Exercise 1.4.43 (Borel-Cantelli lemma)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space, and let $E_1,E_2,\ldots$ be a sequence of $\mathcal{B}$-measurable sets such that
$$\sum_{n=1}^{\infty}\mu(E_n) < \infty.$$
Show that almost every $x\in X$ is contained in at most finitely many of the $E_n$.

Elu Thingol · Accepted Answer

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
Using the equivalence between the measure of a set and the integral of its indicator function (\href{./exercise_1-4-33}{Exercise 1.4.33}-ii) and applying Corollary 1.4.45 (Tonelli's theorem for sums and integrals) we obtain
$$\sum_{n=1}^{\infty}\mu(E_n) = \sum_{n=1}^{\infty} \intx{1_{E_n}} = \intx{\sum_{n=1}^{\infty} 1_{E_n}} < \infty.$$
By \href{./exercise_1-4-35}{Exercise 1.4.35} (vii) this implies that $\sum_{n=1}^{\infty} 1_{E_n}(x)$ is finite for almost every $x\in X$. In other words, almost every $x$ is contained in finitely many $E_n$'s.

Exercise 1.4.43 (Borel-Cantelli lemma)

Answers

Comments

Exercise 1.4.43 (Borel-Cantelli lemma)

Answers

Comments

Add answer