On the Measure of the Midpoints of the Cantor Set in $\mathbb{R}$

TL;DR

This paper investigates the measure-theoretic properties of midpoints in the Cantor set within real numbers, exploring how the size of certain sets constrains the measure of related sets, extending known results in higher dimensions.

Contribution

The paper extends measure results related to midpoint sets of the Cantor set in , providing new insights and examples for the case when n=1, where previous results do not hold.

Findings

Stein's result for n  and spherical sets

Extension of measure results to n=2 by Bourgain and Marstrand

Counterexample demonstrating failure of the result for n=1

Abstract

In this paper, we are going to discuss the following problem: Let $T$ be a fixed set in $\mathbb{R}^n$. And let $S$ and $B$ he two subsets in $\mathbb{R}^n$ such that for any $x$ in $S$, there exists an $r$ such that $x+ r T$ is a subset of $B$. How small can be $B$ be if we know the size of $S$? Stein proved that for $n$ is greater than or equal to 3 and $T$ is a sphere centered at origin, then $S$ has positive measure implies $B$ has positive measure using spherical maximal operator. Later, Bourgain and Marstrand proved the similar result for $n =2$. And we found an example for why the result fails for $n=1$.

Figures

Equations

$$C_i=\frac{1}{3}{C}_{i-1}\cup \left(\frac{2}{3}+\frac{1}{3}{C}_{i-1}\right),\forall i\ge 2.$$

$$M_A:=\left\{\frac{x+y}{2}:x,y\in A,x\ne y\right\}.$$

$$M_{\mathcal{C}}=M_{{\cap }_{n=1}^{\infty }{C}_n}=\bigcap _{n=1}^{\infty }{M}_{C_n}=(0,1).$$

$$P_i=\left\{P_{C_i}^k:P_{C_i}^k=\left(\frac{k-1}{3^i},\frac{k}{3^i}\right),k\in \{1,2,3,\cdots ,3^i\}\right\}$$

$$\max \{z-x_r,y_l-z\}<\delta (z)<\min \{z-x_l,y_r-z\}.$$

$$x_l$$

$$y_l$$

$$z=\frac{1}{2}((z-\delta (z))+(z+\delta (z)))\in M_{P_{C_i}^m\cup P_{C_i}^n}.$$

$$x_s=c_1{x}_t+c_2,and{y}_s=c_1{y}_t+c_2.$$

$$z_s$$

$$=\frac{1}{2}((c_1{x}_t+c_2)+(c_1{y}_t+c_2))$$

$$=c_1\cdot \frac{1}{2}(x_t+y_t)+c_2$$

$$=c_1{z}_t+c_2,$$

$$x_t=\frac{1}{c_1}(x_s-c_2),and{y}_t=\frac{1}{c_1}(y_s-c_2).$$

$$z_t$$

$$=\frac{1}{2}\left(\frac{1}{c_1}(x_s-c_2)+\frac{1}{c_1}(y_s-c_2)\right)$$

$$=\frac{1}{c_1}\left(\frac{1}{2}(x_s+y_s)-c_2\right)$$

$$=\frac{1}{c_1}(z_s-c_2),$$

$$0$$

$$\frac{2}{3}$$

$$z=\frac{1}{2}((z+\delta (z))+(z-\delta (z)))\in M_{C_1}.$$

$$\frac{1}{3}=\frac{1}{2}\left(0+\frac{2}{3}\right),\frac{2}{3}=\left(\frac{1}{3}+1\right),$$

$$M_{C_1}=({\cup }_{i=1}^3{P}_{C_1}^i)\cup \left\{\frac{1}{3},\frac{2}{3}\right\}=(0,1).$$

$$C_{n+1}=\frac{1}{3}{C}_n\cup \left(\frac{2}{3}+\frac{1}{3}{C}_n\right),$$

$$M_{\frac{1}{3}{C}_n}$$

$$M_{\frac{2}{3}+\frac{1}{3}{C}_n}$$

$$\frac{1}{3}=\frac{1}{2}\left(0+\frac{2}{3}\right),\frac{2}{3}=\frac{1}{2}\left(\frac{1}{3}+1\right),$$

