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

I've been asked this question and am not able to construct it.

I am well aware of the construction of a Vitali set, and I know that their outer measure are not known explicitly.

I can construct a set dense in any interval by taking the union of the standard non measurable set with rationals of that interval but that doesn't serve my purpose because measure of countable set is zero and doesn't do anything good to the required result.


Suppose we wish to find a non-Lebesgue-measurable set of outer measure $\alpha > 0$.

Take a Vitali set $V \subseteq [0,1]$. We don't know what its outer measure is, but we know that it has positive outer measure, say $\beta > 0$.

Now consider the set $\{ \frac{\alpha}{\beta} x : x \in V \}$. This set is not Lebesgue measurable, and using the below fact we may conclude that it has outer measure $\alpha$.

Fact. Let $A \subseteq \mathbb R$, and let $\alpha > 0$. Setting $\alpha A = \{ \alpha x : x \in A \}$, then $\lambda^* ( \alpha A ) = \alpha \lambda^* (A)$, where $\lambda^*$ denotes the Lebesgue outer measure.

Proof sketch. Note that if $\{ (a_n,b_n) : n \in \mathbb N \}$ is a family of open intervals covering $A$, then $\{ ( \alpha a_n , \alpha b_n ) : n \in \mathbb N \}$ is a family of open intervals covering $\alpha A$, and clearly $l ( ( \alpha a_n , \alpha b_n ) ) = \alpha (b_n-a_n) = \alpha l ( ( a_n,b_n )$ for all $n$. It will follow that $\lambda^* ( \alpha A ) \leq \alpha \lambda^* ( A )$. The opposite inequality can be similarly demonstrated (or else just use the fact that $\frac{1}{\alpha} ( \alpha A ) = A$).


Use a Bernstein set $A$. Then the outer measure of $A \cap [0,s]$ is $s$, for any $s>0$.

What's application of Bernstein Set?

  • $\begingroup$ Is there not any simpler way for my question....?? M not able to understand the Bernstein set..... $\endgroup$ – user364168 May 24 '17 at 12:08

Start with a nonmeasurable set $A\subset[0,1]$ of outer measure $m^*(A)=a.$

If $b\gt a,$ then $B=A\cup[1,1+b-a]$ is a nonmeasurable set of outer measure $m^*(B)=b.$

If $0\lt b\lt a,$ choose $n\in\mathbb N$ so that $\frac1n\lt b,$ and define $A_k=A\cap[\frac{k-1}n,\frac kn]$ for $1\le k\le n.$ Since $A=A_1\cup\cdots\cup A_n,$ at least one of the sets $A_k$ is nonmeasurable. Since $A_k$ is a nonmeasurable subset of $[0,1]$ with $m^*(A_k)\le\frac1n\lt b,$ we are back in the previous case; i.e., the set $B=A_k\cup[1,1+b-m^*(A_k)]$ is a nonmeasurable set of outer measure $m^*(B)=b.$


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