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LEt $(X, \mathcal{F})$ be measurable space equiped with measure $\mu$. Suppose $\{ F_n \}$ are in $\mathcal{F}$. Put $A = \bigcap_{n=1}^{\infty} ( \bigcup_{i \geq n} F_i ) $. Does it follow that

$$ \mu(A) \leq \mu\left( \bigcup_{i \geq n } F_i \right)$$

Is it required that the $F's$ be pairwise disjoint?

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If $\mu$ is a positive measure, it is monotone; that is, whenever $B \subseteq C$, we have

$$\mu(B) \le \mu(C)$$

This implies the first claim, regardless of any assumption about disjointness.

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To begin with, let's put $E_n=\bigcup_{i\ge n}F_i.$ As a countable union of $\mathcal F$-measurable sets, each $E_n$ is again $\mathcal F$-measurable, so $A=\bigcap_{n=1}^\infty E_n$ is also $\mathcal F$-measurable, as a countable intersection of $\mathcal F$-measurable sets.

Now, since $(X,\mathcal F,\mu)$ is a measure space, then for any $B,C\in\mathcal F$ with $B\subseteq C,$ we have $\mu(B)\le\mu(C).$ Hence, we see that...what?

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For any $n$, you have $$A\subseteq\bigcap_{i\ge n} F_i.$$

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