Constant Intervals and Almost-Everywhere Derivative

Plateaus of the Cantor Function

Removing the middle thirds creates open intervals on which the Cantor function does not change. These plateaus explain its derivative away from the Cantor set: each point in a deleted interval has a whole neighborhood on which the function is constant, so the derivative there is zero. The conclusion is almost everywhere because the deleted intervals make up the complement of the Cantor set, which has Lebesgue measure zero. It does not assert that the derivative is zero at every point of the Cantor set.

Let \(C\subseteq[0,1]\) be the middle-third Cantor set, and let \(F:[0,1]\to[0,1]\) be the Cantor function. We use its ternary-digit description: if a point of \(C\) has ternary digits \(d_j\in\{0,2\}\), its function value is the binary number with digits \(d_j/2\). If a point lies in a deleted interval and its first ternary digit \(1\) occurs in position \(k\), the function value is the binary prefix from positions \(1,\ldots,k-1\), followed by a \(1\) in position \(k\) and zeros thereafter. At endpoints, the two possible ternary expansions give the same value. These are the standard digit rules for \(F\); its continuity and monotonicity were established earlier.

The empty prefix is allowed when \(k=1\), with \(p=0\) and the empty sum equal to zero. For a point in the interval, the first \(k-1\) ternary digits are the specified prefix and the \(k\)-th digit is \(1\). The digit rule therefore assigns the same binary value to every point in that interval. The endpoint computations below also show why that value matches continuously at both ends.

Constancy Gives a Zero Derivative

Proof. Fix a deleted interval at stage \(k\), with prefix \(a_1,\ldots,a_{k-1}\) and \(p=\sum_{j=1}^{k-1}a_j3^{-j}\). Each interior point has the specified prefix followed by ternary digit \(1\) at position \(k\). The digit rule assigns it the value \(\sum_{j=1}^{k-1}a_j2^{-(j+1)}+2^{-k}\), independent of the point. This proves constancy on the whole interval.

Its endpoints have expansions \(0.a_1\cdots a_{k-1}0\,222\ldots_3\) and \(0.a_1\cdots a_{k-1}2\,000\ldots_3\). At the left endpoint, the tail contributes \(\sum_{j=k+1}^{\infty}2^{-j}=2^{-k}\) to the binary value. At the right endpoint, the digit \(2\) at position \(k\) contributes \(2\cdot2^{-(k+1)}=2^{-k}\). Both endpoints therefore have the same value as the interior, consistent with the digit rule at ambiguous expansions.

Now take \(x\in[0,1]\setminus C\). By the construction of \(C\), \(x\) belongs to one of the deleted open intervals, say \((u,v)\). Choose \(r>0\) so that \((x-r,x+r)\subset(u,v)\); for instance, \(r=\min(x-u,v-x)/2\) is positive. For every \(h\) with \(0<|h|<r\), both \(x\) and \(x+h\) lie in that interval, so constancy gives \(F(x+h)-F(x)=0\). Hence \[ \frac{F(x+h)-F(x)}{h}=0 \] for every such \(h\), and the limit as \(h\to0\) is zero. Thus \(F'(x)=0\).

Finally, \(C\) has Lebesgue measure zero. To see the measure bound directly, let \(C_n\) be the union of the \(2^n\) closed intervals remaining after \(n\) stages. Each has length \(3^{-n}\), so \(m(C_n)=(2/3)^n\). Since \(C\subseteq C_n\) for every \(n\), \(0\le m(C)\le(2/3)^n\) for every \(n\), and \((2/3)^n\to0\); therefore \(m(C)=0\). The derivative exists and is zero on \([0,1]\setminus C\), whose complement is this null set. This proves the almost-everywhere assertion. \(\square\)

Almost Everywhere Is Not Everywhere

The derivative conclusion depends on being inside an open deleted interval. A point at an endpoint of a gap lies in \(C\), and a plateau on only one side does not by itself give a two-sided derivative. In fact, the derivative at the endpoint \(1/3\) is not finite. For each integer \(n\ge2\), set \(x_n=1/3-3^{-n}\). Its ternary expansion is \(0.0\underbrace{22\cdots2}_{n-1\text{ digits}}{}_3\), so \(F(x_n)=\sum_{j=2}^{n}2^{-j}=1/2-2^{-n}\), while \(F(1/3)=1/2\). Thus \[ \frac{F(x_n)-F(1/3)}{x_n-1/3} =\frac{-2^{-n}}{-3^{-n}} =\left(\frac32\right)^n. \] As \(n\to\infty\), these quotients tend to \(+\infty\), so they cannot converge to a finite derivative. This does not conflict with the theorem: \(1/3\in C\), and the theorem makes no derivative claim at every point of \(C\).

A common error is to reason that \(F'=0\) everywhere because \(F\) is constant on the deleted intervals and the remaining Cantor set has measure zero. That confuses “outside a null set” with “at every point.” For example, the calculation above shows that the difference quotients at \(1/3\) do not approach zero; indeed, they are \((3/2)^n\) along the displayed sequence. The valid conclusion is exactly that the derivative exists and equals zero at every point of the complement of \(C\), hence almost everywhere, not everywhere.

The plateau description also distinguishes this function from one that is locally constant on all of \([0,1]\): the deleted intervals are its constant regions, while the Cantor set contains all points not covered by those open intervals. The nullity of \(C\) makes the derivative statement an almost-everywhere result without extending it to those exceptional points.