Show that the complement of a non-measurable set is also n.m

1. Sep 25, 2016

cragar

1. The problem statement, all variables and given/known data
Show that the complement of a non-measurable set is also non-measurable.
3. The attempt at a solution
Let A be a set that is non measurable. Let B be the complement of A.
Lets assume for contradiction that B has a measure. Now from the axioms of measure theory they have countable additive of measure. case 1: A+B has a measure.
If B has a measure and A+B has a measure then A+B-B should have a measure. But A has no measure so this is a contradiction therefore B has no measure.
Case 2: A+B has no measure.
Lets assume B has a measure, If B has a measure then A+B has a measure
But this is a contradiction, therefore B has no measure .
Not sure If on case 2 I can talk about A+B having a measure

2. Sep 25, 2016

andrewkirk

It depends on the definition of measurable used. The usual definition is as shown here. Under that definition, all sets in the sigma-algebra $\Sigma$ are measurable and, since a set is in $\Sigma$ iff its complement is, the result follows trivially.

I suspect you are using a different definition of measurable - perhaps a specific measure like Lebesgue measure. There is a list of possible measures here. Do you mean one of them? If so, which one?

3. Sep 26, 2016

cragar

When my teacher told us about the complement of a non-measurable set is non-measurable. I wanted to prove this for myself, he talked about it before Lebesgue measure. And before sigma-algebra. We just had the basic axioms of measure theory like countably additive, the measure of a point is zero. the measure of an interval is the difference between the endpoints. Like for example the measure of the irrationals is infinite.

4. Sep 26, 2016

micromass

Staff Emeritus
A measure is a function. What is the domain of the measure defined like?

5. Sep 26, 2016

cragar

It can be no more than a countable union of disjoint sets. If the measure exists it will be assigned a non-negative real number. And it is closed under compliment.
If its closed under compliment then the set and its compliment will have a measure. Because the set and its compliment is just one partition of the set. thanks for the responses