$$z=\frac{1}{2}(a+b),$$

$$z=\frac{1}{3}+c=\frac{3^n}{3^{n+1}}+c.$$

$${\cup }_{k=2\cdot 3^n+1}^{3^{n+1}}{P}_{C_{n+1}}^k\cap C_{n+1}=\frac{2}{3}+({\cup }_{k=1}^{3^n}{P}_{C_{n+1}}^k)\cap C_{n+1}.$$

$$c=\frac{1}{2}(a+d).$$

$$z=\frac{x_n^k+y_n^k}{2}.$$

$$x_n^0=\min \left\{x_n^k:z=\frac{x_n^k+y_n^k}{2},x_n^k\in C_n\right\}.$$

$$z=\frac{x_{m+1}^0+y_{m+1}^0}{2}$$

$$x=\underset{m\to \infty }{\lim }{x}_n^0.$$

Full text

1 Introduction

The problem in the abstract is included in the paper by Tamás Keleti [3]. The purpose of this paper is to construct a counterexample in $\mathbb{R}$ for the following theorem which holds for dimensions greater than $1$.

Theorem 1.1**.**

Let $S\subset\mathbb{R}^{n}$ $(n\geq 2)$ be a set of positive Lebesgue measure. If $B\subset\mathbb{R}^{n}$ contains a sphere centered at every point of $S$, then $B$ has positive Lebesgue measure.

In 1976, Ellias Stein [4] proved the case for when $n\geq 3$. Then in 1987 and 1986, Marstrand [2] and Bourgain [1] independently proved the theorem for dimension $n=2$. However, the theorem is not true for $n=1.$

$\mathbf{Idea:}$ To construct an counterexample, the idea is to start with a set $B$ with zero length, and then construct a set $S$ with positive measure made up of all midpoints from $B$. Notice that if $S$ contains all the midpoints from $B$, then for any point $z\in S$, it will be the midpoint of a pair of points $x,y$ in $B$. And $\{x,y\}$ will be a sphere centered at $z\in S$ .

Definition 1.2** ($1/3$-Cantor set).**

Let $C_{1}=[0,1]\backslash\left(\frac{1}{3},\frac{2}{3}\right)$, and

[TABLE]

We define the $1/3$-Cantor set to be $\mathcal{C}=\bigcap_{i=1}^{n}C_{i}.$

The 1/3-Cantor set $\mathcal{C}$ will take place of $B$ in the theorem, and the midpoints from $\mathcal{C}$, denoted as $M_{\mathcal{C}}$, will take place of $S$. We define the midpoint set of $A$ in $\mathbb{R}$ as

[TABLE]

Since $\mathcal{C}=\cap_{i=1}^{\infty}C_{i}$, to get $\mathcal{L}^{1}(M_{\mathcal{C}})>0$ we will show that $M_{C_{i}}=(0,1)$ for all $i$. After that, we then prove

[TABLE]

Therefore, by setting $B=\mathcal{C}$ and $S=M_{\mathcal{C}}$, we will have $\mathcal{L}(S)=1$ while $\mathcal{L}(B)=0$, which will be our counterexample.

As a matter of fact, with the same construction, we may make $M_{\mathcal{C}}$ as large as we want by making copies of $\mathcal{C}$ in other intervals!

2 A counterexample in $\mathbb{R}$

Theorem 2.1**.**

There exist $S,B\subset\mathbb{R}$ and $B$ contains a sphere (in $\mathbb{R}$) centered at every point of $S$ such that $\mathcal{L}(S)>0$ but $\mathcal{L}(B)=0.$

In subsection 2.1, we will go over a few definitions and a couple of lemmas which we will use for the proofs of Theorem 2.1. We say proofs, because we will go over two distinct proofs; in subsection 2.2 we will go over an analytic induction proof and in subsection 2.3 we will provide a more geometric argument.

