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Let $\mu$ an almost measure such that $\mu(\Omega) < \infty$. Fix $A \subseteq \Omega$.

I'm required to show that if $\mu(\Omega) = \mu^*(A) + \mu^*(A^C)$, then $A$ is $\mu^*$-measurable, that is equivalent to show that, for every $E \subseteq \Omega$, $\mu^*(E \cap A)+\mu^*(E \setminus A) \leq \mu^*(E)$.

I don't know if I'm missing some result to successfully prove the desired result. Here is what I've done:

Let $E \subseteq \Omega$. Observe that $$\mu^*(E \cap A) + \mu^*(E \setminus A) \le \mu^*(A) + \mu^*(A^C) = \mu(X) = \mu^*(X) = \mu^*(E \cup E^C) \le \mu^*(E) + \mu^*(E^C)$$ since $(E \cap A) \subseteq A$ and $(E \setminus A) \subseteq A^C$ and $\mu^*$ is monotone. Note that $\mu^*(E^C) < \infty$, since $\mu(\Omega) < \infty$.

So I have tried to show that:

$$\mu^*(E \cap A) + \mu^*(E \setminus A) + \mu^*(E^C) \le \mu^*(A) + \mu^*(A^C)$$

Set theoretic, the equality holds, but not necessarily with $\mu^*$.

I would appreciate any hints or other required results. Intuitively I require some result using $\epsilon > 0$. Thanks!

Arturo Magidin
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Blue Tomato
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  • What is an "almost measure"? – copper.hat Mar 10 '24 at 21:18
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    @copper.hat For $\mathscr{A}$ an algebra of $\Omega$, $\mu$ is almost measure if $\mu: \mathscr{A} \to \overline{\mathbb{R}}$ yields $\mu(\emptyset) = 0$, $\mu(A) \ge 0$, for all $A \in \mathscr{A}$ and for a disjoint sequence, if the union is element of $\mathscr{A}$, then $\mu$ is $\sigma-$additive on that specific union. – Blue Tomato Mar 10 '24 at 21:22
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    @copper.hat sorry mate, the correct term is "premeasure". Sorry, not my native language. – Blue Tomato Mar 10 '24 at 23:52

1 Answers1

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I think you have too many $^*$'s in your statement. You want to show $A$ is $\mu-$measurable, otherwise writing $\mu^*(\Omega)=\mu^*(A)+\mu^*(A^c)$ doesn't make sense.

To do this, let $A\subset B,A^c\subset C, B,C\in\mathscr{A}$ be such that $$\mu^*(A)\leq\mu(B)\leq \mu^*(A)+\varepsilon,$$ $$\mu^*(A^c)\leq\mu(C)\leq\mu^*(A^c)+\varepsilon.$$ Then $\mu(\Omega)=\mu^*(A)+\mu^*(A^c)\leq \mu(B)+\mu(C)$. But notice $B\cap C=\emptyset$ by construction, so this gives $\mu(\Omega)=\mu(B\cup C)$.

From this you can show $A,B,$ and $C$ all differ from one another by a set of $\mu^*$-measure 0. Hence $A$ is $\mu-$measurable.

Nick
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