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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.

First intersect $E$ with a sufficiently large bounded interval to obtain a finite positive-measure set. By the Lebesgue density theorem, this set has a density-one point $x$. Hence, for every $\epsilon >0$, some sufficiently small interval $I=(a,b)$ centred at $x$ satisfieswhich is the high-density interval in a positive-measure subset of the real line.

To obtain the final claim, choose $\epsilon <1/2$, put $L=b-a$, and translate $A=E\cap (a,b)$ into $[0,L]$. WritePart (f), after rescaling the interval length, givesThus $E-E$ contains an open interval around zero, which is the Steinhaus theorem.