Exercise 1.4.44 (Borel-Cantelli lemma)

Question

\begin{enumerate}[label=(oman*)]
\item Give an alternative proof of Borel-Cantelli lemma from \href{./exercise_1-4-43}{Exercise 1.4.43} that does not rely on convergence theorems.
\item Give a counterexample that shows that the Borel-Cantelli lemma can fail if the condition $\sum_{i=1}^{\infty}\mu(E_n)<\infty$ is relaxed to $\lim_{n	o\infty}\mu(E_n) = 0$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
\item We look closer at the set $F$ of points in $X$ which are contained in infinitely many $E_n$'s:
\begin{align*}
F &= \left\{x \in X\ |\ \#\{n \in \mathbb{N}: x \in E_n\} = \inftyight\}\
	&= \left\{x \in X\ |\ 	ext{$x\in E_n$ for infinitely many $E_n$} ight\}\
	&= \left\{x \in X\ |\ \forall N \in \mathbb{N}\ \ \exists n\geq N:\ x \in E_n ight\}\
	&= \bigcap_{N=1}^{\infty} \left\{x \in X\ |\ \exists n\geq N:\ x \in E_n ight\}\
	&\subseteq \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} E_n.
\end{align*}
Since $\mu(\bigcup_{n=N}^{\infty}E_n) \leq \sum_{n=1}^{\infty}\mu(E_n)<\infty$, we can apply monotone convergence \href{./exercise_1-4-23}{Exercise 1.4.23}:
$$\mu(F) \leq \mu\left( \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} E_night) = \lim_{N	o\infty} \bigcup_{n=1}^{\infty} E_n \leq \lim_{N	o\infty} \mu(E_N) = 0$$
where the last step is a necessity for the series $\sum_{N=1}^{\infty}\mu(E_N)$ to converge.

\item As for the counterexample, consider $(\mathbb{R}^2,\mathcal{B},m)$ and the shrinking spreading bump sequence 
$$\forall n\in\mathbb{N}: \quad E_n := \left[0,night] 	imes \left[0,\frac{1}{n^2}ight].$$
Then, any point $(0,x)\in\mathbb{R}^2$ is contained in infinitely many of these sets, yet
$$\lim_{n	o\infty} m(E_n) = \lim_{n	o\infty} \frac{1}{n} = 0.$$
\end{enumerate}

Exercise 1.4.44 (Borel-Cantelli lemma)

Answers

Comments

Exercise 1.4.44 (Borel-Cantelli lemma)

Answers

Comments

Add answer