Cantor function has singular distributional derivative

Let $(X_n{)}_{n\in \mathbb{N}}$ be any sequence of nonempty sets. By AC, [F1], the range $\{X_n:n\in \mathbb{N}\}$ has a choice function $s$; then $n\mapsto s(X_n)$ selects from every $X_n$, proving Countable Choice. this discharges precisely the CC assumptions in [F5] (Cantor nullity), [F6] (the Cantor measure and its properties), [F10] (weak-derivative locality/uniqueness), and [F13] (Riemann–Stieltjes/Lebesgue–Stieltjes agreement). The Cantor staircase is a real continuous nondecreasing function on $[0,1]$, with endpoint values $0$ and $1$, and is constant on every closed gap $[a,b]$ of $C$ by [F2]–[F4]. Apart from invoking AC for this reduction, no further family of choices is made; each later CC use is through a named interface.

Put $U=(0,1)\setminus C$. It is open because $C$ is closed by [F22] and $(0,1)$ is open, so [F8] writes it as at most countably many disjoint intervals $J=(a,b)$. Every endpoint lies in $[0,1]$. The endpoints $0$ and $1$ belong to $C$ by [F23]; if an endpoint in $(0,1)$ were outside $C$, closedness would put it in the open set $U$, and a neighborhood would enlarge the component, contradicting maximality. Thus $a,b\in C$ and $(a,b)\cap C=\mathrm{\varnothing }$. By [F3], $c$ is constant on $[a,b]$; by [F17] its restriction is absolutely continuous, and its classical derivative is $0$ throughout $J$. Meanwhile $C$ is null by [F5].

Let $\phi \in C_c^{\mathrm{\infty }}(0,1)$ be real-valued and extend it by zero to $[0,1]$. Then $\phi (0)=\phi (1)=0$. For a partition $0=t_0<\cdots <t_N=1$, the telescoping product identity is $\phi (1)c(1)-\phi (0)c(0)={\sum }_{i=1}^N\phi (t_i)(c(t_i)-c(t_{i-1}))+{\sum }_{i=1}^{N}c(t_{i-1})(\phi (t_i)-\phi (t_{i-1}))\mathrm{.}$ The first sum uses right-endpoint tags for ${\int }_0^1\phi dc$, which exists by [F12]; the second uses left-endpoint tags for ${\int }_0^{1}cd\phi$, which exists by [F11]. As the mesh tends to zero, each sum converges to its Riemann–Stieltjes integral, so the zero boundary term gives ${\int }_0^1\phi dc=-{\int }_0^{1}cd\phi$. By [F11], the right side is $-{\int }_0^{1}c(x){\phi }^{\mathrm{\prime }}(x)dx$. The agreement in [F13] identifies the left side with ${\int }_{(0,1]}\phi d{\mu }_c$; since $\phi$ is supported inside $(0,1)$, this equals ${\int }_{\mathbb{R}}\phi d{\mu }_c$. Thus ${\int }_{\mathbb{R}}\phi d{\mu }_c=-{\int }_0^{1}c(x){\phi }^{\mathrm{\prime }}(x)dx=⟨\mathrm{\partial }{T}_c,\phi ⟩\mathrm{.}$ For a complex test, apply the real identity separately to its real and imaginary parts; complex linearity of the distribution pairing and the measure integral then gives the same identity. The order-zero estimate in [F19] makes $\phi \mapsto \int \phi d{\mu }_c$ a distribution on $(0,1)$. Hence $\mathrm{\partial }{T}_c$ is exactly the distribution induced by ${\mu }_c{\mathrm{\mid }}_{(0,1)}$. Since [F6] makes ${\mu }_c$ singular and concentrated on the null set $C$, and [F7] gives ${\mu }_c((0,1))=1$, this restricted measure is singular and nonzero.

Suppose $c\in W^{1,1}(0,1)$ and let $g\in L^1(0,1)$ be its weak derivative, as supplied by [F9]. For each component interval $J$ of $U$, the constant function $0$ is a weak derivative of $c{\mathrm{\mid }}_J$: for every $\psi \in C_c^{\mathrm{\infty }}(J)$ choose $[a,b]\subset J$ containing its support. By [F21], ${\int }_J{\psi }^{\mathrm{\prime }}={\int }_a^b{\psi }^{\mathrm{\prime }}=\psi (b)-\psi (a)=0$, and constancy of $c$ on $J$ gives ${\int }_{J}c{\psi }^{\mathrm{\prime }}=0$. By [F10], $g=0$ almost everywhere on each $J$. The exceptional sets on these countably many intervals have null union by [F15]; [F5] makes the remaining Cantor set null too. Therefore $g=0$ almost everywhere on $(0,1)$.

The compact intervals $K_m=[1\mathrm{/}m,1-1\mathrm{/}m]$, $m\ge 3$, cover $(0,1)$: for each $x\in (0,1)$, [F20] gives an $m$ with $1\mathrm{/}m<\min \{x,1-x\}$. Since ${\mu }_c((0,1))=1$ by [F7], [F15] implies ${\mu }_c(K_m)>0$ for at least one $m$. Fix one such $m$. The cutoff fact [F16] gives $\chi \in C_c^{\mathrm{\infty }}(0,1)$ with $0\le \chi \le 1$ and $\chi =1$ on a neighborhood of $K_m$. By [F14], $\int \chi d{\mu }_c\ge \int {\mathbf{1}}_{K_m}d{\mu }_c={\mu }_c(K_m)>0.$ But the weak-derivative identity and step 2.1 give $-{\int }_0^{1}c{\chi }^{\mathrm{\prime }}={\int }_0^{1}g\chi =0$, contradicting step 1.3, which identifies the left side with $\int \chi d{\mu }_c>0$. So $c\mathrm{\notin }{W}^{1,1}(0,1)$.

Steps 1.1–1.2 show absolute continuity on every complementary gap away from the null Cantor set, while steps 1.3, 2.1, and 3.1 show that the distributional derivative is the nonzero singular Cantor measure and that no $L^1$ weak derivative exists. This is the claimed counterexample.