Differentiation of Monotone Functions and the Vitali Covering Theorem

Each well-posed Dini derivative is an upper or lower limit of a nonempty family of real difference quotients, so its value exists in $\bar{\mathbb{R}}$ by the definitions of $\mathrm{lim\hspace{0.17em}sup}$ and $\mathrm{lim\hspace{0.17em}inf}$ on the extended line. This proves claim 1.

For every family of real numbers, the liminf is at most the limsup. Applied to the right-hand difference quotients and to the left-hand difference quotients, this gives $D_{+}f(x)\le D^{+}f(x)$ and $D_{-}f(x)\le D^{-}f(x)$. This is claim 2.

Assume first that $f^{\mathrm{\prime }}(x)$ exists as a finite real number $L$. Then the right and left difference quotients both converge to $L$, because a two-sided limit exists exactly when both one-sided limits exist and agree (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree). Hence each one-sided limsup and each one-sided liminf equals $L$, so all four Dini derivatives equal $L$.

Conversely, assume that the four Dini derivatives all equal the same finite real number $L$. Then on the right the limsup and liminf of the difference quotients coincide at $L$, so the right-hand quotient limit exists and equals $L$; the same is true on the left. Therefore the two-sided derivative exists and equals $L$ by If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree.

Steps 2.1 and 2.2 prove claim 3, and steps 1.1 and 1.2 prove claims 1 and 2.

Let $x\in E$. Choose $y>x$ with $F(y)>F(x)$. By continuity at $x$, after shrinking if necessary there is $\epsilon$ with $0<\epsilon <y-x$ such that $F(t)<F(y)$ whenever $\mathrm{\mid }t-x\mathrm{\mid }<\epsilon$ and $t\in [a,b]$. Every $t\in (x-\epsilon ,x+\epsilon )\cap [a,b)$ then satisfies $t<y$ and $F(t)<F(y)$, so it also belongs to $E$. Thus $E$ is open in the subspace $[a,b)$. Therefore $E\cap (a,b)$ is open in $\mathbb{R}$, and Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, namely its order components writes it as a countable disjoint union of open intervals. A component meeting the left endpoint is $[a,d)$ when $a\in E$; if $a\mathrm{\notin }E$, an open component may instead have the form $(a,d)$. Thus every component of $E$ is either $[a,d)$ when $a\in E$, or $(c,d)$ with $a\le c<d\le b$.

Fix a component $I$ of $E$, write its left endpoint as $c$ and its right endpoint as $d$, and let $x\in I$. By the extreme value theorem Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value, choose $t\in [x,b]$ at which $F$ attains its maximum on $[x,b]$. Since $x\in E$, some point to the right of $x$ has value greater than $F(x)$, so $t>x$ and $F(t)>F(x)$. The maximizing point $t$ does not belong to $E$. Since every point of $I$ lies in $E$, this forces $t\ge d$. If $d<b$ and $F(t)>F(d)$, then necessarily $t>d$, which would put $d$ in $E$, contrary to $d$ being the right endpoint of the component. Thus $F(t)\le F(d)$; the reverse inequality holds because $d\in [x,b]$ and $t$ is a maximizer. When $d=b$ one has $t=d$ directly. Hence in all cases $F(x)<F(t)=F(d)$.

Letting $x\downarrow c$ through points of $I$ in step 2.1 and using continuity at $c$ gives $F(c)\le F(d)$. If $c>a$, then $c\mathrm{\notin }E$, so no point to the right of $c$ has value strictly larger than $F(c)$; in particular $F(d)\le F(c)$. Hence $F(c)=F(d)$ when $c>a$.

Steps 1.1 through 3.1 are exactly the claimed conclusions.

