Let $\mu:\mathcal A\to[0,\infty]$ be a measure, where $\mathcal A$ is an algebra and let $\mu^*:\mathcal P(X)\to[0,\infty]$ be the outer measure generated by $\mu$. $\Big($i.e. $\mu^*(E)=\inf\Big\{\sum_{i=1}^\infty \mu(A_i):E\subset \bigcup_{i=1}^\infty A_i,\;\ A_i\in\mathcal A\;\ \forall i\in\Bbb N\Big\}$ $\Big)$

Let $E\in\mathcal A^*$ with $\mu^*(E)<\infty$, so it follows that: $$\forall \epsilon>0\;\ \exists\ A_{\epsilon}\in \mathcal A\;\text{such that}\;\ \mu^*(E\triangle A_{\epsilon})<\epsilon $$

$\big($where $\mathcal A^*=\big\{E\subset X: \mu^*(B)=\mu^*(B\cap E)+\mu^*(B\cap E^C)\;\ \forall B\subset X\big\}$$\big)$

So I started:

Let $\epsilon>0$, so since I have to find or construct some set $A_{\epsilon}\in\mathcal A$ such that $\mu^*(E\triangle A_{\epsilon})<\epsilon$ I managed to get that: (looking for $A_{\epsilon}$)

$$\mu^*(E\triangle A_{\epsilon})=\mu^*(E\setminus A_{\epsilon}\ \cup\ A_{\epsilon}\setminus E)\le \mu^*(E\setminus A_{\epsilon})+ \mu^*(A_{\epsilon}\setminus E)\le \mu^*(E)+ \mu^*(A_{\epsilon}\setminus E)=\mu^*(E)+ \mu^*(A_{\epsilon})-\mu^*(E)=\mu^*(A_{\epsilon})$$

$$\Rightarrow\;\ \mu^*(E\triangle A_{\epsilon})\le \mu^*(A_{\epsilon})$$ And got stuck here since I think I have to construct $A_{\epsilon}\in \mathcal A$ in such a way that $\mu(A_{\epsilon})<\epsilon\;$(since $\mu^*(A_{\epsilon})=\mu(A_{\epsilon})$ by hypothesis), but can't figure out how. Any idea would be appreciated.

  • $\begingroup$ Your equality $\mu^*(A_{\epsilon}\setminus E)=\mu^*(A_{\epsilon})-\mu^*(E)$ is wrong. This only holds for measureable sets where one is contained in the other. And even if your conclusion would hold this does not mean that you have to construct a set with measure less than $\epsilon$. It can also be possible that your inequality is just useless. You should focus on trying to construct a covering of $E$ by sets in $\mathcal{A}$ that are as close to $E$ in measure as possible. Then use the properties of your algebra to contstruct one set that is also close in measure and belongs to $\mathcal{A}$ $\endgroup$ – KoliG Nov 25 '15 at 13:54
  • $\begingroup$ You're right, my bad. Forgot that point. Ok I think I'm getting the idea, but whats the precise meaning of two sets being $close\; in\; measure$. $\endgroup$ – Arnulf Nov 25 '15 at 14:45
  • $\begingroup$ Of course this is not a precise formulation but only an idea. What I mean is that, e.g., from your definition of outer measure, you can construct a sequence $A_n$ such that $\mu^*(E)$ and $\sum_{n\in \mathbb{N}}\mu(A_n)$ differ only by $\epsilon$. Since set differences are in the algebra, you can even construct a sequence of disjoint sets that satisfies this property. The (uncountable) union is unfortunately not anymore in the algebra necessarily but I hope this is still the right approach. $\endgroup$ – KoliG Nov 25 '15 at 15:11
  • $\begingroup$ Yes, actually I was working on that and think I got it. Will post it un a bit. $\endgroup$ – Arnulf Nov 25 '15 at 16:07
  • $\begingroup$ I've post it, what do you think? $\endgroup$ – Arnulf Nov 25 '15 at 21:10

So, since $\epsilon>0$ we can get $(A_i)_{i\in\Bbb N}$ an $\mathcal A-$cover of $E$ such that:

$$\sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$

and it's clear that $\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)$, so we get that:

$$\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$

thus $$0\le \sum_{i=1}^\infty \mu(A_i)-\mu^*(E)<\frac{\epsilon}{2}$$

So, $\exists\;N\in\Bbb N$ such that $\sum_{i=N+1}^\infty\mu(A_i)<\frac{\epsilon}{2}$, so let $A=\bigcup_{i=1}^N A_i\in\mathcal A\Rightarrow\ E\setminus A \subset \bigcup_{i=N+1}^\infty A_i$

$$\Rightarrow \mu^*(E\setminus A)\le \sum_{i=N+1}^\infty \mu(A_i)<\frac{\epsilon}{2}$$

Then, $$\mu^*(A\setminus E)=\mu^*(\bigcup_{i=1}^N A_i\setminus E)\le\mu^*(\bigcup_{i=1}^\infty A_i\setminus E)=\mu^*(\bigcup_{i=1}^\infty A_i)-\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)-\mu^*(E)<\frac{\epsilon}{2}$$

thus $$\mu^*(E\triangle A)\le\mu^*(E\setminus A)+\mu^*(A\setminus E)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

What do you think?


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