a.s. convergence is equivalent to $\forall \epsilon >0\; P(\liminf_n A_n^\epsilon) =1,$ but after that nothing you write makes any sense to me. To start, many of the expressions have obvious problems similar to the issue I pointed out about your iff statement. – spaceisdarkgreen Mar 14 at 11:02
• Thanks, I have updated the RHS of the iff statement. Yes I know that convergence in probability does not imply a.s. convergence. This is why I asked for help in spotting the error. – zeroKnowl Mar 14 at 12:00
• The $\liminf$ of a sequence of sets is not a double intersection, it is the union of an intersection: $\liminf_{n\rightarrow\infty} A_n = \cup_{n=1}^{\infty} \cap_{j=n}^{\infty} A_j$. en.wikipedia.org/wiki/Set-theoretic_limit – Michael Mar 14 at 12:04
• Thanks, I made it more explicit. – zeroKnowl Mar 14 at 12:54