The Cantor Function Has Derivative Zero Almost Everywhere

Zero Derivative Without Constancy

The Cantor function is continuous and nondecreasing on \([0,1]\), with \(F(0)=0\) and \(F(1)=1\). It is therefore not constant. Yet its derivative is zero almost everywhere with respect to Lebesgue measure. The reason is geometric: the function is constant on each open interval removed in constructing the Cantor set, and the Cantor set left over has measure zero.

We work on the finite, complete measure space \(([0,1],\mathcal{L}|_{[0,1]},m)\), where \(m\) is Lebesgue measure. The Cantor function \(F:[0,1]\to[0,1]\) is real-valued. Its derivative is understood at interior points in the usual two-sided sense. To say \(F'=0\) almost everywhere means that \(F'\) exists and equals zero outside a set of Lebesgue measure zero; it makes no assertion about differentiability at every point.

Definition: Set \(C_0=[0,1]\). At each stage, remove the open middle third of every closed interval remaining from the preceding stage, and call the union of the closed intervals left at stage \(n\) the set \(C_n\). The middle-thirds Cantor set is \(C=\bigcap_{n=0}^{\infty}C_n\). The Cantor function \(F\) is the continuous nondecreasing function obtained by the Cantor staircase construction: it takes values \(0\) and \(1\) at the endpoints of \([0,1]\), is constant on each deleted open interval, and its values on the surviving intervals are recursively rescaled copies of the same construction. In particular, on the first surviving intervals the recursion is \(F(x)=\tfrac12F(3x)\) for \(x\in[0,\tfrac13]\), and \(F(x)=\tfrac12+\tfrac12F(3x-2)\) for \(x\in[\tfrac23,1]\); on the middle third it is \(F(x)=\tfrac12\).

The construction gives a useful stage-by-stage description. At stage \(n\), there are \(2^n\) closed intervals, each of length \(3^{-n}\). On each such interval, the restriction of \(F\), after rescaling the horizontal and vertical coordinates, is another copy of the Cantor function. Thus the values of \(F\) at the two endpoints of each stage-\(n\) interval differ by \(2^{-n}\). Each deleted open interval is an interval on which \(F\) is constant. These are the only properties of the construction needed below.

The Cantor Set Has Measure Zero

Worked Example: Measuring the Finite-Stage Sets

Compute the measure of \(C_n\), and use it to determine the measure of \(C\).

There are \(2^n\) disjoint intervals in \(C_n\), each of length \(3^{-n}\). Hence the total length is

m(C_n)=2^n3^{-n}=\left(\frac23\right)^n.

For example, the first three stage measures are \(m(C_1)=2/3\), \(m(C_2)=4/9\), and \(m(C_3)=8/27\). Since \(C\subseteq C_n\) for every \(n\), monotonicity of Lebesgue measure gives \(0\le m(C)\le(2/3)^n\) for every \(n\). The right side tends to zero, so \(m(C)=0\). Check: the stage-\(n\) total length is \(2^n\) intervals times length \(3^{-n}\), which gives the same \((2/3)^n\); it tends to zero as the stages increase.

Proof that \(m(C)=0\). Each \(C_n\) is a finite union of closed intervals and is therefore Lebesgue measurable. The sets are decreasing, with \(C=\bigcap_n C_n\), and \(m(C_0)=1<\infty\). Continuity of measure from above therefore gives \(m(C)=\lim_{n\to\infty}m(C_n)\). The stage calculation yields \(\lim_{n\to\infty}(2/3)^n=0\), so \(m(C)=0\). \(\square\)

Equivalently, the direct bound \(0\le m(C)\le m(C_n)=(2/3)^n\) for every \(n\) already forces the same conclusion. The complement of \(C\) in \([0,1]\) consists of the deleted open intervals. Each point in that complement lies in one of those intervals, not merely arbitrarily near one.

Constancy on the Complement Forces a Zero Derivative

Theorem: For the Cantor function \(F:[0,1]\to[0,1]\), \(F'(x)\) exists and equals \(0\) at every \(x\in(0,1)\setminus C\). Consequently, \(F'=0\) almost everywhere on \([0,1]\) with respect to Lebesgue measure.

Proof. Fix \(x\in(0,1)\setminus C\). Since \(C=\bigcap_n C_n\), there is some stage \(n\) for which \(x\notin C_n\). The point \(x\) belongs to a deleted open interval \((a,b)\) from one of the construction stages. Because \(x\in(a,b)\), the number \(r=\min\{x-a,b-x\}\) is positive. If \(0<|h|<r\), then \(x+h\in(a,b)\). The Cantor function is constant on \((a,b)\), so \(F(x+h)=F(x)\), and therefore \[ \frac{F(x+h)-F(x)}{h}=0. \] Every sufficiently small nonzero \(h\) gives this quotient, so its limit as \(h\to0\) exists and equals zero. This proves the derivative assertion at every such \(x\). The points of \([0,1]\) not covered by this argument lie in \(C\) or at the endpoints; \(m(C)=0\), and the two endpoints also have measure zero. Thus \(F'=0\) almost everywhere. \(\square\)

Worked Example: Differentiating Inside a Stage-Two Gap

Find the derivative of \(F\) at \(x=1/6\) directly from the construction.

