I am working on a problem regarding an extension of the Borel-Cantelli lemma that goes as follows:
Let $E_1, E_2, …$ be an arbitrary sequence of sets. It is known that $\lim_{n \to \infty}P(E_n) = 0$ and $\sum_n P(E_n \cap E_{n+1}^c) < \infty$. Show that $P(E_n \text{ infinitely often})=0$
I have tried a few different attempts at the answer, with the most current one given below:
\begin{align*}
&P(\limsup_n E_n)&&\\
&= P(\cap_{n=1}^\infty \cup_{k=n}^\infty E_n)&&\\
&= \lim_{n \nearrow \infty} P(\cup_{k=n}^\infty E_n)&&\text{intersection is monotone decreasing}\\
&= \lim_{n \nearrow \infty} P\bigg(\cup_{k=n}^\infty \big[[E_n \cap E_{n+1}^c] \cup [E_n \cap E_{n+1}]\big]\bigg)&&\\
&= \lim_{n \nearrow \infty} P\bigg(\big(\cup_{k=n}^\infty [E_n \cap E_{n+1}^c]\big) \cup \big(\cup_{k=n}^\infty [E_n \cap E_{n+1}]\big)\bigg)&&\\
&= \lim_{n \nearrow \infty} P\bigg(\big(\cup_{k=n}^\infty [E_n \cap E_{n+1}^c]\big) \cup \big(\cup_{k=n}^\infty E_n \cap \cup_{k=n}^\infty E_{n+1}]\big)\bigg)&&\\
&= \lim_{n \nearrow \infty} P\bigg(\big(\cup_{k=n}^\infty [E_n \cap E_{n+1}^c]\big) \cup [\cup_{k=n}^\infty E_{n+1}]\bigg)&&\\
&= \lim_{n \nearrow \infty} \bigg[P\big(\cup_{k=n}^\infty [E_n \cap E_{n+1}^c]\big)+ P\big(\cup_{k=n}^\infty E_{n+1}\big)\bigg]&&\text{disjoint additivity}\\
&= \lim_{n \nearrow \infty} P\big(\cup_{k=n}^\infty [E_n \cap E_{n+1}^c]\big)+ \lim_{n \nearrow \infty} P\big(\cup_{k=n}^\infty E_{n+1}\big)&&\\
&\leq \lim_{n \nearrow \infty} \sum_{k=n}^\infty P\big(E_n \cap E_{n+1}^c\big)+ \lim_{n \nearrow \infty} P\big(\cup_{k=n}^\infty E_{n+1}\big)&&\text{subadditivity}\\
\end{align*}
The main problem with this answer, as well as the other ones I have drafted, is that I can't seem to get rid of the union over $E_n$. Because of this I can't make any statements about the limiting probability of the union going to $0$ (e.g. if $P(E_n) = 1/n$ the conditions of the problem are fulfilled but the probability of the union does not converge).
Am I missing something in the way I am breaking up the sets? I have been banging my head against this for days with no success – any resources you all could suggest/ direction that you could give would be highly appreciated.
Best Answer
Hint:
Show that for any $\omega \in \bigcup_{n \geq k} E_n$ one of the following conditions is satisfied:
a) there exists $N \geq k$ such that $\omega \in E_n$ for all $n \geq N$,
b) there exists $n \geq k$ such that $\omega \in E_{n}$ and $\omega \notin E_{n+1}$.
This gives $$\bigcup_{n \geq k} E_n \subseteq \left( \bigcup_{N \geq k} \bigcap_{n \geq N} E_n \right) \cup \bigcup_{n \geq k} (E_n \cap E_{n+1}^c).$$
It follows from Step 1 that $$\mathbb{P} \left( \bigcup_{n \geq k} E_n \right) \leq \mathbb{P} \left( \bigcup_{N \geq k} \bigcap_{n \geq N} E_n \right) + \sum_{n \geq k} \mathbb{P}(E_n \cap E_{n+1}^c).$$
Since $\bigcap_{n \geq N} E_n$ is increasing in $N$, it holds that $$\mathbb{P} \left( \bigcup_{N \geq k} \bigcap_{n \geq N} E_n \right) = \lim_{N \to \infty} \mathbb{P} \left( \bigcap_{n \geq N} E_n \right)$$ and so $$\mathbb{P} \left( \bigcup_{N \geq k} \bigcap_{n \geq N} E_n \right) \leq \lim_{N \to \infty} \mathbb{P}(E_N)=0.$$
Combine Step 2 and 3 to conclude that $$\limsup_{k \to \infty} \mathbb{P} \left( \bigcup_{n \geq k} E_n \right) \leq \limsup_{k \to \infty} \sum_{n \geq k} \mathbb{P}(E_n \cap E_{n+1}^c)=0$$ which gives $$\mathbb{P}\left( \limsup_{n \to \infty} E_n \right)=0.$$