2.1 Definitions and lemmas

Remark 2.2**.**

In $\mathbb{R}$, a sphere of radius $r$ centered at a point $x$ will be two points $x-r$ and $x+r.$

Definition 2.3** ($i$th Cantor partition set for $[0,1]$).**

Define

[TABLE]

to be the $i$th Cantor partition set of $(0,1).$

In order to prove Theorem 2.1, let’s first prove the following lemmas:

Lemma 2.4** (Average properties of indexes set).**

If $s=\frac{1}{2}(m+n)$, then $P^{s}_{C_{i}}\subset M_{P^{m}_{C_{i}}\cup P^{n}_{C_{i}}}$.

Proof.

Without the loss of generality, let’s assume that $m<s<n.$ Denote $x_{l},x_{r},y_{l},y_{r}$ to be left and right end points for $P^{m}_{C_{i}}$ and $P^{n}_{C_{i}}$ respectively. Take any point $z\in P^{s}_{C_{i}},$ there exists a $\delta(z)$ such that

[TABLE]

And

[TABLE]

therefore $z-\delta(z)\in P^{m}_{C_{i}}$ and $z+\delta(z)\in P^{n}_{C_{i}}$, and

[TABLE]

Hence $P^{s}_{C_{i}}\subset M_{P^{m}_{C_{i}}\cup P^{n}_{C_{i}}}$. ∎

Lemma 2.5** (Scaling and translation property of the average set).**

If $S=c_{1}T+c_{2}$ ($c_{1}\neq 0$) and $M_{T}=(a,b)$, then $M_{S}=c_{1}(a,b)+c_{2}.$

Proof.

Pick any point $z_{s}\in M_{S}$, there exist $x_{s},y_{s}\in S$ such that $z_{s}=\frac{1}{2}(x_{s}+y_{s}).$ Since $S=c_{1}T+c_{2}$, there exist $x_{t},y_{t}\in T$ such that

[TABLE]

Therefore

[TABLE]

where $z_{t}=\frac{1}{2}(x_{t}+y_{t})\in M_{T}$. So $z_{s}\in c_{1}M_{T}+c_{2}$ and then $M_{S}\subset c_{1}M_{T}+c_{2}.$

On the other hand, pick any point $z_{t}\in M_{T}$, there exists a $x_{t},y_{t}\in M_{T}$ such that $z_{t}=\frac{1}{2}(x_{t}+y_{t}).$ As $S=c_{1}T+c_{2}$, $T=\frac{1}{c_{1}}(S-c_{2}).$ So there exist $x_{s},y_{s}\in S$ such that

[TABLE]

Therefore

[TABLE]

where $z_{s}=\frac{1}{2}(x_{s}+y_{s})\in M_{S}.$ So $z_{t}\in\frac{1}{c_{1}}(M_{S}-c_{2})$ and then $M_{T}\subset\frac{1}{c_{1}}(M_{S}-c_{2})$, i.e. $c_{1}M_{T}+c_{2}\subset M_{S}$. Hence $M_{S}=c_{1}M_{T}+c_{2}.$ ∎

2.2 An analytic argument

Now we will use the lemmas above to prove Theorem 2.1.

Proof.

Claim 1: $M_{C_{1}}=(0,1).$

Proof of Claim 1: Let $P^{1}_{C_{1}}=\left(0,\frac{1}{3}\right)$, $P^{2}_{C_{1}}=\left(\frac{1}{3},\frac{2}{3}\right)$ and $P^{3}_{C_{1}}=\left(\frac{2}{3},1\right)$ be the $1$st Cantor partition set in Definition 2.3.

