$m^*(A \cup B) = m^*(A) + m^*(B) $ implies $A$ and $B$ are measurable?

Question

In the proof, it says that for disjoint $A$ and $B$, $m^*(A \cup B) = m^*(A) + m^*(B) \implies A$ and $B$ are measurable.

The caratheodory definition of measurable set is that for some $A$, if $m^*(A) = m^*(A \cup B) + m^*(A \setminus B)$, then $B$ is measurable. But I can't see the relation between these two. Could you explain this?

Thank you in advance.

Answer 1

For every pair of disjoint sets $A$ and $B$, assume the following holds:

$m^*(A\cup B) = m^*(A) + m^*(B).$

Now, let $A$ and $B$ be arbitrary sets. Then $A = (A \cap B) \cup (A\setminus B)$ where $A\cap B$ is disjoint from $A\setminus B$.

By our assumption, this means that $m^*(A) = m^*(A\cap B) + m^*(A\setminus B)$, and so by Caratheodory we have that $B$ is measurable since $A$ was arbitrary. Unfortunately, out choice of $B$ was also arbitrary, so every set is measurable!

Answer 2

That holds. You get to be so careful about where outer measure is defined on. We start with a measurable space (X,F) and let Γ be an outer measure on (X,F) and let μ be an measure on it. Let's go through what we've learned.

Γ: 2^X→[0,+∞] holds Γ(∪i E_i) ≦ ΣiΓ(E_i) (subadditivity) as for E_i∈F i=1,2,..., regardless of whether E_i and E_i' (i≠i') are disjoint or not.

μ: F→[0,+∞] holds: μ(∪i E_i) = ΣiΓ(E_i) (σ-additivity) as for E_i∈F i=1,2,..., obly E_i and E_i' (i≠i') are disjoint.

There are a lot of constructions on measure, but one defines measure μ as a restriction of outer measure Γ from 2^X to F. As we already see Γ has subadditivity, but measure μ which is a restriction of outer measure Γ has σ-additivity. So as long as an outer measure Γ holds finite additivity and we pick up disjoint sets E_i and E_i' (i≠i') in F, E_i and E_i' are measurable.

Sorry for being illegible since I am a newbie here.