# Thread: limits superior and inferior

1. ## limits superior and inferior

suppose
$\lim sup_{n\to\infty}A_n=\bigcap_{n=1}^{\infty}\bigcup_ {m\ge n}A_m$
and
$\lim inf_{n\to\infty}A_n=\bigcup_{n=1}^{\infty}\bigcap_ {m\ge n}A_m$

where $\bigcup_{m\ge n}A_m=A_n\cup A_{n+1}\cup ....$ and $\bigcap_{m\ge n}A_m=A_n\cap A_{n+1}\cap ....$

Then given that $\lim inf_{n\to\infty}A_n\in \mathcal{A}$ and $\lim sup_{n\to\infty}A_n\in \mathcal{A}$

If $A_n\rightarrow A$, i.e. $\lim_{n\to\infty}1_{A_n}=1_A$, so $1_{\text{lim sup}_{n\to\infty}A_n}=\lim sup_{n\to\infty}1_{A_n}=1_A=\lim inf_{n\to\infty}1_{A_n}=1_{\text{lim inf}_{n\to\infty}A_n}$

Then how do I show that $\lim sup_{n\to\infty}1_{A_n}-\lim inf_{n\to\infty}1_{A_n}=1_{(\text{lim sup}_nA_n\setminus \text{lim inf}_nA_n)}$

where $A\setminus B=A\cap B^c \text{ whenever } B\subset A$.

I was wondering since $\lim sup_{n\to\infty}1_{A_n} \text{ and } \lim inf_{n\to\infty}1_{A_n}$ are the same, why wouldn't the end result goes to 0?

2. Hello,

There's no element telling you that limsup and liminf are the same. The equality you want to prove isn't dependent on the fact that $A_n\to A$.

As to prove the equality, when does $1_{(\limsup_n A_n\setminus \liminf_n A_n)}$ equal 1 and equal 0 ? What does it imply for the LHS ? It's not that hard.

3. Originally Posted by Moo
Hello,

There's no element telling you that limsup and liminf are the same. The equality you want to prove isn't dependent on the fact that $A_n\to A$.

As to prove the equality, when does $1_{(\limsup_n A_n\setminus \liminf_n A_n)}$ equal 1 and equal 0 ? What does it imply for the LHS ? It's not that hard.
Ok. It is obvious that: $\lim inf_{n\to\infty}A_n\subset \lim sup_{n\to\infty}A_n$

So when $\lim sup_{n\to\infty}1_{A_n}-\lim inf_{n\to\infty}1_{A_n}$, I will get $A \cap B^c$ letting $A= \lim sup_{n\to\infty}1_{A_n} \text{ and } B= \lim inf_{n\to\infty}1_{A_n}$.

Is this the correct interpretation?