“Buried” non-density points in the fat Cantor set

A point $p$ of a Lebesgue measurable set $A \subset \mathbb{R}^n$ is called the Lebesgue density point of $A$ if $\lim_{\epsilon \to 0} \frac{\mu(A \cap B(p,\epsilon))}{\mu(B(p,\epsilon))}=1$ where $B(p, \epsilon)$ denotes the open ball of radius $\epsilon$ about $p$ ; in some sense this means that $A$ takes 100% of the space about $p$ . In general, if the limit above exists (regardless of whether it’s equal to 1 or not), we call it’s value the Lebesgue density of $A$ at $p$ , denoted by $\mathrm{dens}_A(p)$ . Fixing any Lebesgue measurable set $A$ , there’s a theorem called Lebesgue density theorem which states that for almost every $x \in \mathbb{R}$ we have either $\mathrm{dens}_A(x)=1$ or $\mathrm{dens}_A(x)=0$ . The theorem looks reasonable if we think of simple shapes in space, for example if $A$ is a closed filled square in $\mathbb{R}^2$ , then the interior points have density $1$ while the points in the complement have density 0, and the points on the boundary have measure density $\frac{1}{2}$ , but the boundary of the square is a measure zero in $\mathbb{R}^2$ . Restricting our attention to $\mathbb{R}$ and thinking about it for the favorite set in real analysis, Cantor set, it’ll still make sense, because the Cantor set after all has measure zero itself, so those fractions are zero for all of the radii $\epsilon>0$ and point $x \in \mathbb{R}$ . Now one might wonder to twist the situation more by thinking about the fat Cantor set $C$ ; where the set now has some mass in contrast with the Cantor set. Then how do the density and non-density points (i.e. points where that limit is either not equal to 1 or it doesn’t exist) of $C$ look like? The fat Cantor set contains some easy to find Lebesgue non-density points: the boundary points of the open intervals that we remove at each step have density $\frac{1}{2}$ , let such points in $C$ be denoted by $\mathrm{Ends}(C)$ .

Question: Is there any point in $C \setminus \mathrm{Ends}(C)$ that is a non-density point of $C$ ?

The answer is Yes.

Let’s use a number system $\{ R,L \}^\omega$ for representing the points of the fat Cantor set, which is similar to the representation $\{ 0,2 \}^\omega$ for points of the normal Cantor set in the ternary system, in the sense that we make a tree out of our fat Cantor set and each digit $L$ or $R$ will correspond to moving left or right in the tree. Then in this way each sequence $\{ l_k \} \in \{ R,L \}^\omega$ represents exactly one point of our Cantor set, because the points in the set described by $\{ l_k \}$ can be obtained by intersecting a sequence of nested closed (hence compact) intervals, which is non empty, and further that the intervals are shrinking, hence denoting exactly a point. So for example, in our system, the number $R R R\dots$ is the right-most point of $C$ . And it’s easy to see that the boundary points of $C$ correspond to eventually-constant sequences of $\{ L,R \}^\omega$ .

We now begin the construction of a desired point $p \in C\setminus \mathrm{Ends}(C)$ . We consider the natural numbers $k\in \{ 2,3,4, \dots \}$ , and $\delta _x (\epsilon)=\frac{\mu((x-\epsilon,x+\epsilon)\cap C)}{2\epsilon}$ and we want to construct a point

$\displaystyle p=\underbrace{RR\dots R}_{a_1}\underbrace{L L\dots L}_{a_2}\underbrace{RR\dots R}_{a_3}\dots\in C \setminus \mathrm{Ends}(C)$

such that $\delta _p\left( \frac{1}{k_{i}} \right)<51 \%$ for a subsequence $\{ k_{i} \}_{i \in \mathbb{N}}$ of $k$ ‘s. Start with $k_{1}=k=2$ and at first add $a_1$ -many $R$ ‘s in the beginning of the name of $p$ so that we can be sure that any point in $C$ whose first $a_1$ -many digits are $R$ (will be using the notation $\underbrace{RR\dots R}_{a_1\text{-many}}\bigstar$ for them), is close enough to the right-most possible point of $C$ (call it $q$ ) which ensures $\delta _p\left( \frac{1}{k} \right)<51\%$ (this is possible because the right half of the interval $(q- 1 /k, q+ 1 /k)$ contains no points in $C$ , meaning that $\delta _q\left( \frac{1}{k} \right) \leq 50\%$ , so if $p$ is close enough to $q$ , we can have the intended inequality). So any number of the form $\underbrace{RR\dots R}_{a_1\text{-many}}\bigstar$ is working for $k=2$ . Let’s increment $k$ to $3$ . Now it’s possible that any number of the aforementioned form also works for $k=3$ , meaning that for any such $p$ we have $\delta _p(\frac{1}{3})<51 \%$ ; if that happened at any $k$ , increment $k$ another time. If $\delta _p\left( \frac{1}{k} \right)<51\%$ for all of the $k\geq {3}$ , meaning that we can increment $k$ indefinitely, then already any point in $\underbrace{RR\dots R}_{a_1\text{-many}}\bigstar$ that is not a boundary point is an ideal point; so as boundary points correspond to eventually-constant $\{ R,L \}^\omega$ -numbers, we can for example set the rest of the digits of $p$ as alternating between $R$ and $L$ every other time. So now assume that we can’t increment $k$ indefinitely and there is some $k_{2}^\mathrm{temp} \in \mathbb{N}$ such that not for all $p \in \underbrace{RR\dots R}_{a_1\text{-many}}\bigstar$ we have $\delta _p\left( \frac{1}{k_{2}^\mathrm{temp}} \right)<51\%$ . Surely we can remedy that by adding more $R$ ‘s at the beginning of our current general form for $p$ , but if we keep doing that with all $k$ ‘s, then the point $p$ we are constructing would become the right-most boundary point, which we have wanted to avoid. So whenever $\delta _p\left( \frac{1}{k} \right)$ is not always $<51\%$ , it’s a good time to make an alteration in our general form. Now if we let $q$ to be the left-most point of $\underbrace{RR\dots R}_{a_1\text{-many}}\bigstar$ , (which is just the left boundary point of the right-most interval at the $a_2$ -level of the construction of the fat Cantor set where $2^{a_1}$ intervals are present), we check whether $(q-1 /k_{2}^\mathrm{temp}, q)\cap C$ is empty or not; if it’s non empty then increment $k_{2}^\mathrm{temp}$ again and evaluate $(q-1 /k_{2}^\mathrm{temp}, q)\cap C$ again; this incrementation eventually stops at some $k$ , calling it $k_{2}$ , because $q$ is an endpoint. When $(q-1 /k_{2}, q)\cap C = \varnothing$ , we proceed to add $a_2$ -many $L$ ‘s to our general form so that (by a similar argument as above) for any $p \in \underbrace{RR\dots R}_{a_1\text{-many}}\underbrace{L L\dots L}_{a_2\text{-many}}\bigstar$ , we have $p$ close enough to $q$ and hence $\delta _p\left( \frac{1}{k_{2}} \right)<51\%$ . And we proceed in the same manner, adding $a_3$ -many $R$ ‘s, $a_4$ -many $L$ ‘s, etc.

So after iterating once through the numbers $\frac{1}{k}$ , we have built a subsequence $\{ \frac{1}{k_{i}} \}$ and a point $p \in C\setminus\mathrm{Ends}(C)$ such that $\liminf \delta _p\left( \frac{1}{k_{i}} \right) \leq \frac{51}{100}$ , showing that either $\mathrm{dens}_C(p)$ does not exists or it’s $\leq 51\%$ .