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Given the Lebesgue measure on $\mathbb{R}$, I am wondering if a non-measurable set can always be included in a null set? More precisely, let $A$ be a set in the Borel σ-algebra $\mathcal{B}(\mathbb{R})$, to which we assign a strictly positive Lebesgue measure. Given a theorem in measure theory, it contains non-measurable sets.

Can we say, according to some theorem (please state the theorem) that all the non-measurable sets contained in the set $A$ are also contained in some measurable subset $B$ of $A$ to which we can assign the value zero? I know that this is possible for null sets by defining the negligeable sets and eventually the completion of the measurable space.

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  • $\begingroup$ No, because if $A$ is non-measurable, then $A^c$ is non-measurable too $\endgroup$ – Tryss Apr 29 '15 at 8:58
  • $\begingroup$ A is in fact measurable as a borel set. $\endgroup$ – ivo Apr 29 '15 at 9:30
  • $\begingroup$ My $A$ is not the Borel set $A$, I should have called this set $C$. Sorry for the confusion $\endgroup$ – Tryss Apr 29 '15 at 12:46
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No, it is not possible to say this. In fact, the Lebesgue measure is a complete measure in that all subsets of measure zero sets are already measurable (and assigned measure 0). You shouldn't think of non-measurable sets as being "small," but as being "nasty enough that we can't reason about their size."

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  • $\begingroup$ Thanks Mike. Can we generalize your statement ? Given any measure on a Sigma-Algebra, one can assign the measure zero only to subsets of measure zero sets through the concept of completion but not to all subsets of a non zero measurable set ? $\endgroup$ – ivo Apr 29 '15 at 9:47
  • $\begingroup$ I am also wondering, if it is possible to find a subset A of a non-measurable set B such that A can be assigned the measure zero or more particularly a non zero value ( in the case of the Lebesgue measure then a strictly positive value ) ? $\endgroup$ – ivo Apr 29 '15 at 9:52
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Every subset $A$ of a null set $N$ is itself null. This can be shown directly by observing that every covering of $N$ also covers $A$.
Note. Every measurable set of positive outer measure has a non measurable subset.

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