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edited Apr 13, 2017 at 12:58

This question arose a few years back when I was an assistant teacher on a course of basic (Lebesgue) measure theory, but I didn't find an answer or anyone able to solve the problem. The setting of the problem is as follows:

We say that $A$ has full outer measure in the unit interval $[0,1]$ if $$m^\ast(A) = m^\ast([0,1]).$$ How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable. In this MO post this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question.

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?

We say that $A$ has full outer measure in the unit interval $[0,1]$ if $$m^\ast(A) = m^\ast([0,1]).$$ How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable. In this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question.

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?

We say that $A$ has full outer measure in the unit interval $[0,1]$ if $$m^\ast(A) = m^\ast([0,1]).$$ How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable. In this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question.

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?

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asked Jan 23, 2013 at 11:22

Rami Luisto

Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?

We say that $A$ has full outer measure in the unit interval $[0,1]$ if $$m^\ast(A) = m^\ast([0,1]).$$ How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable. In this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question.

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?