# Is outer measure a measure?

In my text book Lebesgue measure is shown to have countable additivity on disjoint sets, i.e. if $A_k$ is a countable sequence of disjoint measurable sets, then $\mu(\bigcup A_k)=\sum\mu(A_k)$.

Thus Lebesgue measure is a measure with an additional property that the empty set has zero measure.

But for outer measure, my text book never says outer measure is a measure, and it does not say anything if outer measure has countable additivity on disjoint sets. So my question is does outer measure has such countable additivity? If not, is there a counter example?

• An outer measure is usually defined on all subsets of the underlying space. In order to obtain a measure, one typically restricts to measurable subsets that are defined using Caratheodory's condition: $E$ is measurable if and only if $\mu(A) = \mu(A \cap E) + \mu(A \cap E^c)$ for all $A \subset X$. – J. J. Jul 17 '15 at 13:28

## 2 Answers

The following seems to be a perfect match to this question. In the following, $|*|$ denotes a Lebesgue measure, and $|*|_e$ denotes an outer measure. The outer measure does not satisfy the countable additivity on disjoint sets because of the existence of non-measurable sets. As a result, outer measure is not a measure. • would you please write down the book name containing this page!? – mwomath Nov 18 '17 at 17:56

You need to be careful about whether you're talking about measurable sets or any sets at all! There's confusion about this already in your first sentence Lebesgue measure is shown to have countable additivity on disjoint sets, i.e. if $A_k$ is a countable sequence of disjoint measurable sets...; you need to add the word "measurable" to the first clause.

You ask whether outer measure is countably additive, and precisely because of this confusion I'm not sure whether you're asking about additivity for any disjoint sets or just for measurable sets. The answer to your question is yes or no, depending. Outer measure is countably additive on measurable sets. It is not countably additive on sets in general.

You ask about a counterexample. It's not possible to give a simple example of non-additivity in general. Because the example has to involve non-measurable sets, and even proving that a non-measurable set exists is not trivial...