• (i)

For any $z\in P^{1}_{C_{1}}$, since $P^{1}_{C_{1}}$ is open, there exists an open interval $I_{z}$ centered at $z$ such that $I_{z}\subset P^{1}_{C_{1}}$ . Therefore, $P^{1}_{C_{1}}\in M_{C_{1}}$. Similarly, $P^{3}_{C_{1}}\in M_{C_{1}}.$

• (ii)

For any point $z\in P^{2}_{C_{1}},$ without the loss of generality, $\frac{1}{3}<z\leq\frac{1}{2}$ (i.e. the other half $\frac{1}{2}\leq z<\frac{2}{3}$ can be solved by symmetry), there exists a $\delta(z)$ satisfying $\frac{1}{3}<\delta(z)<z$ such that

[TABLE]

Therefore $z-\delta(z)\in P^{1}_{C_{1}}\subset C_{1}$ and $z+\delta(z)\in P^{3}_{C_{1}}\subset C_{1}$, and

[TABLE]

Hence $P^{2}_{C_{1}}\subset M_{C_{1}}.$

• (iii)

Since

[TABLE]

and $\{0,\frac{1}{3},\frac{2}{3},1\}\in C_{1}$, $\{\frac{1}{3},\frac{2}{3}\}\in M_{C_{1}}.$

• (iv)

There’s no way to get [math] or $1$ for $M_{C_{1}}$.

Finally,

[TABLE]

Claim 2: $M_{C_{i}}=(0,1)$ for all $i$.

Proof of Claim 2: Let $\{P^{k}_{C_{i}}\}$ be the $i$th Cantor partition set of $[0,1]$. Assume when $i=n$, $M_{C_{n}}=(0,1).$

• (a)

$M_{C_{i}}\supset\left(0,\frac{1}{3}\right)\cup\left(\frac{1}{3},\frac{2}{3}\right)$ for all $i$.

Since $M_{C_{n}}=(0,1),$ then $M_{C_{n}}\supset\left(0,\frac{1}{3}\right)\cup\left(\frac{1}{3},\frac{2}{3}\right)$. When $i=n+1$, since

[TABLE]

and by Lemma 2.5,

[TABLE]

$M_{C_{n}+1}\supset\left(0,\frac{1}{3}\right)\cup\left(\frac{2}{3},1\right).$ Hence by induction $M_{C_{i}}\supset\left(0,\frac{1}{3}\right)\cup\left(\frac{1}{3},\frac{2}{3}\right)$ for all $i$.

• (b)

$M_{C_{i}}\supset\{\frac{1}{3},\frac{2}{3}\}$ for all $i$.

For each $i$,

[TABLE]

and $\{0,\frac{1}{3},\frac{2}{3},1\}\subset C_{i}$, therefore $M_{C_{i}}\supset\{\frac{1}{3},\frac{2}{3}\}$ for all $i.$

• (c)

$M_{C_{i}}\supset\left(\frac{1}{3},\frac{2}{3}\right)$ for all $i$.

Since $M_{C_{n}}=(0,1)$, then $M_{C_{n}}\supset\left(\frac{1}{3},\frac{2}{3}\right).$ For $i=n+1$, we will use the assumption $M_{C_{n}}=(0,1)$ and the following trick to show $M_{C_{n+1}}\supset\left(\frac{1}{3},\frac{2}{3}\right).$

Take any $z\in\left(\frac{1}{3},\frac{2}{3}\right)=\left(\frac{3^{n}}{3^{n+1}},\frac{2\cdot 3^{n}}{3^{n+1}}\right)=\cup_{k=3^{n}+1}^{2\cdot 3^{n}}P^{k}_{C_{n+1}},$ we want to show

[TABLE]

where $a\in(\cup_{k=1}^{3^{n}}P^{k}_{C_{n+1}})\cap C_{n+1}$ and $b\in(\cup_{k=2\cdot 3^{n}+1}^{3^{n+1}}P^{k}_{C_{n+1}})\cap C_{n+1}.$ Notice that since $z\in\left(\frac{1}{3},\frac{2}{3}\right)$, there exists some $c\in\left(0,\frac{1}{3}\right),$ such that

[TABLE]

Similarly since $C_{n+1}=\frac{1}{3}C_{n}\cup\left(\frac{2}{3}+\frac{1}{3}C_{n}\right)$, we know

[TABLE]

