# An additivity property of outer measure

By definition, as set $E$ is measurable if for any set $C$ $$\mu^\ast(C) = \mu^\ast(C\cap E) + \mu^\ast(C \cap E^c).$$

Is it true that if either $A$ or $B$ is measurable, then we can't have

$$\mu^\ast(A\cup B) \lt \mu^\ast(A) + \mu^\ast(B)?$$

Please if it is true can you offer a proof? I've tried but failed.

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Take a non-measurable subset $B$ of a measurable set $A$... Or even easier, let $A = B$ be of positive outer measure. – t.b. Mar 30 '12 at 19:23
Of course it's possible: take $A=[0,1]$ and $B=[0.5,1.5]$. Then $\mu^*(A\cup B) = 1.5$, $\mu^*(A)=1$, and $\mu^*(B)=1$. – Arturo Magidin Mar 30 '12 at 19:28
@ArturoMagidin: So the claim is false... – Dave Mar 30 '12 at 19:34
@Dave: Yes, it's quite false. Or as t.b. suggested, any measurable $A$ of nonzero outer measure will give an example that the strict inequality can indeed hold. What made you think it couldn't? – Arturo Magidin Mar 30 '12 at 19:39
I guess you want the sets to be disjoint. – Michael Greinecker Mar 30 '12 at 19:39

Without assuming that $A$ and $B$ are disjoint, this is wrong for silly reasons, as was pointed out in the comments: if $A$ has positive (outer) measure and $B$ overlaps with $A$ in a non-trivial way, we always have $\mu^\ast (A \cup B) \lt \mu^\ast (A) + \mu^\ast(B)$. For a specific example, we may take $A = [0,2]$ and $B=[1,2]$ so that $2 = \mu^\ast(A \cup B) \lt 3 = \mu^\ast[0,2]+\mu^\ast[1,2]$ with respect to Lebesgue outer measure.
However, if $A$ is measurable and $B$ is disjoint from it, we can take $C = A \cup B$ and $E = A$ and use Carathéodory's criterion (the definition of measurability you gave) together with $C \cap A = A$ and $C \cap A^c = B$ to see that $$\mu^\ast (A \cup B) = \mu^\ast(C) = \mu^\ast (C \cap A) + \mu^\ast(C \cap A^c) = \mu^\ast(A) + \mu^\ast(B),$$ which is the desired equality.