# Connection between almost sure convergence and lim sup and lim inf

I noticed other similar questions about this, but I'm going in circles. Am I correct that the following are equivalent ways of showing $X_n$ a sequence of random variables converges almost surely to zero? Assume $X_n$ greater than or equal to $0$ for all $n$ for ease of notation, and limsup's and liminf's are taken as n goes to infinity.

1. Show that $P(\limsup \{X_n > \epsilon\}) = 0$
2. Show that $P(\liminf {X_n < \epsilon}) = 1$, this being the complement of (1), correct?

Is it correct to interpret this as: if the limit of $X_n$ exists, say $0$, then $$\liminf X_n = \limsup X_n = \lim X_n = 0$$ so statements (1) and (2) above are really just saying that, for sufficiently large $n$, the probability of $X_n$ exceeding arbitrarily small $\epsilon$ is zero, or, equivalently, that the probability of $X_n$ being less than or equal to $\epsilon$ is one. But this sounds exactly like the definition of convergence in probability, but proving this will not guarantee almost sure convergence, hence, my issue.

I appreciate any help!

• I'm a bit confused. "But this sounds exactly like the definition of convergence in probability, but proving this will not guarantee almost sure convergence, hence, my issue." Are you saying that proving convergence will not guarantee almost sure convergence? – Omnomnomnom Nov 17 '13 at 21:24
• I mean that proving convergence in probability does not imply convergence almost surely. – shoeburg Nov 18 '13 at 2:59

The complement of $\limsup \{X_n > \epsilon\}$ is $\liminf \{X_n \leqslant \epsilon\}$, which contains $\{\limsup X_n\lt\epsilon\}$ and is included in $\{\limsup X_n\leqslant\epsilon\}$. Equivalently, $$\{\limsup X_n \gt \epsilon\}\subseteq\limsup \{X_n > \epsilon\}\subseteq\{\limsup X_n \geqslant \epsilon\}.$$ An example to show that the first inclusion may be strict is $\{\forall n,X_n=\epsilon+1/n\}$. An example to show that the second inclusion may be strict is $\{\forall n,X_n=\epsilon-1/n\}$.
• @Did : In $\limsup\{X_n>\epsilon\}$, the subtlety does not arise from the fact that $\mathbb N$ is countable ? – Al Bundy Nov 16 '16 at 7:19
• @Did : Do you mean that $n$ tend to $+\infty$($\notin \mathbb N$) in the version of $\limsup$ for sets ? – Al Bundy Nov 16 '16 at 9:34