So there exists a $d\in\cup_{k=1}^{3^{n}}P^{k}_{C_{n+1}}\cap C_{n+1}\subset\left(0,\frac{1}{3}\right).$ such that $b=\frac{2}{3}+d.$ Therefore Eq. (1) is reduced to for any $c\in\left(0,\frac{1}{3}\right)$, find $a,d\in(\cup_{k=1}^{3^{n}}P^{k}_{C_{n+1}}\cap C_{n+1})\subset\left(0,\frac{1}{3}\right)$ such that

[TABLE]

Note that $\cup_{k=1}^{3^{n}}P^{k}_{C_{n+1}}\cap C_{n+1}=(\cup_{k=1}^{3^{n+1}}P^{k}_{C_{n+1}}\cap C_{n+1})\cap\left(0,\frac{1}{3}\right)=\frac{1}{3}C_{n}.$ And in (a), it is already shown that $M_{\frac{1}{3}C_{n}}=\left(0,\frac{1}{3}\right)$. So the existence for $a,d$ are guaranteed. Hence Eq. (1) is true and therefore $M_{C_{n+1}}\supset\left(\frac{1}{3},\frac{2}{3}\right)$ is true for all $i$.

Therefore $M_{C_{i}}=(0,1)$ for all $i$.

Claim 3: $M_{\mathcal{C}}=\bigcap_{i=1}^{\infty}M_{C_{i}}=(0,1).$

• (d)

One direction $M_{\mathcal{C}}\subset\bigcap_{i=1}^{\infty}M_{C_{i}}=(0,1)$ is trivial, since $\mathcal{C}\subset C_{i}$ for all $i$, then $M_{\mathcal{C}}\subset M_{C_{i}}$ for all $i$. Hence $M_{\mathcal{C}}\subset\bigcap_{i=1}^{\infty}M_{C_{i}}=(0,1).$

• (e)

For the other direction, pick any $z\in(0,1),$ that exists at least one pair $(x_{n}^{k},y_{n}^{k})\in C_{n}$ such that

[TABLE]

Define

[TABLE]

Claim: $\{x_{n}^{0}\}$ converges to $x$ and $x\in\mathcal{C}.$

Proof of claim: First we will show $\{x^{0}_{n}\}$ is nondecreasing. Assume it is not true, then there exists an $m$ such that $x_{m+1}^{0}\geq x_{m}^{0}.$ However, $x_{m+1}^{0}\in C_{m+1}\subset C_{m}$, and

[TABLE]

it contradicts with the construction of $x_{m}^{0}.$ Therefore $x^{0}_{n}$ is nondecreasing. Moreover, $0\leq x_{n}^{0}<z$, by monotone convergence theorem, $\{x_{n}^{0}\}$ converges. Denote

[TABLE]

For any $N$, $\{x_{n}\}_{n\geq N}\subset C_{N}$. Since $C_{N}$ is compact, $x^{0}_{n}\rightarrow x\in C_{N}$. Therefore $x\in\bigcap_{n=1}^{\infty}C_{n}=\mathcal{C}.$ Similarly, $\{y^{0}_{n}\}$ is nonincreasing since $y_{n}^{0}=2z-x_{n}^{0}$, and $z<y_{n}^{0}<1.$ Again, by monotone convergence theorem and the same argument, $y_{n}^{0}\rightarrow y\in\mathcal{C}.$ Hence $z=\frac{x+y}{2}\in M_{C}$. Therefore $(0,1)\subset M_{C}.$

Finally, let $B=\mathcal{C}$ and $S=M_{C}$, then for any point $x\in S$, $B$ contains a sphere around $x$, and $\mathcal{H}^{1}(S)=1$ but $\mathcal{H}^{1}(B)=0$. In fact, we can have countably many copies of $B$ such that $\mathcal{H}^{1}(B)=0$ while $\mathcal{H}^{1}(S)=\infty.$ ∎

2.3 A geometric argument

Theorem 2.6**.**

Suppose $C$, $C_{i}$ and $M_{C_{i}}$ are defined as above. Then $M_{C_{i}}=(0,1)$ for all integers $i\geq 1$.

Proof.

We will show that $M_{C_{i}}=(0,1)$ for all integers $i\geq 1$ by inducting on $i$.

