Construction of a set with density of half at $0$.

Question

If we define for a given set $A \subset \Bbb{R}$ and $x\in \Bbb{R}$ the density of $A$ at $x$ being the limit as $[I]$ goes to zero of the ratio $[I \cap A]/[I]$ wherever the limit exists for intervals $I$ containing $x$ with $[.]$ denoting the Lebesgue outer measure, is it possible to construct an $A$ with density of $1/2$ at the point $0$,say?

Answer 1

Consider the following set $$ A = x \in \Bbb{R} - \{0\} : \left\{ 2^{ \left \lfloor \left|\frac{1}{x} \right| \right \rfloor} \sqrt{2} \right\} > \frac{1}{2} $$ where we use the brace notation used by Knuth for the fractional part of a real number $\left\{ r \right\} \equiv r - \left \lfloor r \right \rfloor$. (Note that the expression in the power of two is the absolute value of the floor of $\frac1x$.)

The density of this set at zero is $\frac12$. You can see this in an intuituve (though not rigorous) was by noting that a point $z$ close to zero is in $A$ iff the $n$-th bit after the binary point in the binary expansion of $\sqrt{2}$ is 1, where $n$ is the integer part of $\frac1z$. Thus by getting sufficiently close to zero you can always find an interval $I$ which has $[A \cap I]/[I]$ arbitrarily close to $\frac12$. This, in turn, implies a density of $\frac12$ at zero.

This example does not violate the Lesbegue density theorem because the density is in the interior of $[(0,1)$ only at the isolated point $x=0$, not on a set of non-zero measure.

Answer 2

A more correct definition of the density would be that the density is $d$ if $$[I_n\cap A]/[I_n]$$ for every sequence of intervals $I_n$ such that $0\in I_n$ and $[I_n]\to 0$.

That may well be what you meant. By the definition you gave, if $A$ has the property that the limit never exists then the density is $42$.

Yes, a set with density $1/2$ at the origin exists. Notation: If $J=[a,b]$ is an interval let $$h(J,n) = \bigcup_{j=0}^{n-1}\left[a+\frac{2j}{2n}(b-a),a+\frac{(2j+1)}{2n}(b-a)\right].$$ That is, take $J$, divide it into $2n$ equal subintervals, and let $h(J,n)$ be the union of the first, third, fifth, etc of these subintervals.

Now let $J_n=[2^{-(n+1)},2^{-n}]$, define $$B=\bigcup_{n=1}^\infty h(J_n,n),$$and let $$A=B\cup(-B).$$

Hint why that density is $1/2$: Say $I$ is an interval with $0\in I$. Just to simplify the explanation, assume that $I=[0,\alpha]\subset[0,1/2]$. Now there are infinitely many $n$ with $J_n\subset I$; for such $n$ we have $$[I\cap h(J_n,n)] = \frac12[J_n].$$And there may be $n$ so that $I\cap J_n=\emptyset$; ignore them. There's at most one problematic value of $n$, where $I\cap J_n$ is a nonempty proper subset of $J_n$. For this $n$ you can write $I\cap h(J_n,n)=E\cup F$, where $E\subset J'$ (for a certain interval $J'$) and $$\frac{[E]}{[J']}=\frac12$$and $$[F]\le2^{-(n+1)}/(2n)\le c[I]/n.$$Since $n\to\infty$ as $I\to\{0\}$ the measure of $F$ is small enough it doesn't matter.