Convergene of a sequence if and only if limsup and liminf agree This is in terms of sets (within the context of probability measure theory) if that matters. The book says "this is equivalent" without proof, I want to prove the gap!
Indicator function of a set $A$ is defined as: $\mathbb{I}_A(\omega)=1$ if $\omega\in A$  else $=0$ defined over all of $\Omega$, basically it's a function that is 1 over $A$ and 0 elsewhere.
Let $\{A_n\}$ be a sequence of subsets of $\Omega$ inside the $\sigma$-algebra $\mathcal{A}$ (this just means it is closed under countable union and intersections)
$A_n\rightarrow A$ ($A_n$ is said to converge) if $\lim_{n\rightarrow\infty}(\mathbb{I}_{A_n})=\mathbb{I}_A$
Limsup and liminf are defined as follows
$\limsup_{n\rightarrow\infty}(A_n)=\cap^\infty_{n=1}(\cup_{i=n}^\infty A_i)$
$\liminf_{n\rightarrow\infty}(A_n)=\cup^\infty_{n=1}(\cap_{i=n}^\infty A_i)$ 
The claim is $A_n\rightarrow A\iff A=\limsup(A_n)=\liminf(A_n)$
I've tagged this "Real analysis" because that's where I learnt about limits, and consider the definition of them to be under its umbrella.
What have I tried?
I'm struggling to get started, I'm not sure if I should be looking at the limit of the indicator function pointwise or as a function, and if so how to consider it in those separate lights, and then use the information from limsup and liminf to show the result.
So I need to more carefully define the limit of the indicator function (does it start: $\forall\omega\in\Omega\forall\epsilon>0$ or does it start $\forall\epsilon>0\forall\omega\in\Omega$) then get that in terms of the limsup and liminf.
I can see glimpses of the proof but evidently cannot formulate it entirely. I'd love some clearing up.
I've spotted similar questions just now and I shall read those, I would none the less like to know how I should be treating convergence of a function, point-wise or not.
 A: Assume $A_n \to A$ in the sense that $\mathbb{I}_{A_n}(\omega) \to \mathbb{I}_{A}(\omega) \quad \forall \omega \in \Omega$.
The sequence which has only zeros and ones as its terms will converge if and only if the sequence eventually becomes either zeros (that is, after some $n$ all the terms will be zeros) or ones 
Now, assume $\omega \in A$. Then $\mathbb{I}_{A}(\omega) = 1$ which means there exists $n$ such that $\omega \in A_k \quad \forall k > n$ which implies that $A \subset \liminf A_n$.
Now assume that $\omega \in A^c$. Then $\mathbb{I}_{A}(\omega) = 0$ which means there exists $n$ such that $\omega \in A_k^c \quad \forall k > n$ which implies that $A^c \subset \liminf A_n^c$ which means $\limsup A_n \subset A$.
Using the above two results, we have $A = \limsup A_n = \liminf A_n$
The other direction of the result follows using similar ideas.
A: Suppose that your $A_i$, $A$ are all subsets of a set $X$. Suppose first that for each $x\in X$, we have that $1_{A_n}(x)\longrightarrow 1_A(x)$. Since $1_{A_n}$ takes the value $0$ or $1$, this means that for each $x\in X$ there is $N(x)=N$ such that $1_{A_n}(x)$ is constant for $n>N$ with value $1$ or $0$, and it equals $1_A(x)$. It is always the case that $A_*\subseteq A^*$ (I'm being lazy to denote the limit inferior and limit superior). Thus, it suffices to show that that $A^*\subseteq A_*$, that is, it suffices we show that if there is an infinite subset $S\subseteq \Bbb N$ such that $x\in \bigcap_S A_{n}$; then $x\in A_*$.   
Now, since $S$ is infinite, we can pick $M\in S$ larger that $N(x)$. This means that $1=1_{A_n}(x)=1_{A_N}(x)$ for every $n>N$ (because we know that the sequence is constant!), so that $x\in \bigcap_{n\geqslant N}A_n$, so $x\in A_*$. In fact $A=A*=A_*$.
For the converse, it should be immediate to see that if $A=A*=A_*$ then $\lim\limits_{n\to\infty} A_n= A$ 
