High-density interval for outer measure ($m^{\ast }(E\cap I)>\lambda |I|,0<\lambda <1$)

For any $E\subseteq \mathbb{R}$ with positive Lebesgue outer measure and any $0<\lambda <1$, some bounded nonempty open interval $I$ satisfies the displayed inequality. The set $E$ need not be a Lebesgue measurable set.

To prove this, first choose a bounded portion $F=E\cap [-N,N]$ with $0<a=m^{\ast }(F)<\infty$, using countable subadditivity. Choose a countable open-interval cover $(I_j)$ of $F$ with ${\sum }_j|I_j|<a/\lambda$. If $m^{\ast }(E\cap I_j)\le \lambda |I_j|$ for every $j$, thena contradiction. For measurable $E$ this gives the high-density interval in a positive-measure subset of the real line without requiring the full Lebesgue density theorem.