outer-measure and $\mu^∗$-measurable

Question

Let $\mu^∗$ be an outer-measure on $X$ induced by a pre-measure $µ_0$ on an algebra $A$. Suppose that $E$ is $\mu^∗$-measurable, and there exist $A, B \subseteq X$, such that $E = A \cup B$ and $\mu^∗(E) = \mu^∗(A) + \mu^∗(B)$. I have to prove that $A, B$ are both $\mu^∗$-measurable.

Answer 1

When $\mu^*(E) = +\infty$, your claim fails, since you can let $A = E$ and $B$ be an arbitrary nonmeasurable set in $E$. However, if $E$ of finite measure, you can use the following facts:

1. For every $S \subset X$, there is a measurable $E$ such that $S \subset E$ and $\mu^*(S) = \mu^*(E)$.

2. If $E \subset S \subset F$ where $E$ and $F$ are both measurable and $\mu^*(E) = \mu^*(F) < +\infty$, then $S$ is also measurable.

Let $C, D \subset E$ be measurable sets such that $A \subset C$, $B \subset D$, and $\mu^*(A) = \mu^*(C)$, $\mu^*(B) = \mu^*(D)$. Then $E\setminus C \subset B$, $E\setminus D \subset A$ and $\mu^*(E \setminus C) = \mu^*(B)$, $\mu^*(E \setminus D) = \mu^*(A)$.

From $E \setminus C \subset B \subset D$ and $\mu^*(E\setminus C) = \mu^*(D)$ we can see $B$ is measurable. Similarly $A$ is measurable.