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Given a positive real number construct a non measurable set of that outer measure

There are infinitely many non measurable sets but we cannot find the Lebesgue outer measure of a non-measurable set unless it is known explicitly.

Then how can we construct a non-measurable set of given Lebesgue outer measure? I don't want to use Bernstein sets since I'm not aware of them, but somebody suggested them to me.

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marked as duplicate by Mirko, Batman, Matthew Conroy, user642796 May 25 '17 at 3:35

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It's easy to show that for every positive real $r$, there is a non-measurable set $M$ of outer measure $r$. Fix some non-measurable set $S\subset [0, 1]$. Since all sets with null outer measure are measurable, the outer measure of $S$ is positive - call it "$\epsilon$".

The important observation is that outer measure scales: for any set $A$ and any positive real $t$, the outer measure of $tA=\{ta: a\in A\}$ is exactly $t$ times the outer measure of $A$. (This is a good exercise - think about how to scale an open cover of $A$ . . .) So the set $S_r=\{{rs\over\epsilon}: s\in S\}$ has outer measure $r$.

This gives an explicit construction relative to a starting $S$. This is arguably the best we can do, though, since there are no explicit descriptions of non-measurable sets in the first place (they all involve invoking the axiom of choice in a nontrivial way - indeed, without the axiom of choice it is consistent that all sets are measurable).

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