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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.

user642796
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user364168
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    Your question is a little confused. If a set is nonmeasurable it can't be assigned a measure. As a totally wild guess, are you thinking of generalized Cantor sets? https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set – user4894 May 25 '17 at 01:57
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    If you want to ask a follow up question I thing it would be best to give a link to your previous question https://math.stackexchange.com/questions/2294723/non-measurable-set-of-given-outer-measure – clark May 25 '17 at 02:12

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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).

Noah Schweber
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