Differentiation of Monotone Functions and the Vitali Covering Theorem — Examples

By The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, every point of $[0,1]\setminus C$ lies in an open interval on which $c$ is constant. Hence $c^{\mathrm{\prime }}(x)=0$ for all $x\mathrm{\notin }C$. Since $C$ is Lebesgue null by The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero, this proves $c^{\mathrm{\prime }}=0$ almost everywhere.

Fix $x\in C$. Write the ternary expansion of $x$ using only digits $0$ and $2$, and let $u_n\le x\le v_n$ be the two points of $C$ obtained by freezing the first $n$ ternary digits of $x$ and filling the remaining digits with all $0$'s and all $2$'s. Then $v_n-u_n=3^{-n}$ and $c(v_n)-c(u_n)=2^{-n}$ by the digit description of The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$ and the definition of the Cantor function. At least one of the two numerator differences $c(v_n)-c(x)$ and $c(x)-c(u_n)$ is at least $2^{-n-1}$. For that choice, the corresponding denominator is positive and at most $v_n-u_n=3^{-n}$, so one of the two secant slopes is at least $2^{-n-1}\mathrm{/}{3}^{-n}=\frac{1}{2}(3\mathrm{/}2{)}^n$. These lower bounds are unbounded, so $c$ cannot have a finite derivative at $x$.

The endpoint values $c(0)=0$ and $c(1)=1$ are part of The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, so the function still rises by one.

Steps 1.1, 2.1, and 2.2 prove the example.

Every summand $x\mapsto {\mathbf{1}}_{\{q_n\le x\}}{2}^{-n}$ is nondecreasing, so $J$ is nondecreasing. If $x<y$, choose $q_m\in \mathbb{Q}\cap (x,y]$; then the $m$th summand contributes $0$ at $x$ and $2^{-m}$ at $y$, so $J(y)-J(x)\ge 2^{-m}>0$. Hence $J$ is increasing. At a rational point $q_m$, the value of the $m$th summand jumps by $2^{-m}$, so $J$ is discontinuous at $q_m$. Thus the discontinuity set contains $\mathbb{Q}\cap (0,1]$, hence is dense in $[0,1]$.

Let $x\in [0,1)$ be irrational. Given $\epsilon >0$, choose $N$ so large that ${\sum }_{n>N}{2}^{-n}<\epsilon \mathrm{/}3$. Because $x\mathrm{\ne }{q}_n$ for $n\le N$, there is a neighborhood of $x$ containing none of the finitely many rationals $q_1,\dots ,q_N$, so the first $N$ partial sums are constant on that neighborhood. The tail contributes less than $\epsilon \mathrm{/}3$ on either side, so $J$ is continuous at $x$. Also $J(0)=0$ because every $q_n$ is positive. Therefore the discontinuity set is exactly $\mathbb{Q}\cap (0,1]$.

The function $J$ has no endpoint defect at $0$, and because the enumeration has no repetitions, at each $q_n$ its jump size is exactly $2^{-n}$. Therefore claim 2 of A nondecreasing function splits uniquely into a jump part and a continuous part identifies the jump function of $J$ with $J$ itself. Countable Choice is assumed, so the jump-function theorem A jump function has derivative zero almost everywhere gives $J^{\mathrm{\prime }}=0$ almost everywhere.

Steps 1.1 through 3.1 prove the example.

Enumerate all closed rational intervals $I_n=[u_n,v_n]$ with $0\le u_n<v_n\le 1$. For each $n$, let $c_n$ be the function that is $0$ on $[0,u_n]$, is $1$ on $[v_n,1]$, and on $[u_n,v_n]$ is the affine rescaling of the Cantor function. The Cantor function is continuous by The Cantor function is continuous on $[0,1]$, so each $c_n$ is continuous, nondecreasing, and takes values in $[0,1]$. Define $S(x):={\sum }_{n\ge 1}{2}^{-n-1}{c}_n(x)$. The series converges uniformly because each summand is bounded by $2^{-n-1}$, so $S$ is continuous and nondecreasing.

If $x<y$, choose a rational interval $I_n$ with $x<u_n<v_n<y$. Then $c_n(x)=0$ and $c_n(y)=1$, so $S(y)-S(x)\ge 2^{-n-1}>0$. Hence $S$ is strictly increasing. For each $n$, the derivative of $c_n$ is $0$ almost everywhere because off the scaled Cantor set inside $I_n$ the function is locally constant by The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, and that scaled Cantor set is null by The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points. Since each $c_n$ is nondecreasing and Countable Choice is assumed, the term-by-term differentiation theorem Fubini's theorem on term-by-term differentiation for pointwise sums of nondecreasing functions applies and gives $S^{\mathrm{\prime }}(x)={\sum }_{n\ge 1}{2}^{-n-1}{c}_n^{\mathrm{\prime }}(x)=0$ almost everywhere.

