measure satisfies countable additivity

[[concept]]

Topics

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Theorem

Suppose that $\{E_n\}$ is a countable collection of disjoint, measurable sets. Then $m\left({\bigcup }_n{E}_n\right)={\sum }_{n}m(E_n)$

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

We already showed

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 $m\left({\bigcup }_n{E}_n\right)=m^{\ast }\left({\bigcup }_n{E}_n\right)\le {\sum }_n{m}^{\ast }(E_n)={\sum }_{n}m(E_n)$ So we just need to show the other way. For any $N\in \mathbb{N}$, since measure of finite disjoint measurable sets is the sum of the measures, we have

m\left( \bigcup_{n=1}^N E_{n} \right) &= m^*\left( \mathbb{R} \cap \bigcup_{n=1}^N E_{n} \right) \\ &= \sum_{n=1}^N m^*(\mathbb{R} \cap E_{n}) \\ &= \sum_{n=1}^N m^*(E_{n}) \\ &= \sum_{n=1}^N m(E_{n}) \\ (*)&= m\left( \bigcup_{n=1}^N E_{n}\right) \\ &\leq m\left( \bigcup_{n=1}^\infty E_{n} \right) \end{align}$$ Where $(*)$ is because of the disjointness of the $E_{n}$. Now we have a bound over all $N$, and taking $N \to \infty$ we get the desired result. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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