we can always find an open set with outer measure slightly more

[[concept]]

Topics

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Theorem

for every $A\subset \mathbb{R}$ and $ϵ>0$, there exist an open set $O$ such that $A\subset O$ and $m^{\ast }(A)\le m^{\ast }(O)\le m^{\ast }(A)+ϵ$

Proof

The result is obvious if $m^{\ast }(A)=\infty$ (just take $O=\mathbb{R}$), so we assume $m^{\ast }(A)<\infty$.

Let $\{I_n\}$ be a collection of open intervals that cover $A$ and have total length at most $m^{\ast }(A)+ϵ$. Then $O={\bigcup }_n{I}_n$ is open and obviously $A\subset O$. Thus by subadditivity ( theorem from Lecture 6 )

m^*(O) &= m^*\left( \bigcup_{n}I_{n} \right) \\ &\leq \sum_{n} m^*(I_{n}) \\ &\leq \sum_{n} \ell(I_{n}) \\ & \leq m^*(A) + \epsilon \end{align}$$

References

References

See Also

Mentions

Mentions

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