Exact Critical Measure Of The Middle-Third Cantor Set

The Exact-Measure Question

The Hausdorff dimension of the middle-third Cantor set identifies the exponent at which its Hausdorff measure changes from infinite to zero. It does not, by itself, determine the measure at that exponent. Here we compute that critical value using the unnormalized diameter convention: the cost of a set in a cover is its diameter raised to the exponent, with no dimensional constant in front.

Let $C\subset[0,1]$ be the middle-third Cantor set, obtained by repeatedly deleting the open middle third from every interval remaining. At stage $n$, it is contained in $2^n$ closed intervals, called level-$n$ cylinders, each of length $3^{-n}$. Set $s=\log(2)/\log(3)$, so $3^s=2$ and $0<s<1$.

Definition: For $A\subset\mathbb{R}$, $s>0$, and $\delta>0$, define $$ \mathcal{H}^s_\delta(A)=\inf\left\{\sum_{i=1}^{\infty}(\operatorname{diam}U_i)^s:\ A\subseteq\bigcup_{i=1}^{\infty}U_i,\ \operatorname{diam}U_i\le\delta\right\}. $$ The unnormalized $s$-dimensional Hausdorff measure is $\mathcal{H}^s(A)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(A)$. The infimum is over countable covers by arbitrary subsets of $\mathbb{R}$; a set of diameter zero has cost zero.

The limit exists in $[0,\infty]$, since reducing $\delta$ restricts the available covers and therefore cannot decrease the infimum. Our goal is to prove $\mathcal{H}^s(C)=1$.

A Coded Probability Measure

We first build a probability measure that records how much of the Cantor set lies in an interval. For $t\in[0,1)$, use its binary expansion that is not eventually all ones, and write its digits as $\varepsilon_j(t)\in\{0,1\}$, $j\ge1$. At the point $t=1$, set every digit equal to one. Define $$ X(t)=\sum_{j=1}^{\infty}2\varepsilon_j(t)3^{-j}. $$ The partial sums are measurable functions of $t$, so their pointwise limit $X$ is measurable. Its values lie in $C$. Define a Borel probability measure on $\mathbb{R}$ by $\mu(E)=m(\{t\in[0,1]:X(t)\in E\})$ for Borel sets $E$, where $m$ is Lebesgue measure. In particular, $\mu(C)=1$.

A binary prefix of length $n$ corresponds, apart from endpoints of measure zero, to a dyadic interval of length $2^{-n}$. Its image lies in the associated level-$n$ Cantor cylinder. Also, any point has at most one sequence of Cantor digits, while a real number has at most two binary expansions. Thus the preimage of a singleton under $X$ has Lebesgue measure zero, and $\mu(\{x\})=0$ for every $x\in\mathbb{R}$.

For a binary sequence $\varepsilon=(\varepsilon_j)_{j\ge1}$, write $T(\varepsilon)=\sum_{j\ge1}\varepsilon_j2^{-j}$ and $Z(\varepsilon)=\sum_{j\ge1}2\varepsilon_j3^{-j}$. The first quantity is its binary value and the second its Cantor-coded point. We need the following sharp estimate between these two coordinates.

Lemma: If two binary sequences $\alpha,\beta$ are ordered lexicographically, with $\alpha$ no larger than $\beta$, then $$ Z(\beta)-Z(\alpha)\ge \bigl(T(\beta)-T(\alpha)\bigr)^q,\qquad q=\frac{\log 3}{\log 2}. $$ Consequently, for $x,y\in C$ with $x\le y$, $\mu([x,y])\le(y-x)^s$.

Proof. First consider sequences that are zero after some finite position. We prove the inequality by induction on the number of remaining positions. If the sequences agree throughout, both differences are zero. Otherwise let $k$ be their first differing position. The digits there are $0$ for $\alpha$ and $1$ for $\beta$. Let $a,b\in[0,1]$ be the binary values of their respective tails after position $k$, and let $A,B\in[0,1]$ be the Cantor-coded values of those tails. The tails are lexicographically ordered, so the induction hypothesis gives $B-A\ge(b-a)^q$. Directly separating the $k$th digit from the tails gives $$ T(\beta)-T(\alpha)=2^{-k}(1+b-a),\qquad Z(\beta)-Z(\alpha)=3^{-k}(2+B-A). $$ For $0\le d\le1$, the function $h(d)=2+d^q-(1+d)^q$ has $h(1)=3-2^q=0$ and $h'(d)=q(d^{q-1}-(1+d)^{q-1})<0$ for $0<d\le1$. Hence $h(d)\ge0$ throughout $[0,1]$. Taking $d=b-a$, and using $3^{-k}=2^{-kq}$, proves the desired inequality for these finite sequences. For arbitrary sequences, truncate both after $N$ digits. The finite inequalities hold for every $N$, and the truncated binary and Cantor values converge to the original values. Passing to the limit proves the first assertion.

To obtain the measure assertion, the Cantor coding preserves lexicographic order, and the binary value of the digits of $t$ is $t$, apart from the immaterial binary-expansion ambiguity at dyadic points. Therefore, for a coded point $x=Z(\alpha)$, the measure of coded points at most $x$ is $T(\alpha)$. There are no atoms, so for $x=Z(\alpha)\le y=Z(\beta)$, $\mu([x,y])=T(\beta)-T(\alpha)$. The proved inequality gives $\mu([x,y])\le(y-x)^{1/q}=(y-x)^s$. $\square$

Corollary: For every bounded interval $J\subset\mathbb{R}$, with any choice of open or closed endpoints, $\mu(J)\le |J|^s$, where $|J|$ denotes its length.

