 Solution

This requires an argument using Lebesgue outer measure, since $E$ need not be a Lebesgue measurable set. Some bounded portion $F=E\cap [-N,N]$ has $a=m^{\ast }(F)>0$: otherwise countable subadditivity would give $m^{\ast }(E)=0$. Also $a\le 2N<\infty$.

Suppose, for a contradiction, that every nonempty bounded open interval $I$ satisfies $m^{\ast }(E\cap I)\le \lambda |I|$. Because $a/\lambda >a$, the definition of Lebesgue outer measure gives a countable cover $F\subseteq {\bigcup }_j{I}_j$ by nonempty bounded open intervals withUsing countable subadditivity again,a contradiction. Hence at least one interval in such a cover satisfiesThis high-density interval for outer measure applies even to nonmeasurable sets; the argument does not require the Lebesgue density theorem.