measure of union of nested sets converges to measure of limiting set

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Theorem (continuity of measure)

Suppose ${E_{k}}$ is a countable collection of measurable sets such that $E_{1} \subset E_{2} \subset \dots$ Then

m (⋃_{k = 1}^{\infty} E_{k}) = lim_{n \to \infty} m (⋃_{k = 1}^{n}) = lim_{n \to \infty} m (E_{n})

Proof

The second equality is because $E_{n} = ⋃_{k = 1}^{n} E_{k}$ . So it is enough to just show $m (⋃_{k} E_{k}) = lim_{n \to \infty} m (E_{n})$ .

We can do this by showing the countable union as the countable disjoint union (recall that algebras have closure under finite disjoint countable unions).

So define $F_{1} = E_{1}$ and $F_{k} = E_{k} ∖ E_{k - 1}$ . Each $F_{k}$ is measurable since $F_{k} = E_{k} \cap E_{k - 1}^{c}$ and the collection ${F_{k}}$ is disjoint. Then for all $n \in N$ , we have

\begin{aligned} ⋃_{k = 1}^{n} F_{k} = E_{n} & ⋃_{k = 1}^{\infty} F_{k} = ⋃_{k = 1}^{\infty} E_{k} \\ ⟹ m (⋃_{k = 1}^{\infty} E_{k}) & = \sum_{k = 1}^{\infty} m (F_{k}) \\ = lim_{n \to \infty} \sum_{k = 1}^{n} m (F_{k}) \\ = lim_{n \to \infty} m (⋃_{k = 1}^{n} F_{k}) \\ = lim_{n \to \infty} m (E_{n}) \end{aligned}

Since measure of finite disjoint measurable sets is the sum of the measures. Thus we have shown the desired equality.

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References

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measure of union of nested sets converges to measure of limiting set

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