# Property of $\mu*$-measurable sets

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.

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.
• 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 $A\subset \cup A_n$ – SlyxBrd Feb 12 at 10:02
• @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. – Song Feb 12 at 15:42