outer measure has countable subadditivity

[[concept]]

Topics

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Theorem

Let $\{A_n\}$ be any countable collection of subsets of $\mathbb{R}$. Then $m^{\ast }\left({\bigcup }_n{A}_n\right)\le {\sum }_n{m}^{\ast }(A_n)$

Proof

If there is any $n$ such that $m^{\ast }(A_n)=\infty$ (ie, if any of the $A_n$ are unbounded) or ${\sum }_n{m}^{\ast }(A_n)=\infty$ then the inequality is true.

So consider only when all $m^{\ast }(A_n)<\infty$ and ${\sum }_n{m}^{\ast }(A_n)<\infty$.

So let $\{I_{n_k}{\}}_{k\in \mathbb{N}}$ be a collection of open intervals with

A_{n} &\subset \bigcup_{k \in \mathbb{N}} I_{n_{k}} \\ \sum_{k=1}^\infty \ell(I_{n_{k}}) &< m^*(A_{n}) + \frac{\epsilon}{2^n} \end{align}$$ Then we have $$\begin{align} \bigcup_{n \in \mathbb{N}} A_{n} & \subset \bigcup_{n \in \mathbb{N}, k \in \mathbb{N}} I_{n_{k}} \\ \implies m^*\left( \bigcup_{n} A_{n} \right)&\leq \sum_{n,k} \ell(I_{n_{k}}) \\ &= \sum_{n} \sum_{k} \ell(I_{n_{k}}) \\ &< \sum_{n} m^*(A_{n}) + \frac{\epsilon}{2^n} \\ &= \left[ \sum_{n} m^*(A_{n}) \right] + \epsilon \end{align}$$ So letting $\epsilon \to 0$, we get the desired result. $$\tag*{$\blacksquare$}$$

References

References

• Boole’s Inequality

See Also

• outer measure

Mentions

Mentions

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