Measuring the Middle-Third Cantor Set

From the Construction to a Measure Estimate

The middle-third Cantor set is built by repeatedly removing open middle thirds from the intervals that remain. At each finite stage, the retained set is a finite union of closed intervals. The limiting Cantor set lies inside every one of these finite-stage sets. The key measure estimate is therefore a covering estimate: count the stage intervals, multiply by the length of each, and show that the resulting total tends to zero.

We work on \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, and use \(m^*\) for Lebesgue outer measure on \(\mathbb{R}\). Let \(C_0=[0,1]\). For each integer \(n\geq 0\), form \(C_{n+1}\) by removing the open middle third of every interval making up \(C_n\). Thus endpoints of the removed intervals are retained. The middle-third Cantor set is

In particular, \(C\subseteq C_n\) for every \(n\). We will establish exactly how many intervals make up \(C_n\) and how long they are. These two counts, taken together, determine the useful cover bound.

Proof. At stage \(n=0\), \(C_0=[0,1]\) consists of one closed interval, and its length is \(1=3^0\). Suppose the assertion holds at stage \(n\). Each interval of length \(3^{-n}\) is divided into three equal closed thirds, with the open middle third removed. The two retained closed thirds each have length \(3^{-(n+1)}\). Each of the \(2^n\) intervals therefore produces two retained intervals, so \(C_{n+1}\) has \(2\cdot2^n=2^{n+1}\) intervals. Their length is \(3^{-(n+1)}\), proving the assertion at stage \(n+1\). The total length at stage \(n\) is the number of intervals times their individual length, namely \(2^n3^{-n}=(2/3)^n\). \(\square\)

The intervals at any fixed stage are disjoint and separated by gaps: the construction removes an open interval between the two retained thirds at each step. In particular, adding their lengths gives the measure of \(C_n\), although for the null-set proof we need only that they cover \(C\).

Worked Covers at Finite Stages

The Cantor Set Has Lebesgue Measure Zero

Proof. Each \(C_n\) is a finite union of closed intervals, hence is closed. An arbitrary intersection of closed sets is closed, so \(C=\bigcap_{n=0}^{\infty}C_n\) is closed. Closed subsets of \(\mathbb{R}\) are Borel sets, and Borel sets are Lebesgue measurable; thus \(C\in\mathcal{L}\).

For each \(n\), the \(2^n\) stage intervals cover \(C\), because \(C\subseteq C_n\). The definition of outer measure as the infimum of total lengths of countable interval covers therefore gives \(0\leq m^*(C)\leq 2^n3^{-n}=(2/3)^n\). The sequence \((2/3)^n\) tends to zero. To see that this forces the outer measure to vanish, let \(\varepsilon>0\). Choose \(n\) large enough that \((2/3)^n<\varepsilon\). Then \(0\leq m^*(C)<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(m^*(C)=0\). Because \(C\) is Lebesgue measurable, \(m(C)=m^*(C)=0\). \(\square\)

The argument uses the total length of a cover, not the length of one component interval. It also does not infer the measure of an infinite intersection merely by taking a limit of the measures of the finite-stage sets. Instead, the covers give a direct outer-measure bound at every stage. In this setting the finite-stage sets are measurable and decreasing, and \(m(C_0)=1<\infty\), so continuity of measure from above would also yield \(m(C)=\lim_n m(C_n)=0\). The explicit covers show precisely why the bound tends to zero.

A Common Counting Error

A tempting but incorrect calculation is: “At stage three, each interval has length \(1/27\), so the Cantor set has measure at most \(1/27\).” The proposed bound does not follow: the single interval of length \(1/27\) is not a cover of \(C\). At that stage, \(C\) is contained in eight intervals, not one. The correct total covering length is \(8(1/27)=8/27\). More generally, the interval length \(3^{-n}\) must be multiplied by the number \(2^n\) of intervals, giving \((2/3)^n\). It is this total, rather than the length of one stage interval, that tends to zero and proves the measure claim.

One should also distinguish “small at a chosen stage” from “zero.” A stage-\(n\) cover of total length \((2/3)^n\) gives a positive upper bound for each finite \(n\); it does not by itself say that any particular finite stage has length zero. The conclusion \(m(C)=0\) follows because for every positive \(\varepsilon\), a possibly later stage supplies a cover with total length less than \(\varepsilon\).

Why the Covering Argument Matters

The Cantor set is a useful example of a set that is uncountable yet has Lebesgue measure zero. The proof here establishes the measure statement without needing to count its points: it uses the geometry of the construction and a sequence of finite covers. In later measure arguments, the same style of reasoning is valuable whenever a set can be placed inside covers whose total lengths (or, in higher dimensions, total volumes) can be made arbitrarily small.

The logic has three distinct parts. First, the construction gives a cover of \(C\) at every stage. Second, the number and lengths of those intervals give a total length of \((2/3)^n\). Third, the limit of these upper bounds is zero, so the outer measure is zero. Measurability follows here also from closedness. Keeping these steps separate prevents an invalid inference from a shrinking component interval or from a finite-stage picture.