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Let $\mu$* be an outer measure on X. Let $E$ be $\mu$*-measurable set so that there exist disjoint $A,B$ such that $E=A\cup B$ and $\mu$ * $(E)$ = $\mu$ * $(A)$ + $\mu$ * $(B)$. Show that $A,B$ are $\mu$*-measurable.

I've tried playing with the identities but I can't seem to get anywhere.

for any $Y\subset X$ we know that $\mu$ * $(Y)$ = $\mu$ * $(Y\cap E)$ + $\mu$ * $(Y\cap E^c)$. how do I continue from here?

Putting $A^c$ in gives: $\mu$ * $(A^c)$ = $\mu$ * $(A^c\cap E)$ + $\mu$ * $(A^c\cap E^c)=\mu$ * $(B)$ + $\mu$ * $(E^c)$ but I still don't know how to continue.

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Let $\epsilon>0$ be given. Then by the definition of outer measure, there exist measurable sets $ U\supset A$ and $V\supset B$ such that $$ \mu^*(U)<\mu^*(A)+\frac{\epsilon}2,\quad \ \ \mu^*(V)<\mu^*(B)+\frac{\epsilon}2. $$ Using $E\subset U\cup V$, we can find that $$\begin{align*} \mu^*(U\cap V)&=\mu^*(U)+\mu^*(V)-\mu^*(U\cup V)\\&\le\mu^*(U)+\mu^*(V)-\mu^*(E)\\&<\mu^*(A)+\mu^*(B)+\epsilon-\mu^*(E)\\&=\epsilon. \end{align*}$$

Assume arbitrary $Y\subset X$ is given. Because $E$ is measurable, we may write $$ \mu^*(Y\cap A)+\mu^*\left(Y-A\right)=\mu^*(Y\cap A)+\mu^*\left((Y\cap E)-A\right)+\mu^*\left(Y-E\right). $$ Let $Z = Y\cap E\subset E$. Then, we obtain the following: $$\begin{align*} \mu^*(Y\cap A)+\mu^*\left((Y\cap E)-A\right)&=\mu^*(Z\cap A)+\mu^*(Z-A)\\&\le \mu^*(Z\cap U) +\mu^*(Z \cap B)\\ &=\mu^*(Z)-\mu^*(Z-U)+\mu^*(Z\cap V)\\&\le \mu^*(Z)+\mu^*\left((Z\cap V)-(Z-U)\right)\\ &\le \mu^*(Y\cap E) +\mu^*(V\cap U)\\&< \mu^*(Y\cap E) +\epsilon. \end{align*}$$ This gives $$ \mu^*(Y\cap A)+\mu^*\left(Y-A\right)<\left( \mu^*(Y\cap E)+\epsilon\right)+\mu^*\left(Y-E\right)= \mu^*(Y)+\epsilon. $$ Since $\epsilon>0$ is arbitrary, we get for every $Y\subset X$, $$ \mu^*(Y\cap A)+\mu^*\left(Y-A\right)\le \mu^*(Y). $$ The other direction ($\ge$) is obvious from countable subadditivity of $\mu^*$. Thus this establishes measurability of $A$, and that of $B=E-A$ as a difference of measurable subsets.

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  • $\begingroup$ This is amazing, Thank you. I'm having a bit of trouble understanding the first part where you derive that from the definition of outer measure there is a measurable U so that $\mu*(U) < \mu*(A) + \epsilon$, the definition talks about a sum of functions so that $\sum p(A_n) < \mu*(A) + \epsilon$ where $A\subset \cup A_n$ $\endgroup$ – SlyxBrd Feb 12 at 10:02
  • $\begingroup$ @SlyxBrd Then, by letting $U = \cup_{n\ge 1}A_n$, we get $\mu^*(U)\le \sum_{n\ge 1}\mu^*(A_n)<\mu^*(A)+\epsilon$ as wanted. I hope this makes it clear. $\endgroup$ – Song Feb 12 at 15:42
  • $\begingroup$ It does, thank you! $\endgroup$ – SlyxBrd Feb 12 at 15:44

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