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

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Homework Help Overview

The discussion revolves around the properties of non-measurable sets in the context of measure theory, specifically addressing whether the complement of a non-measurable set is also non-measurable.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • The original poster attempts to prove that the complement of a non-measurable set is also non-measurable, exploring contradictions based on the axioms of measure theory. Some participants question the definition of measurability being used, suggesting that different definitions may lead to different conclusions. Others reflect on foundational concepts of measure theory and the implications of countable additivity.

Discussion Status

The discussion is ongoing, with participants exploring various definitions of measurability and the implications of those definitions on the original poster's argument. There is a recognition of the complexity surrounding the topic, and some guidance has been offered regarding the foundational axioms of measure theory.

Contextual Notes

Participants note that the discussion may be influenced by the specific measure being referenced, such as Lebesgue measure, and the basic axioms of measure theory that have been introduced in their studies.

cragar
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Homework Statement


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

The Attempt at a Solution


Let A be a set that is non measurable. Let B be the complement of A.
Let's 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.
Let's 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
 
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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?
 
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.
 
cragar said:
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.

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

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