Can $\limsup = \liminf$ only almost surely?

Question

Might be a dumb question once you get to Kolmogorov but whether or not $A_n$'s are independent, is it possible

$$\liminf A_n \ne \limsup A_n$$

but $$P(\liminf A_n) = P(\limsup A_n)$$?

Possible cases to consider:

• $A_n$'s are/aren't independent.
• $\liminf A_n$ is/isn't empty.
• $\limsup A_n$ is/isn't empty.
• $\limsup A_n$ has probability 0/probability 1.
• $\liminf A_n$ has probability 0/probability 1.

Answer 1

Both events are 0,1. Notice that by definition if $\omega\in \liminf_nA_n$, then $\omega\in \limsup A_n$. So if $\liminf A_n\neq \limsup A_n$, then this implies there's some set $R$ which belongs to one but not the other. If $P(R)=0$ then everything is fine. But if $P(R)>0$, then this implies $P(\mbox{liminf}A_n)=0$ and $P(\mbox{limsup}A_n)=1$

Answer 2

This is easily possible -- for example, let $X$ be a (single) normal distributed variable and $$ A_n = \begin{cases} X\in(0,1] & \text{when $n$ is odd} \\ X \in[0,1) & \text{when $n$ is even} \end{cases} $$ These $A_n$s are far from independent, so the 0-1 law does not apply here.