# Open sets are $\mu*$-measurable in a metric space with a given condition

I have a homework problem that I'm very stuck on. The problem statement is as follows:

"Suppose that $X$ is a metric space, and that for any sets $E,F \subseteq X$, if dist$(E,F) > 0$ then $\mu^*(E \cup F) = \mu^*(E) + \mu^*(F)$. Prove that every open set is a splitting set. (Recall that the distance between subsets $E$ and $F$ of a metric space is defined to be dist$(E,F) = \inf \{ d(x,y) : x \in E, y \in F \}$.)"

Our professor defines a "splitting set" as follows: Let $\mu^*$ be an outer measure on a nonempty set $X$. A set $A \subseteq X$ is called a splitting set if, for all $E \subseteq X$, $\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c)$, where $A^c = X \backslash A$. (This is what Folland's Real Analysis calls a $\mu^*$-measurable set.)

Here are a couple of my (failed) attempts:

My first try was to let $U$ be an arbitrary open set in $X$, let $G \subseteq X$ be arbitrary, and define $E = G \cap U$, $F = G \cap U^c$. If I could somehow show that dist$(E,F) > 0$ in this case, then the result would follow, but in general, this is not true (take $\mathbb{R}$ with the standard metric, let $G = [0,1]$, $U = (-1/2,1/2)$).

My next attempt was by contradiction: suppose there is an open set $U$ such that $U$ is not a splitting set. Then there is some $E \subseteq X$ such that $\mu^*(E) \neq \mu^*(E \cap U) + \mu^*(E \cap U^c)$... and by monotonicity this means that $\mu^*(E) < \mu^*(E \cap U) + \mu^*(E \cap U^c)$. But then I only have one set to work with, and with the assumptions, I need two sets $E,F$ to work with in order to get anywhere.

I also tried exploring what I could do with closed sets, since if dist$(E,F) > 0$, then the closures of $E$ and $F$ respectively are disjoint. But I'm still stuck. Any hints would be appreciated!!! Thanks in advance.

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Here's a strategy for making your second attempt work: since $E = (E\cap U) \cup (E \cap U^c)$, you can conclude that $d(E\cap U, E \cap U^c)=0$. Now you should be able to shrink $U$ just a little, yielding $U'$ so that $d(E \cap U', E \cap U^c)>0$ but still $\mu^{*}(E)<\mu^{*}(E\cap U')+ \mu^{*}(E\cap U^c)$.
I'm confused as to how you arrived at that last inequality... when I try your suggestion, I get that $$\mu^*(E \cap U') + \mu^*(E \cap U^c) = \mu^*((E \cap U') \cup (E \cap U^c)) = \mu^*(E \cap (U' \cup U^c)) \leq \mu^*(E \cap X) = \mu^*(E)$$ ...??? – phaiakia Feb 5 '14 at 4:50
Notice that it's from your second approach, where you strive for contradiction. You already have that $\mu^{*}(E) < \mu^{*}(E \cap U) + \mu^{*}(E \cap U^{C})$, so the suggestion was to find a $U' \subseteq U$ that is just big enough to maintain the inequality $\mu^{*}(E) < \mu^{*}(E \cap U') + \mu^{*}(E \cap U^{C}) \leq \mu^{*}(E \cap U) + \mu^{*}(E \cap U^{C})$, and just small enough to also yield $d(E \cap U', E \cap U^{C}) > 0$. From there, your calculation shows that the inequality cannot be strict, and therein lies your contradiction. – JoeDub Feb 5 '14 at 6:48
Ah ha! So, I can assume without loss of generality that $U$ is an open ball of radius $r$, and then just "shrink" $U$ by $\epsilon$ to get the $U'$ that I want. I think I can go from here. Thanks to both of you! – phaiakia Feb 5 '14 at 15:34
Felt like I should come back and add the correct answer. Let $U$ be an open set in $X$ and consider $U^c$. For each $n$, let $U_n = \{ x \in U : d(x,y) > 1/n \; \forall y \in U^c \}$. Then $\mu^*(E \cap (U_n \cup U^c)) = \mu^*(E \cap U_n) + \mu^*(E \cap U^c)$. Then taking the limit of both sides, continuity from below of the outer measure gives the result.