Let \((X, \mathcal M, \mu)\) be a measure space and suppose \(\theset{E_n} \subset \mathcal M\) satisfies \[ \lim _{n \rightarrow \infty} \mu\left(X \backslash E_{n}\right)=0. \]
Define \[ G \definedas \theset{x\in X \suchthat x\in E_n \text{ for only finitely many } n}. \]
Show that \(G \in \mathcal M\) and \(\mu(G) = 0\).
Solution
Set \[ F_N:=\bigcap_{n=N}^\infty E_n^c. \] A point belongs to only finitely many \(E_n\) exactly when it belongs to \(E_n^c\) for all sufficiently large \(n\), so \[ G=\bigcup_{N=1}^\infty F_N. \] Thus \(G\in\mathcal M\).
Fix \(N\). For every \(m\ge N\), \[ F_N\subseteq E_m^c, \] and hence \[ 0\le \mu(F_N)\le \mu(E_m^c)=\mu(X\setminus E_m). \] Letting \(m\to\infty\) gives \(\mu(F_N)=0\). Therefore \[ \mu(G)\le \sum_{N=1}^\infty\mu(F_N)=0. \] Hence \[ \boxed{G\in\mathcal M\quad\text{and}\quad \mu(G)=0.} \] No summability hypothesis on the numbers \(\mu(X\setminus E_n)\) is needed.