# What would be an example?

1. Jul 16, 2009

### symbol0

I read that the outer measure is not countably additive, but I can only think of cases where the outer measure is countably additive.
Can anybody give me an example where the outer measure is not countably additive?

Thanks

2. Jul 16, 2009

### quasar987

Take A= [-1,1] and B=[0,2]. Then AuB=[-1,2], so mes(AuB)=3. But mes(A)+mes(B) = 2+2=4.

Indeed, you will always have mes(A_1 u ... u A_n)< mes(A_1)+...+mes(A_n) as soon as some pair of A_i intersect in a set of positive measure...

3. Jul 16, 2009

### symbol0

mmm?
the definition of countably additive only considers disjoint sets.

4. Jul 16, 2009

### JCVD

I think the OP wants a countable collection of pairwise disjoint sets such that the outer measure of their union is strictly less than the sum of their outer measures. As for Lebesgue Outer Measure, you can create a Vitali set V contained in [0,1] (Vitali sets are non-measurable, and no two of their members differ by a rational number). Since V is non-measurable subset of [0,1], its outer measure d must be strictly greater than 0 but no greater than 1. The set {r_k} of rational numbers in [0,1] is countable, so {V+r_k} is a countable collection of pairwise disjoint sets, with V+r_k (translating every element of V by r_k to the right) a subset of [0,2] with outer measure d for each k. Thus the outer measure of the union of any collection of V+r_k cannot exceed 2. If it were the case, however, that the outer measure of any such union equaled the sum of the outer measures, then if we chose N>2/d and looked at a collection of N different sets V+r_k, we would have the outer measure of their union equal to Nd, which is strictly greater than 2, providing our contradiction.

This counterexample would be horrendously difficult to think up on your own, and the construction of Vitali sets requires the Axiom of Choice.

5. Jul 16, 2009

### symbol0

I see.
Thank you Mr. Van Damme