Computing the Measure of the Cantor Set

From Finite Stages to the Limiting Set

The middle-third Cantor set is built by repeatedly deleting open middle thirds from the intervals that remain. Each finite stage still has positive length, but the number of remaining intervals grows while their individual lengths shrink. To find the measure of the limiting set, the key is not to add the measures of the stages: they are nested sets. Instead, we compute each stage’s measure and apply continuity of measure from above.

Throughout, we work in \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. Let \(C_0=[0,1]\). Given \(C_n\), form \(C_{n+1}\) by removing the open middle third of every interval component of \(C_n\). The limiting middle-third Cantor set is \(C=\bigcap_{n=0}^{\infty}C_n\). Each \(C_n\) is a finite union of closed intervals, so it is Borel and Lebesgue measurable. The sets are decreasing: \(C_{n+1}\subseteq C_n\).

Proof. We use induction on \(n\). At \(n=0\), \(C_0=[0,1]\) consists of one interval of length \(1=3^0\), and its measure is \(1=(2/3)^0\). Suppose the claim holds at stage \(n\). Each of the \(2^n\) intervals has length \(3^{-n}\). Removing its open middle third leaves two closed intervals, each of length \(3^{-n}/3=3^{-(n+1)}\). The resulting intervals from different components remain separated by gaps, and the two pieces from each component are separated by the removed open interval. Thus \(C_{n+1}\) has \(2\cdot2^n=2^{n+1}\) intervals. Their interiors are disjoint, so finite additivity of Lebesgue measure gives \(m(C_{n+1})=2^{n+1}3^{-(n+1)}=(2/3)^{n+1}\). This proves the induction step and the proposition. \(\square\)

Continuity from Above

The measure rule needed here applies to a decreasing sequence, provided its first set has finite measure. We include its proof to identify exactly where finiteness enters.

Proof. Put \(A_n=E_1\setminus E_n\). Since the \(E_n\) decrease, the \(A_n\) increase, and \(\bigcup_n A_n=E_1\setminus\bigcap_n E_n\). Define \(B_1=A_1\) and \(B_k=A_k\setminus A_{k-1}\) for \(k\ge2\). The \(B_k\) are pairwise disjoint and \(\bigcup_{k=1}^n B_k=A_n\). By countable additivity, \(\mu(A_n)=\sum_{k=1}^n\mu(B_k)\); taking \(n\to\infty\) gives \(\lim_n\mu(A_n)=\sum_{k=1}^{\infty}\mu(B_k)=\mu(\bigcup_n A_n)\). Each \(A_n\subseteq E_1\), so these measures are finite. Also, \(E_n\) and \(A_n\) partition \(E_1\), hence \(\mu(E_n)=\mu(E_1)-\mu(A_n)\). Subtracting the limit from the finite number \(\mu(E_1)\) proves the asserted identity. \(\square\)

Apply the theorem to \(E_n=C_{n-1}\) for \(n\ge1\). The first set \(C_0\) has measure \(1\), so the finiteness hypothesis holds. The intersection of this sequence is \(C\), and the stage formula gives

The last limit is zero because \(0<2/3<1\). Thus the middle-third Cantor set is Lebesgue measurable and has Lebesgue measure zero. This conclusion concerns measure; it does not say that the set is empty.

A Second Check by Adding the Deleted Intervals

At stage \(k\ge1\), exactly \(2^{k-1}\) open middle thirds are removed, each of length \(3^{-k}\). These deleted intervals are pairwise disjoint across all stages: once a point is removed, it is no longer in any later stage interval from which another middle third is deleted. Their total length through stage \(N\) is therefore \[ \sum_{k=1}^{N}2^{k-1}3^{-k} =\frac{1}{3}\sum_{j=0}^{N-1}\left(\frac{2}{3}\right)^j =1-\left(\frac{2}{3}\right)^N. \] Here the finite geometric-sum identity follows by multiplying the sum by \(1-2/3\) and cancelling the intermediate terms. The original interval has length \(1\), so the length not yet removed at stage \(N\) is \((2/3)^N\), again the measure of \(C_N\). As \(N\) tends to infinity, the total deleted length tends to \(1\). This calculation corroborates the limit, while continuity from above directly establishes the measure of the intersection.

A Common Pitfall: Adding Measures of Nested Stages

A tempting but incorrect calculation is to add \(m(C_1)+m(C_2)+m(C_3)\) and treat the result as the measure of part of the Cantor set. The values are \(2/3\), \(4/9\), and \(8/27\), so this sum is \(18/27+12/27+8/27=38/27\). It cannot be the measure of a subset of \([0,1]\), whose measure is \(1\). The error is that the stages are nested, not disjoint: every point of \(C_3\) is already counted in \(C_2\) and \(C_1\). Countable additivity permits summing measures of disjoint measurable sets, not of these overlapping stages.

The valid approaches keep the distinction clear. Use the stage formula to compute \(m(C_n)\), then use continuity from above; or sum the lengths of the disjoint intervals removed at successive stages. In the continuity argument, the condition \(m(C_0)<\infty\) matters. For example, in \((\mathbb{R},\mathcal{L},m)\), the sets \(E_n=[n,\infty)\) decrease to the empty set, but \(m(E_n)=\infty\) for every \(n\). Thus their measures do not converge to the measure of their intersection. Continuity from above cannot be applied without its finite-first-set hypothesis.