Nonmeasurable Set with Finite Outer Measure

In summary, for a nonmeasurable set ##E## with finite outer measure, there exists a ##G_\delta## set ##G## containing ##E## for which ##m^*(G)=m^*(E)## and ##m^*(G-E) > 0##, contradicting the first theorem and proving that an analogous theorem for ##F_\sigma## sets is also true.
  • #1
Bashyboy
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5

Homework Statement


Let ##E## be a nonmeasurable set of finite outer measure. Show that there is a ##G_\delta## set ##G## that contains ##E## for which ##m^*(E)=m^*(G)##, while ##m^*(G-E) > 0##.

Homework Equations



##E## is a measurable set if and only if there is a ##G_\delta## set ##G## containing ##E## for which ##m^*(G-E)=0##.

Let ##E## have finite outer measure. Show that there is an ##F_\sigma## set ##F## and a ##G_\delta## set ##G## such that ##F \subseteq E \subseteq G## and ##m^*(F)=m^*(E)=m^*(G)##.

The Attempt at a Solution



I pretty certain it wouldn't follow immediately from the two theorems I quoted in section 2, right? In fact, we could prove an analogous theorem for ##F_\sigma## sets, right?
 
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  • #2
Since E is nonmeasurable, if we can find a measurable set G that contains it and has the same outer measure, the result will follow from the first theorem you quote in section 2.

Are you allowed to use the axiom of choice? If so, I suggest the following strategy.
  1. prove that, for any positive integer ##n##, there exists a measurable set ##G_n\supset E## such that ##m^*(G_n)-m^*(E)<1/n##.
  2. find a way to construct a measurable set ##G\supset E## such that ##m^*(G)-m^*(E)=0##.
Part 2 is easier than part 1. But I have a feeling that 1 is provable, given some key properties of outer measures and measurable sets.

If you're not allowed to use axiom of choice, I don't know whether it's still true. If it is, a more subtle proof may be needed, as the axiom is needed for the approach I had in mind for step 2. IIRC, the axiom is usually assumed in measure theory, as the construction of the most famous nonmeasurable sets, the Vitali sets, requires it.
 
  • #3
I don't understand why the axiom of choice is needed. Here is proof I came up with:

Suppose that ##E## is a nonmeasurable set of finite outer measure. Then for every ##G_\delta## set ##G## containing ##E## is such that ##m^*(G-E) > 0## (negation of first theorem. Moreover, since ##E## has finite out measure, there exists a ##G_\delta## set ##G## such that ##m^*(G)=m^*(E)## (second theorem). Hence, we have found a ##G_\delta## set ##G## containing ##E## for which ##m^*(G)=m^*(E)## and ##m^*(G-E) > 0##.
 

Related to Nonmeasurable Set with Finite Outer Measure

What is a nonmeasurable set with finite outer measure?

A nonmeasurable set with finite outer measure refers to a set in a measure space that cannot be assigned a well-defined measure or size, even though it has a finite outer measure. This means that the set cannot be broken down into smaller measurable subsets, making it impossible to accurately determine its size.

Why is it important to study nonmeasurable sets with finite outer measure?

Studying nonmeasurable sets with finite outer measure is important because it challenges our understanding of measure theory and the concept of size. It also has applications in areas such as probability theory and geometric measure theory.

How are nonmeasurable sets with finite outer measure identified?

Nonmeasurable sets with finite outer measure are identified through the use of the Vitali-Hahn-Saks theorem, which states that a set is nonmeasurable if and only if it has a non-measurable subset that is not too large or too small.

What are some examples of nonmeasurable sets with finite outer measure?

One example of a nonmeasurable set with finite outer measure is the Vitali set, which is constructed by taking one representative from each equivalence class in the real numbers under the relation x~y if x-y is a rational number. Another example is the Bernstein set, which is a subset of the real numbers that contains exactly one element from each equivalence class under the relation x~y if x-y is an irrational number.

What are the implications of nonmeasurable sets with finite outer measure?

The existence of nonmeasurable sets with finite outer measure has implications in the foundations of mathematics and the notion of size. It also has practical implications in areas such as probability and integration theory, where the existence of nonmeasurable sets can lead to paradoxes and difficulties in defining certain concepts.

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