I have the following exercise:
Let $(\Omega,\mathcal A, P)$ be a probability space and let $A_n \in \mathcal A$ ($n \in \mathbb N$) be events with $\sum_{n\in \mathbb N} P(A_n) < \infty$. Show that we can find a random variable $n^*:\Omega \rightarrow \mathbb N_0$, such that almost surely none of the events $A_n$ with $n \ge n^*$ will occur.
So I think it follows directly with Borel-Cantelli:
Because of $\sum_{n\in \mathbb N} P(A_n) < \infty$, we know that
\begin{align}
P(\limsup\limits_{n\rightarrow \infty} A_n)=P(\bigcap\limits_{n\in \mathbb N} \bigcup\limits_{m\ge n} A_m)& =P(\{ \omega \in \Omega \mid \forall n \in \mathbb N: \exists m \ge n: \omega \in A_m \})=0\\ &\Leftrightarrow P(\{ \omega \in \Omega \mid \exists n \in \mathbb N: \forall m \ge n: \omega \notin A_m \})=1
\end{align}
With $n^*(\omega)=n$, this solves the exercise.
So, am I right here? Because it seems a bit to easy. But it is true that $n$ depends on $\omega \in \Omega$ (i.e. $n=n(\omega)=:n^*(\omega)),$ isn't it?
Maybe someone can look over it and say if I think wrong or if it is correct the way I wrote it here.
Thanks.
As pointed out in the comment you have to make sure that $n^*$ is a random variable, that is, $n^* : \Omega \to \mathbb{N}_0$ is measurable. By Borel-Cantelli lemma we know that if we define $\Omega_0 := \{ \omega \in \Omega | \exists n \in \mathbb{N}: \forall m \geq n :\omega \notin A_m\}$ then $P(\Omega_0) = 1$. Now set $n^*(\omega) := \min \{ n \in \mathbb{N} | \forall m \geq n, \, \omega \notin A_m \}$ if $\omega \in \Omega_0$, and $n^*(\omega) := 0$ if $\omega \notin \Omega_0$. Then $n^*: \Omega \to \mathbb{N}_0 $ is measurable. In fact $\{ n^* = 0 \} = \Omega \setminus \Omega_0$ and for $k \in \mathbb{N}$, $\{n^* = k \} = A_{k-1} \cap \left( \cap_{l=k}^{\infty} A_l^c \right)$ where $A_0 := \Omega$.