Fix $R>0$ and put $G(x):=F(x)-Rx$ on $[u,v]$. If $x\in E_R^{+}(u,v)$, then some $y\in (x,v]$ satisfies $\frac{F(y)-F(x)}{y-x}>R$, hence $G(y)>G(x)$. Therefore $E_R^{+}(u,v)$ is contained in the rising-sun set of $G$ on $[u,v]$. By Riesz's rising sun lemma with the correct endpoint conclusion, the components of that set are intervals $I_n$ whose left and right endpoints we call $c_n,d_n$, with each $I_n$ either $[u,d_n)$ or $(c_n,d_n)$, and with $G(c_n)\le G(d_n)$ for every $n$. Thus $R(d_n-c_n)\le F(d_n)-F(c_n)$ for every $n$. Summing over finitely many components and using that the disjoint ordered intervals $I_n$ lie in $[u,v]$ gives $R{\sum }_{n\le N}(d_n-c_n)\le {\sum }_{n\le N}(F(d_n)-F(c_n))\le F(v)-F(u)$, so $R\lambda (E_R^{+}(u,v))\le F(v)-F(u)$.

Fix $0<r<R$ and put $H(x):=Rx-F(x)$ on $[u,v]$. If $x\in E_{r,R}^{-}(u,v)$, then for some $y\in [u,x)$ one has $\frac{F(x)-F(y)}{x-y}<r$, hence $H(x)-H(y)>(R-r)(x-y)>0$. Reflecting $H$ across the midpoint of $[u,v]$ turns this into the right-hand rising-sun situation on a continuous function, so the same argument as in step 1.1 yields a disjoint family of intervals whose total length bounds $\lambda (E_{r,R}^{-}(u,v))$ and on each such interval $H(d)-H(c)\ge (R-r)(d-c)$. Summing gives $(R-r)\lambda (E_{r,R}^{-}(u,v))\le H(v)-H(u)=R(v-u)-F(v)+F(u)$.

Steps 1.1 and 2.1 are the two asserted inequalities.

Replacing $F$ by $-F$ if necessary, we may assume that $F$ is nondecreasing. For each integer $N\ge 1$, the first inequality of One-sided Hardy-Littlewood inequalities for the Dini derivatives of a continuous monotone function with $u=a$, $v=b$, and $R=N$ shows that the sets $\{x\in [a,b):D^{+}F(x)>N\}$ and, after reflecting the interval, $\{x\in (a,b]:D^{-}F(x)>N\}$ have measure at most $(F(b)-F(a))\mathrm{/}N$. Hence $D^{+}F$ and $D^{-}F$ are finite almost everywhere.

Fix rationals $0<r<R$ and put $E_{r,R}:=\{x\in (a,b):D^{+}F(x)>R\text{ and }{D}_{-}F(x)<r\}\mathrm{.}$ For each rational $h>0$, the maps $x\mapsto \frac{F(x+h)-F(x)}{h}\text{on }(a,b-h),x\mapsto \frac{F(x)-F(x-h)}{h}\text{on }(a+h,b)$ are continuous because $F$ is continuous. Therefore $\{D^{+}F>R\}={\bigcap }_{m\ge 1}{\bigcup }_{\begin{array}{c}h\in \mathbb{Q}\\ 0<h<1\mathrm{/}m\end{array}}\{x\in (a,b-h):\frac{F(x+h)-F(x)}{h}>R\},$ and similarly $\{D_{-}F<r\}={\bigcap }_{m\ge 1}{\bigcup }_{\begin{array}{c}h\in \mathbb{Q}\\ 0<h<1\mathrm{/}m\end{array}}\{x\in (a+h,b):\frac{F(x)-F(x-h)}{h}<r\}\mathrm{.}$ Hence both sets are Borel, so $E_{r,R}$ is Lebesgue measurable. For any subinterval $[u,v]\subseteq [a,b]$, the two inequalities of One-sided Hardy-Littlewood inequalities for the Dini derivatives of a continuous monotone function give $R\lambda (E_{r,R}\cap [u,v])\le F(v)-F(u)$ and $(R-r)\lambda (E_{r,R}\cap [u,v])\le R(v-u)-F(v)+F(u)$. Adding them yields $(2R-r)\lambda (E_{r,R}\cap [u,v])\le R(v-u)$, so $\lambda (E_{r,R}\cap [u,v])\le \frac{R}{2R-r}(v-u)\mathrm{.}$ The constant $\frac{R}{2R-r}$ is strictly less than $1$. Therefore no point of $E_{r,R}$ can be a density-one point of $E_{r,R}$. By the Lebesgue density theorem Lebesgue density theorem, $E_{r,R}$ is null.

