Show that if \(\sum \mu(E_k) < \infty\) then almost every \(x\in X\) is in at most finitely many \(E_k\).
Solution
Let \((X, \mu)\) be a measure space and \(E_k\) measurable. The set of points lying in infinitely many \(E_k\) is \(\limsup_k E_k = \bigcap_{n\ge1}\bigcup_{k\ge n} E_k\), which is measurable.
1. \(\mu(\limsup_k E_k) \le \sum_{k\ge n}\mu(E_k)\) for every \(n\).
\(\limsup_k E_k \subseteq \bigcup_{k\ge n} E_k\), and countable subadditivity gives \(\mu(\bigcup_{k\ge n} E_k) \le \sum_{k\ge n}\mu(E_k)\).
2. Q.E.D.
Since \(\sum_k \mu(E_k) < \infty\), the tails \(\sum_{k\ge n}\mu(E_k)\) tend to \(0\), so step 1 gives \(\mu(\limsup_k E_k) = 0\).