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.