The additivity of outer measure for disjoint sets, one of these is Borel set.

Question

Denote Lebesgue outer measure $\mu^\ast$.

I want to show

if $A,B\subset \mathbb R$ are disjoint and one of these sets is Borel set, then $\mu^\ast(A\cup B)=\mu^\ast(A)+\mu^\ast(B)$ holds. $\cdots \bigstar$

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I know two facts that

if $A,B\subset \mathbb R$ are disjoint and one of these sets is open, then $\mu^\ast(A\cup B)=\mu^\ast(A)+\mu^\ast(B)$

and

if $A,B\subset \mathbb R$ are disjoint and one of these sets is closed, then $\mu^\ast(A\cup B)=\mu^\ast(A)+\mu^\ast(B)$.

I thinks these facts seem to work for $\bigstar$, but I didn't go well at present.

This is the step for showing that $\mu^\ast$ is a measure on $(\mathbb R, \mathcal L)$ so I cannot use the additivity of $\mu^\ast.$

Thanks for any idea.

Answer 1

This is proven in Theorem 2.66 of Sheldon Axler's Measure, Integration and Real Analysis. In addition to the facts you listed, he uses the additional one that for any Borel set $B$, there exists a closed set $F \subset B$ such that $\mu^{*}(B\backslash F) < \varepsilon$ for all $\varepsilon >0$. The latter he proves in Theorem 2.65. In his book $|A|$ denotes the outer measure of $A$. Hope it helps!