Proof. Let $[a,b]$ be the closed interval hull of $J$; it has length $|J|$, and $\mu(J)\le\mu([a,b])$. If $[a,b]\cap C$ is empty, then $\mu([a,b])=0$, since $\mu(C)=1$. Otherwise, the compact set $[a,b]\cap C$ has a minimum $x$ and a maximum $y$. Since $\mu$ is concentrated on $C$, $\mu([a,b])=\mu([x,y])$. The lemma gives $\mu([a,b])\le(y-x)^s\le(b-a)^s=|J|^s$. This also covers a one-point intersection, whose measure is zero. $\square$

The Upper Bound from Cylinder Covers

At level $n$, the $2^n$ cylinders cover $C$, and each has diameter $3^{-n}$. Their total cost at exponent $s$ is $$ 2^n(3^{-n})^s=2^n(3^s)^{-n}=1. $$ For any $\delta>0$, choose $n$ large enough that $3^{-n}\le\delta$. This is an admissible cover in the definition of $\mathcal{H}^s_\delta(C)$, so $\mathcal{H}^s_\delta(C)\le1$. Letting $\delta\downarrow0$ yields $\mathcal{H}^s(C)\le1$.

Worked Example: Checking the First Three Cylinder Costs

At levels $n=1,2,3$, the cylinder counts are $2,4,8$, and their lengths are $1/3,1/9,1/27$, respectively. Since $3^s=2$, the costs are $$ 2(1/3)^s=2/2=1,\qquad 4(1/9)^s=4/2^2=1,\qquad 8(1/27)^s=8/2^3=1. $$ For level two, the cylinders are $[0,1/9]$, $[2/9,1/3]$, $[2/3,7/9]$, and $[8/9,1]$. Their four lengths are each $1/9$, so their total cost is $4(1/9)^s=4/4=1$. Check: the level-three cost is also $8(1/27)^s=8/8=1$, in agreement with the general calculation.

The Lower Bound for Every Cover

The interval estimate applies to arbitrary sets in a Hausdorff cover by enclosing each set in an interval of the same diameter. More precisely, for any bounded $U\subset\mathbb{R}$, its closed interval hull has length $\operatorname{diam}U$. The outer measure $\mu^*(U)$, defined by taking the infimum of $\mu$ over Borel supersets of $U$, therefore satisfies $\mu^*(U)\le(\operatorname{diam}U)^s$, by the interval corollary. This remains true for a singleton, whose $\mu$-measure is zero. Unbounded sets have infinite diameter and do not weaken the inequality.

Theorem: For the middle-third Cantor set $C$ and $s=\log(2)/\log(3)$, the unnormalized Hausdorff measure satisfies $\mathcal{H}^s(C)=1$.

Proof. The cylinder covers show that $\mathcal{H}^s_\delta(C)\le1$ for every $\delta>0$. For the reverse inequality, take any countable cover $C\subseteq\bigcup_{i=1}^{\infty}U_i$ by sets of diameter at most $\delta$. Countable subadditivity of the outer measure $\mu^*$ and $\mu^*(C)=\mu(C)=1$ give $$ 1=\mu^*(C)\le\sum_{i=1}^{\infty}\mu^*(U_i) \le\sum_{i=1}^{\infty}(\operatorname{diam}U_i)^s. $$ Taking the infimum over all such covers yields $\mathcal{H}^s_\delta(C)\ge1$. Thus $\mathcal{H}^s_\delta(C)=1$ for every $\delta>0$, and taking the limit as $\delta\downarrow0$ proves the theorem. $\square$

Worked Example: A Cover That Attains Cost One

The four level-two cylinders listed above cover $C$. Each has diameter $1/9$, and $3^s=2$ gives $(1/9)^s=(3^{-2})^s=2^{-2}=1/4$. The total cost is therefore $$ \frac14+\frac14+\frac14+\frac14=1. $$ This verifies an explicit cover of cost one. The theorem’s lower bound says that no countable cover, including covers by sets that are not intervals, can have smaller total $s$-cost. Check: the four pieces have total cost $4\cdot(1/4)=1$, as required.

Endpoint Pitfalls and the Role of the Estimate

A tempting but incorrect shortcut is to bound the mass between two points whose Cantor codes first differ at position $k$ by $2^{-k}$. The two points can enclose mass from both branches at that split, so that bound is not generally valid. For instance, $\mu((0,1))=1$, whereas $2^{-1}=1/2$. The coding estimate avoids this error: for $x=0$ and $y=1$, their binary values differ by $1$, their Cantor-coded points differ by $1$, and the interval estimate reads $1\le1^s=1$.

There is a separate endpoint issue when an interval $J$ is open or half-open: $J\cap C$ need not have a minimum or maximum. The corollary instead passes to the closed interval hull $[a,b]$, which has the same length and whose intersection with compact $C$ is compact. Monotonicity gives $\mu(J)\le\mu([a,b])$, so the closed-interval estimate applies without assuming that $J\cap C$ has extreme points.

The exact value depends on the normalization. With the diameter convention stated here, the cylinder costs are exactly one. A Hausdorff measure defined with an additional multiplicative normalization would have a correspondingly different numerical value; the result proved here is specifically for the unnormalized definition.

Check Your Understanding

Use the definitions and the interval-mass estimate to check each step of the exact-measure argument.

Why does a level-$n$ cylinder cover have total $s$-cost $2^n(3^{-n})^s=1$?
What property of the coded measure ensures that a singleton has measure zero?
Explain why an open bounded interval can be replaced by its closed interval hull when applying the mass estimate.
How does the inequality $\mu^*(U)\le(\operatorname{diam}U)^s$ imply that every countable cover of $C$ has total cost at least one?
Why would the bound $2^{-k}$ for mass between points whose codes first differ at position $k$ fail for the interval $(0,1)$?

Exact Critical Measure Of The Middle-Third Cantor Set

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