$\mathbf{Base\ Case:}$ Let $i=1$, and consider the partition set $P_{i}$. To show that $M_{C_{1}}$ contains $P_{C_{1}}^{1}$, notice that for any point $m\in P_{C_{1}}^{1}$, there exists an $r>0$ dependent on $m$, such that $m-r$ and $m+r$ are contained in $P_{C_{1}}^{1}$. Similarly, $M_{C_{1}}$ contains $P_{C_{1}}^{3}$.

Now, to show that $P_{C_{1}}^{2}\cup\left\{\frac{1}{3},\frac{2}{3}\right\}\subset M_{C_{1}}$ we have that for any point $m\in P_{C_{1}}^{2}$, $m+\frac{1}{3}$ and $m-\frac{1}{3}$ are both contained in $C_{1}$. Furthermore, $\frac{1}{3}=\frac{1}{2}\left(0+\frac{2}{3}\right)$ and $\frac{2}{3}=\frac{1}{2}\left(\frac{1}{3}+1\right)$ in which case we have shown our base case.

$\mathbf{Induction:}$ By the inductive hypothesis, suppose the theorem holds for $i=n$ for some $n\in\mathbb{N}$. We will show that the theorem holds true for $i=n+1$. That is, we will show that $M_{C_{n+1}}$ contains $(0,1)$.

First consider the scaled version of $C_{n}$, i.e. $\frac{1}{3}C_{n}$. By the scaling property 2.5, since $M_{C_{n}}$ contains $(0,1)$, $M_{\frac{1}{3}C_{n}}$ contains $\left(0,\frac{1}{3}\right)$ and hence $M_{C_{n+1}\cap\left(0,\frac{1}{3}\right)}\subset M_{C_{n+1}}$ contains $\left(0,\frac{1}{3}\right)$. By the same scaling property, we get that that $M_{\frac{1}{3}C_{n}+\frac{2}{3}}$ contains $\left(\frac{2}{3},1\right)$ and hence, $M_{C_{n+1}}$ contains $\left(\frac{2}{3},1\right)$.

Lets take a quick break here to notice that $\frac{1}{3}$ and $\frac{2}{3}$ are both contained in $M_{C_{n+1}}$ because $0,\frac{1}{3},\frac{2}{3}$, and $1$ are all in $C_{n+1}$ just as we showed in the base case. So in the following, we just need to argue that $M_{C_{n+1}}$ contains $\left(\frac{1}{3},\frac{2}{3}\right)$.

Consider the partitions in $P_{n+1}\cap\left(\frac{1}{3},\frac{2}{3}\right)$ and $P_{n}\cap(1/3,2/3)$, which are exactly $\mathcal{P}_{n+1}=\{P_{C_{n+1}}^{3^{n}+1},P_{C_{n+1}}^{3^{n}+2},...,P_{C_{n+1}}^{2\cdot 3^{n}}\}$, and $\mathcal{P}_{n}=\{P_{C_{n}}^{3^{n-1}+1},...,P_{C_{n}}^{2\cdot 3^{n-1}}\}$ respectively, see Fig 3.

Pick $\bar{P}_{C_{n}}\in\mathcal{P}_{n}$ and partition $\bar{P}_{C_{n}}$ into thirds to get $\{\bar{P}_{C_{n}}^{1},\bar{P}_{C_{n}}^{2},\bar{P}_{C_{n}}^{3}\}$. It is important to notice that $\bar{P}_{C_{n}}^{m}\in\mathcal{P}_{n+1}$ for each $m=1,2,3$.

By the inductive hypothesis, there exists distinct partitions $P_{C_{n}}^{k}$ and $P_{C_{n}}^{j}$ contained in $C_{n}$ for which the midpoint set of $P_{C_{n}}^{k}\cup P_{C_{n}}^{j}$ contains $\bar{P}_{C_{n}}$.

