I am having troubles in understanding a part of the proof of the part $1$ Borel Cantelli lemma.
Let $(\Omega, \mathcal{F}, P)$ a probability space and $(A_n)$ a sequence in $\mathcal{F}$. $$\sum_{n = 1}^{+\infty} P(A_n) < +\infty \implies P(\limsup_{n\to +\infty} A_n) = 0$$
Proof
According to my notes:
We recall that $\limsup_n A_n = \cap_k \cup_{n\geq k} A_n$. Thus we have $\omega \in\limsup_n A_n$ iff $\omega$ appears in infinitely many of the $A_n$. This shows we have $\limsup_n A_n = \{ \sum_{n = 1}^{+\infty} \mathbb{1}_{A_n} = +\infty\}$.
Via Beppo Levi Theorem we have:
$$\int\sum_{n = 1}^{+\infty} \mathbb{1}_{A_n} dP = \sum_{n = 1}^{+\infty} \int \mathbb{1}_{A_n} dP = \sum_{n = 1}^{+\infty} P(A_n) < +\infty$$
Now since $\sum_{n = 1}^{+\infty} \mathbb{1}_{A_n} < +\infty$ almost everywhere, then $P(\limsup_{n\to +\infty} A_n) = 0 follows.
What I don't understand
Principally, this: "since $\sum_{n = 1}^{+\infty} \mathbb{1}_{A_n} < +\infty$ almost everywhere, then..."
It's not clear to me why this is true and how we got this.
Also, some clarification about the statement
"$\omega \in\limsup_n A_n$ iff $\omega$ appears in infinitely many of the $A_n$. This shows we have $\limsup_n A_n = \{ \sum_{n = 1}^{+\infty} \mathbb{1}_{A_n} = +\infty\}$"
what is $\omega$? What does this really tell me?
Thank you!
Since $\int \sum A_n dP < \infty $ that means the integrand is finite almost everywhere (otherwise we would have $\int \sum A_n dP = \infty$).
$\omega$ is one of the possible out comes from the sample space $\Omega$. The fact that $\omega \in \limsup A_n = \cap_n \cup_{k \geq n} A_k$ means that for every $n \in \mathbb{N}$ (the intersection part) there is some $k \geq n$ so that $\omega \in A_k$ (the union part). This is equivalent to that $\omega \in A_n$ for infinitely many $n \in \mathbb{N}$.
Now if $\omega \in A_n$ for infinitely many (i.e. $\omega \in \limsup A_n)$ then $\sum_n 1_{A_n}(\omega) = \infty$ since $1_{A_n} (\omega) = 1$ for everytime $\omega \in A_n$, which is infinitely many times. So $\limsup A_n = \{ \omega \in \Omega : \sum_n 1_{A_n}(\omega) = \infty \}$