The interval \((1/9,2/9)\) is removed at stage two: it lies inside the left stage-one interval \([0,1/3]\), whose middle third is \((1/9,2/9)\). The recursion gives \(F(1/9)=\tfrac12 F(1/3)=\tfrac12\cdot\tfrac12=\tfrac14\), and \(F(2/9)=\tfrac12 F(2/3)=\tfrac12\cdot\tfrac12=\tfrac14\). The function is constant throughout this deleted interval, so \(F(x)=1/4\) there. The point \(x=1/6\) is inside it. Its distances to the endpoints are \(1/6-1/9=1/18\) and \(2/9-1/6=1/18\). Thus, whenever \(0<|h|<1/18\), both \(x\) and \(x+h\) lie in the same gap and \(F(x+h)=F(x)=1/4\). It follows that

\frac{F(1/6+h)-F(1/6)}{h} =\frac{1/4-1/4}{h}=0, \qquad 0<|h|<\frac1{18},

so \(F'(1/6)=0\). Check: the difference quotient is exactly zero for every allowed nonzero \(h\), not just in the limit.

Worked Example: The Derivative Integral Does Not Give the Endpoint Change

Compare the integral of an almost-everywhere representative of \(F'\) with the change in \(F\) from \(0\) to \(1\).

Define the real-valued measurable function \(g:[0,1]\to\mathbb{R}\) by \(g(x)=0\) for every \(x\). The theorem shows that \(g\) agrees with \(F'\) at every point of \((0,1)\setminus C\), and hence agrees with the derivative almost everywhere. Its Lebesgue integral is \(\int_0^1g\,dm=0\). In contrast, the endpoint change is \(F(1)-F(0)=1-0=1\). Thus the integral of the almost-everywhere derivative does not recover the endpoint change for this function. Check: the two quantities are \(0\) and \(1\), respectively, so they are unequal.

Why This Does Not Contradict the Fundamental Theorem

The fundamental theorem for absolutely continuous functions has an additional hypothesis. In particular, an absolutely continuous function on \([0,1]\) is recovered from its derivative by integrating that derivative. The Cantor function is not absolutely continuous, so that conclusion cannot be applied to it. The almost-everywhere derivative statement alone is not a replacement for absolute continuity.

The stage intervals give a direct verification that \(F\) is not absolutely continuous. Recall the definition: \(F\) is absolutely continuous if for every \(\varepsilon>0\) there is a \(\delta>0\) such that, for every finite collection of pairwise disjoint intervals \((a_j,b_j)\) in \([0,1]\) with \(\sum_j(b_j-a_j)<\delta\), one has \(\sum_j|F(b_j)-F(a_j)|<\varepsilon\). At stage \(n\), take the \(2^n\) surviving closed intervals. They are pairwise disjoint and separated by gaps. Their total length is \((2/3)^n\), while the sum of the endpoint changes of \(F\) is \(2^n\cdot2^{-n}=1\). For example, at stage five the total length is \(2^5\cdot3^{-5}=32/243<1/3\), but the sum of the endpoint changes is \(2^5\cdot2^{-5}=1\). Taking \(\varepsilon=1/2\), no \(\delta\) required by absolute continuity can work: choose \(n\) large enough that \((2/3)^n<\delta\), and the stage-\(n\) intervals have total length below \(\delta\) but endpoint-change sum \(1\), which is not below \(1/2\). This contradicts the definition, proving that \(F\) is not absolutely continuous.

Worked Example: Testing the Absolute-Continuity Condition

Use the fifth-stage intervals to test the condition with \(\varepsilon=1/2\) and \(\delta=1/3\).

There are \(2^5=32\) stage-five intervals, each of length \(3^{-5}=1/243\). Their total length is \(32/243\), and \(32/243<1/3\) because \(96<243\). The endpoint values differ by \(2^{-5}=1/32\) on each interval, so the sum of their absolute changes is \(32\cdot(1/32)=1\). This is not less than \(\varepsilon=1/2\), even though the total interval length is less than \(\delta=1/3\). Check: the length and endpoint-change sums are computed separately from the stage count and interval lengths; the former is below \(\delta\), while the latter is above \(\varepsilon\). Thus this particular \(\delta\) fails the absolute-continuity condition. The all-stages argument above shows that every proposed \(\delta>0\) fails when \(\varepsilon=1/2\).

Common mistake. It is incorrect to argue that \(F(1)-F(0)=\int_0^1F'(x)\,dx=0\) because \(F'=0\) almost everywhere. The calculation of the integral is valid for the almost-everywhere representative \(g=0\), but identifying that integral with the endpoint change requires absolute continuity. The Cantor function fails that hypothesis, as the stage-interval calculation shows. The correct conclusion is precisely that \(F'\) exists and equals zero off the null set \(C\), while \(F\) itself remains nonconstant.

Check Your Understanding

Use the stage construction and the distinction between differentiability almost everywhere and absolute continuity to answer these questions.

How many intervals of what length make up \(C_n\), and what does this imply about \(m(C)\)?
Why does a point in a deleted open interval have a neighborhood on which the Cantor function is constant?
At a point inside a deleted interval, write the difference quotient and explain why its limit is zero.
What does the statement \(F'=0\) almost everywhere say about points in the Cantor set?
Why can the integral of an almost-everywhere representative of \(F'\) fail to equal \(F(1)-F(0)\) in this example?