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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 suppose $m(E)<\infty$. The interval-cover definition of Lebesgue outer measure, together with $m^{\ast }(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that ${\sum }_j|I_j|<m(E)+\epsilon$. If the definition uses intervals that are not open, enlarge each by an amount whose total is smaller than the available error. For the open set $U={\bigcup }_j{I}_j$, countable subadditivity and additivity of Lebesgue measure giveThis subtraction is valid because both measures are finite.

For arbitrary $E$, including $m(E)=\infty$, set $E_n=E\cap [n,n+1)$ for $n\in \mathbb{Z}$, and choose positive weights $w_n=2^{-|n|}/3$, whose sum is one. Each $E_n$ has finite Lebesgue measure, so the preceding argument gives an open set $U_n\supseteq E_n$ with $m(U_n\setminus E_n)<\epsilon w_n$. Then $U={\bigcup }_{n\in \mathbb{Z}}{U}_n$ is open, contains $E$, andThus the required form of regularity of Lebesgue measure does not require $E$ itself to have finite measure.

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.

A set $E\subseteq \mathbb{R}$ is Lebesgue measurable if it satisfies the Caratheodory criterionfor every $A\subseteq \mathbb{R}$, where $m^{\ast }$ is Lebesgue outer measure. Equivalently, $E$ belongs to the completion of the Borel sigma-algebra with respect to Lebesgue measure.