As we did with $\bar{P}_{C_{n}}$, we may partition $P_{C_{n}}^{k}$ and $P_{C_{n}}^{j}$ into thirds, to get $\{P_{C_{n}}^{k_{1}},P_{C_{n}}^{k_{2}},P_{C_{n}}^{k_{3}}\}$ and $\{P_{C_{n}}^{j_{1}},P_{C_{n}}^{j_{2}},P_{C_{n}}^{j_{3}}\}$. However, only $\{P_{C_{n}}^{k_{1}},P_{C_{n}}^{k_{3}}\}$ and $\{P_{C_{n}}^{j_{1}},P_{C_{n}}^{j_{3}}\}$ are in $C_{n+1}$.

Hence, we get

[TABLE]

Since this argument holds holds for arbitrary $\bar{P}_{C_{k-1}}$, we may therefore iterate through $\mathcal{P}_{n+1}$ by iterating through $\mathcal{P}_{n}$ and showing (2); therefore showing that $\left[\frac{1}{3},\frac{2}{3}\right]\subset M_{C_{k}}$.

By the principle of mathematical induction we have shown that $(0,1)\subset M_{C_{i}}$ for all $i\geq 1$. ∎

Theorem 2.7**.**

Suppose $\mathcal{C}$, $C_{i}$ and $M_{C_{i}}$ are defined as above. Then

[TABLE]

Proof.

To show $\cap_{i=1}^{\infty}M_{C_{i}}\subset M_{\mathcal{C}}$, we just need to notice that $\mathcal{C}$ being contained in all $C_{i}$, implies $M_{\mathcal{C}}$ is contained in all $M_{C_{i}}$.

To show that $\cap_{i=1}^{\infty}M_{C_{i}}\subset M_{\mathcal{C}}$, let $z\in\cap_{i=1}^{\infty}M_{C_{i}}$. For each $i=1,2,...$ there exists $p_{i}:=(x_{i},y_{i})\in C_{i}\times C_{i}$ such that $(x_{i}+y_{i})/2=z$. Now, for the bounded sequence $(p_{i})$, there exists a subsequence $(p_{i_{j}})$ converging to $p:=(x,y)$. To show that $p\in\mathcal{C}\times\mathcal{C}$, assume by way of contradiction that $p$ is in $\mathbb{R}^{2}\setminus\mathcal{C}\times\mathcal{C}$.

First notice that since $\mathcal{C}\times\mathcal{C}$ is closed, $\mathbb{R}^{2}\setminus\mathcal{C}\times\mathcal{C}$ is open, and hence there exists $\epsilon>0$ for which the open ball $B(p,\epsilon)\subset(\mathbb{R}^{2}\setminus\mathcal{C}\times\mathcal{C})$.

Now, since $\mathcal{C}\times\mathcal{C}=\cap_{i=1}^{\infty}(C_{i}\times C_{i})$, and since $C_{i+1}\subset(C_{i}\times C_{i})$ for all $i$, there exists $N_{1}\in\mathbb{N}$ such that $\left(C_{i}\times C_{i}\right)\cap B(p,\epsilon)=\emptyset$ for all $i>N_{1}$. However, since $p_{i_{j}}\to p$ as $j\to\infty$, there exists $N_{2}\in\mathbb{N}$ for which $p_{i_{j}}\in B(p,\epsilon)$ for all $j>N_{2}$.

Hence for any $j>\mathrm{max}\{N_{1},N_{2}\}$, we have that $p_{i_{j}}$ is in both $C_{i_{j}}\times C_{i_{j}}$ and $B(p,\epsilon)$; contradicting the fact that $(C_{i}\times C_{i})\cap B(p,\epsilon)=\emptyset$ for all $i>N_{1}$.

We therefore get that, $p\in\mathcal{C}\times\mathcal{C}$, $(x+y)/2=z$ and hence $z\in M_{C}$ thus concluding the proof. ∎

By the previous two theorems, we get that the midpoint set of the $1/3$-Cantor set is the interval $(0,1)$. We therefore get that the positive measure set, $S=M_{\mathcal{\mathcal{C}}}=(0,1)$ contains spheres centered about the zero measure set, $\mathcal{C}$.