For increasing $E_n$, $\mu(E_n)\to\mu(\bigcup E_n)$; for decreasing $E_n$ with $\mu(E_1)<\infty$, $\mu(E_n)\to\mu(\bigcap E_n)$.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space and let \(E_n\in\mcm\) for \(n\geq1\) [@Fol13].
If \(E_1\subseteq E_2\subseteq\cdots\) and \(E\coloneqq\bigcup_{n\geq1}E_n\), then \(\mu(E_n)\to\mu(E)\).
If \(E_1\supseteq E_2\supseteq\cdots\), \(\mu(E_1)<\infty\), and \(E\coloneqq\bigcap_{n\geq1}E_n\), then \(\mu(E_n)\to\mu(E)\).
The hypothesis \(\mu(E_1)<\infty\) in (b) is needed: for Lebesgue measure \(m\) on \(\RR\) and \(E_n\coloneqq[n,\infty)\), \(m(E_n)=\infty\) for all \(n\) but \(\bigcap_n E_n=\emptyset\).
State continuity of measure from above and from below.