Let $A\subseteq \mathbb{R}$ be a Lebesgue-measurable set. We say that $A$ is *locally $\varepsilon$-dense* if for any $\varepsilon > 0$, there are $x<y\in\mathbb{R}$ such that $$\frac{\mu(A\cap[x,y])}{y-x} \geq 1-\varepsilon,$$ where $\mu$ denotes the Lebesgue measure on $\mathbb{R}$.
Clearly, if $A$ has positive measure, then $A$ is locally $\varepsilon$-dense. Does the converse hold?



Does a subset of positive measure in $\mathbb{R}$ locally "almost" have density $1$?
$A=[0,1]$ has positive measure and is not locally $\varepsilon$-dense for $(x,y)=(3,4)$ ...


For the converse, yes and even more $\mathbf{R}\setminus A$ has measure $0$. And you only need the estimate for one single $\varepsilon$ :


$\mathbf{1}\_A$ is locally integrable si almost every $x\in \mathbf{R}$ is a Lebesgue point, that is  

\begin{align\*}
\lim\_{r\rightarrow 0^+} \frac{1}{2r}\int\_{[x-r,x+r]} \mathbf{1}\_A = \mathbf{1}\_A(x).
\end{align\*}
The local $\varepsilon$-density imposes that $\mathbf{1}\_A>0$ almost everywhere.


