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Let $\mu$ be the Lebesgue measure.

Let $A$ be a Lebesgue-nonmeasurable set.

Define $S:=\{E\subset A : E\text{ is Lebesgue measurable}\}$.

Does there exists a nonmeasurable set $A$ satisfying $\forall E\in S, \mu(E)=0$?

John. p
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  • let $B$ be a measurable set of measure 3, say disjoint from $A$. Then $A\cup B$ is not measurable but has a measurable subset of positive measure – user126154 Mar 05 '14 at 13:07
  • @user126154 Edited:) – John. p Mar 05 '14 at 13:10
  • It seems to me that the argument used by user126154 is still applicable, am I wrong? – alex Mar 05 '14 at 13:22
  • @alex How so? $A\cup B$ has a measure 3 subset – John. p Mar 05 '14 at 13:26
  • In other words, I'm asking whether there exists a nonmeasurable boundary between null sets and positive measures – John. p Mar 05 '14 at 13:28
  • This question was according to be satisfactorily answered in another thread: http://math.stackexchange.com/questions/88449/the-subset-of-non-measurable-set . The name Vitali set might not be familiar but the standard construction of a nonmeasurable set using countably many disjoint Cantor sets is probably familiar can be taken a more concrete example (see Stein Real analysis p. 25) – Squid Mar 05 '14 at 13:48

1 Answers1

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An arbitrary non-measurable subset $A$ with inner Lebesgue measure zero has the property indicated by John in his question. Such are for example, Vitali and Bernstein sets.

George
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  • Assuming the continuum hypothesis you can do even better (or worse, depending on your point of view), namely, you can construct a nonmeasurable set $A$ such that every measurable subset of $A$ is countable. – bof Mar 06 '14 at 05:33