The function $S$ is continuous, nondecreasing, strictly increasing, and has derivative $0$ almost everywhere, so it is a singular function by A singular function on a compact interval.

Steps 1.1 through 3.1 prove the example.

For $h\mathrm{\ne }0$, $\frac{f(h)-f(0)}{h}=\sin (1\mathrm{/}h)$. Choose $h_n=\frac{1}{\pi \mathrm{/}2+2\pi n}$ and $k_n=\frac{1}{3\pi \mathrm{/}2+2\pi n}$, both positive. Then $\sin (1\mathrm{/}{h}_n)=1$ and $\sin (1\mathrm{/}{k}_n)=-1$, so the upper and lower right Dini derivatives are at least $1$ and at most $-1$ respectively. Since $\mathrm{\mid }\sin (1\mathrm{/}h)\mathrm{\mid }\le 1$ always, we get $D^{+}f(0)=1$ and $D_{+}f(0)=-1$.

Choose instead $h_n^{-}=-\frac{1}{3\pi \mathrm{/}2+2\pi n}$ and $k_n^{-}=-\frac{1}{\pi \mathrm{/}2+2\pi n}$. Then $\sin (1\mathrm{/}{h}_n^{-})=1$ and $\sin (1\mathrm{/}{k}_n^{-})=-1$, so the upper and lower left Dini derivatives are $D^{-}f(0)=1$ and $D_{-}f(0)=-1$.

Therefore the four Dini derivatives are exactly the four stated values and are not all equal.

Each summand $x\mapsto {\mathbf{1}}_{\{q_n\le x\}}{2}^{-n}$ is nondecreasing, and the geometric tail ${\sum }_{n>N}{2}^{-n}$ tends to $0$. Consequently $J(x):={\sum }_{q_n\le x}{2}^{-n}$ is well defined and nondecreasing. Since $x\mapsto x$ is continuous and increasing, $F=x+J$ is nondecreasing.

Fix $N$. Away from the finite set $\{q_1,\dots ,q_N\}$, the first $N$ summands defining $J$ are locally constant, while the remaining summands have total size at most ${\sum }_{n>N}{2}^{-n}$. Letting $N\to \mathrm{\infty }$ shows that $J$ is continuous at every point outside the enumeration. At $q_m$, take $N\ge m$: the same tail estimate shows that the left limit differs from $J(q_m)$ by exactly $2^{-m}$ and that the right limit equals $J(q_m)$. It also shows that $J(x)\to 0=J(0)$ as $x\downarrow 0$. Thus $J$ has no endpoint defect at $0$, has an interior left jump of size $2^{-n}$ at each $q_n\in (0,1)$, has the left jump $2^{-n}$ at the unique $q_n=1$, and has no right jumps.

The continuous summand $x\mapsto x$ does not change these jump data. Claim 2 of A nondecreasing function splits uniquely into a jump part and a continuous part therefore computes the jump function of $F$ on $[0,1]$ as precisely $J$. The continuous remainder is $F-J=x$.

This is the claimed decomposition.

Consider the family $\mathcal{V}:=\{[0,t]:0<t<1\}\cup \{[t,1]:0<t<1\}$. It covers $[0,1]$, but it is not a fine cover at any interior point.

Exactly as in FALSE: the Vitali covering theorem holds for arbitrary covers, any disjoint subfamily of $\mathcal{V}$ has at most one left interval and at most one right interval, hence leaves a nonempty open gap. So this cover has no disjoint subfamily with null uncovered remainder.

Therefore the fine-cover hypothesis is genuinely necessary.

For every partition of $[0,1]$, all endpoint increments vanish except possibly the one crossing $1\mathrm{/}2$, and that increment has absolute value $1$. Hence the total variation of $H$ is $1$, so $H$ is of bounded variation.

On each side of $1\mathrm{/}2$ the function is locally constant, so $H^{\mathrm{\prime }}(x)=0$ for every $x\mathrm{\ne }1\mathrm{/}2$. Thus $H$ is differentiable almost everywhere. This is exactly the phenomenon asserted abstractly by Every function of bounded variation is differentiable almost everywhere.

Steps 1.1 and 2.1 prove the example.