Outer measure, Caratheodory measure - proof

Question

Let $m^*$ be an outer measure on a set $X:=[a, b]$. $A \subset X$ is a null set, i.e., $m^*(A) =0$. If $E \subset X$ is measurable, show that

$m^*(E\cup A)+m^*(E\cap A) = m^*(E)+m^*(A)$

I'm pretty sure I need to use the Caratheodory extension to show this, but not sure. Any help would be great!

You know the outer measure is subadditive, right? And monotone? Won't that be enough? — GEdgar, Apr 24 '13 at 12:44

score 2 · Accepted Answer · answered Apr 24 '13 at 12:00

Note that the equation is equivalent to $m^*(E \cup A) = m^*(E)$, because $A$ and thus $E \cap A$ are null sets. But this equation holds, since $m^*(E \cup A) \leq m^*(E) + m^*(A) = m^*(E) \leq m^*(E \cup A)$, using subadditivity and monotonicity. In particular, $E$ need not be measurable for this.

Outer measure, Caratheodory measure - proof

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