Suppose you have a sequence of measurable sets $A_n$ and a measure $\mu$. Then I believe, $\mu(\bigcap_n A_n)=\mu(\inf_{n\in\mathbb N}A_n)\le \inf_{n\in\mathbb N}\mu(A_n)$ holds true.
How does one verify that inequality properly?
2026-03-28 07:16:36.1774682196
Infimum of a measure
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If $A=\bigcap_{n\in\mathbb N} A_n$ then $A\subseteq A_n$ for every $n\in\mathbb N$, whence $\mu A\leq \mu A_n$ for every $n\in\mathbb N$.
That means that $\mu A$ is a lower bound of the set $\{\mu A_n\mid n\in\mathbb N\}$ so that: $$\mu A\leq\inf_{n\in\mathbb N}\mu(A_n)$$