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.


  • 1
    $\begingroup$ You would have to show that the $n^\star$ defined as you did is a random variable. For this you should probably consider choosing a bit more carefully, which would allow you to conclude $\endgroup$ – Max May 20 '17 at 10:48
  • $\begingroup$ So I have to show here, that $n^*$ is measurable ,i.e. $(n^*)^{-1}(B) \in \mathcal A$ $\forall B \in \mathcal P (\mathbb N_0)$. I really don't know how to show it in this setting. Can someone maybe help me and show me how I have to choose $n^*$ to be measurable here? $\endgroup$ – Frodo361 May 20 '17 at 12:35

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 $A_0 := \Omega$.

  • $\begingroup$ Your choice of $n^*$ makes sense. Thank you for answering me and helping me out. $\endgroup$ – Frodo361 May 20 '17 at 13:29

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