measure satisfies countable additivity

Theorem

Suppose that ${E_{n}}$ is a countable collection of disjoint, measurable sets. Then

m (⋃_{n} E_{n}) = \sum_{n} m (E_{n})

Note

We already showed countable subadditivity of measure, but now we are showing equality for our nice sets.

Proof

We know that the set $\cup_{n} E_{n}$ is measurable since measurable sets form a sigma algebra and are therefore closed under countable unions. So we already have (from countable subadditivity of measure) that

m (⋃_{n} E_{n}) = m^{*} (⋃_{n} E_{n}) \leq \sum_{n} m^{*} (E_{n}) = \sum_{n} m (E_{n})

So we just need to show the other way. For any $N \in N$ , since measure of finite disjoint measurable sets is the sum of the measures, we have

\begin{aligned} m (⋃_{n = 1}^{N} E_{n}) & = m^{*} (R \cap ⋃_{n = 1}^{N} E_{n}) \\ = \sum_{n = 1}^{N} m^{*} (R \cap E_{n}) \\ = \sum_{n = 1}^{N} m^{*} (E_{n}) \\ = \sum_{n = 1}^{N} m (E_{n}) \\ (*) & = m (⋃_{n = 1}^{N} E_{n}) \\ \leq m (⋃_{n = 1}^{\infty} E_{n}) \end{aligned}

Where $(*)$ is because of the disjointness of the $E_{n}$ . Now we have a bound over all $N$ , and taking $N \to \infty$ we get the desired result.

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References

See Also

Mentions

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