Applying step 2.1 to the reflected function $x\mapsto F(a+b-x)$ shows that for every rational $0<r<R$ the set $E_{r,R}^{\mathrm{\prime }}:=\{x\in (a,b):D^{-}F(x)>R\text{ and }{D}_{+}F(x)<r\}$ is null as well. Taking the countable union over rational pairs and using A countable union of measure-zero sets has measure zero, by countable choice, we conclude that outside a null set one has $D^{+}F\le D_{-}F$ and $D^{-}F\le D_{+}F$. Combined with the sidewise inequalities $D_{+}F\le D^{+}F$ and $D_{-}F\le D^{-}F$ from The four Dini derivatives always exist in the extended reals, satisfy the one-sided order inequalities, and detect finite differentiability, this forces $D_{+}F=D^{+}F=D_{-}F=D^{-}F$ at almost every point where the four Dini derivatives are finite.

Steps 1.1 and 3.1 show that all four Dini derivatives agree finitely almost everywhere on $(a,b)$. Therefore The four Dini derivatives always exist in the extended reals, satisfy the one-sided order inequalities, and detect finite differentiability implies that $F^{\mathrm{\prime }}$ exists almost everywhere. Since a null set in the Lebesgue sense is the same as elementary measure zero on the line (A subset of $\mathbb{R}$ has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers), this is exactly the claimed almost-everywhere statement.

If $a=b$, the definition gives $J_F(a)=0$ and $C_F(a)=F(a)$, so all five claims are immediate on the singleton interval. Hence assume $a<b$ from now on.

If $a<x<y\le b$, every finite pair $(S,T)$ with $S\subseteq (a,x]$ and $T\subseteq (a,x)$ is also admissible for $y$, so $J_F(x)\le J_F(y)$. The comparison with $x=a$ follows from $J_F(a)=0$ and the nonnegative definition of $J_F(y)$. Thus $J_F$ is nondecreasing on $[a,b]$.

Fix $a\le x<y\le b$. First suppose $x>a$, and split any finite pair $(S,T)$ admissible in the definition of $J_F(y)$ into its old part $S_0\subseteq (a,x]$, $T_0\subseteq (a,x)$ and its new part $S_1\subseteq (x,y]$, $T_1\subseteq [x,y)$. The old contribution, including ${\beta }_a$, is at most $J_F(x)$. Order the distinct points of $S_1\cup T_1$. Monotonicity of $F$ makes the new jump contributions telescope through disjoint successive value intervals from $F(x)$ to $F(y)$; this includes the possible right jump $F(x^{+})-F(x)$ when $x\in T_1$, and gives a total at most $F(y)-F(x)$. Taking the supremum over $(S,T)$ yields $J_F(y)-J_F(x)\le F(y)-F(x)$. If $x=a$, the endpoint contribution ${\beta }_a=F(a^{+})-F(a)$ followed by the jumps in $(a,y]$ and $(a,y)$ telescopes in the same way, giving more precisely $0\le J_F(y)-{\beta }_a\le F(y)-F(a^{+})$ and hence $J_F(y)\le F(y)-F(a)$. Therefore the increment inequality holds for every $a\le x<y\le b$, and $C_F(y)-C_F(x)=F(y)-F(x)-(J_F(y)-J_F(x))\ge 0.$ Thus $C_F$ is nondecreasing, proving claim 1.

By Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used, the discontinuity set of $F$ in $(a,b)$ is at most countable; index it without repetitions as $(s_n{)}_{n\in A}$ for some $A\subseteq \mathbb{N}$. At points outside that set both interior one-sided jump sizes vanish. The only remaining possible contribution in the defining supremum is the left jump at $b$, which occurs exactly when $x=b$. Thus The jump function of a nondecreasing function on a compact interval agrees with the nonnegative series $J_F(x)={\beta }_a+{\sum }_{n\in A,s_n\le x}(F(s_n)-F(s_n^{-}))+{\sum }_{n\in A,s_n<x}(F(s_n^{+})-F(s_n))+{\mathbf{1}}_{\{b\}}(x)(F(b)-F(b^{-}))(x>a)\mathrm{.}$ This is claim 2.

Fix $c\in (a,b)$. Step 3.1 shows that $J_F(c)-J_F(c^{-})=F(c)-F(c^{-}),J_F(c^{+})-J_F(c)=F(c^{+})-F(c)\mathrm{.}$ Therefore $C_F(c)-C_F(c^{-})=0,C_F(c^{+})-C_F(c)=0.$ Since $C_F$ is nondecreasing, One-sided limits of a monotone function always exist: for $f$ nondecreasing on an interval $I$ and $c\in I$, ${\lim }_{x\to c^{-}}f(x)=\sup \{f(x):x\in I,x<c\}$ whenever $I$ has points below $c$, ${\lim }_{x\to c^{+}}f(x)=\inf \{f(x):x\in I,x>c\}$ whenever it has points above $c$, and these satisfy ${\lim }_{x\to c^{-}}f(x)\le f(c)\le {\lim }_{x\to c^{+}}f(x)$ gives the one-sided limits at $c$, and the displayed equalities force $C_F(c^{-})=C_F(c)=C_F(c^{+})$. Hence $C_F$ is continuous at every interior point. Moreover, the displayed equalities show that $J_F$ has exactly the same left and right jumps as $F$. This proves claim 4 except for the tautological identity $F=J_F+C_F$.

To treat the endpoints, note first that step 2.1 with $x=a$ gives $0\le J_F(x)-{\beta }_a\le F(x)-{\lim }_{t\to a^{+}}F(t)$ for every $x>a$. Because the right-hand side tends to $0$ as $x\downarrow a$, $J_F(x)\to {\beta }_a$. Hence $C_F(x)=F(x)-J_F(x)⟶{\lim }_{x\to a^{+}}F(x)-{\beta }_a=F(a)=C_F(a),$ so $C_F$ is right-continuous at $a$. At $b$, the endpoint term in step 3.1 gives $J_F(b)-J_F(b^{-})=F(b)-F(b^{-}),$ and therefore $C_F(b^{-})=C_F(b)$. Thus $C_F$ is continuous on all of $[a,b]$. This proves claim 3.

Claim 4 contains the identity $F=J_F+C_F$ by definition of $C_F$. For claim 5, step 3.1 expresses $J_F(x)$ pointwise in terms of ${\beta }_a$, the interior left and right jump sizes of $F$, and the left jump at $b$, so those data determine $J_F$ uniquely. Once $J_F$ is fixed, the remainder is forced by $C_F=F-J_F$.

Steps 1.1 through 5.1 prove the theorem.

Choose an open set $U\supseteq E$ with finite length and $\lambda (U)<{\lambda }^{\ast }(E)+1$. Because $\mathcal{V}$ is fine, every $x\in E$ lies in arbitrarily short intervals of $\mathcal{V}$, so after shrinking we may restrict to the subfamily ${\mathcal{V}}_U:=\{I\in \mathcal{V}:I\subseteq U\},$ which is still a fine cover of $E$. Using dependent choice, choose a sequence $(I_n{)}_{n\ge 1}$ in ${\mathcal{V}}_U\cup \{\mathrm{\varnothing }\}$ as follows: as long as there exists an interval of ${\mathcal{V}}_U$ disjoint from the previously chosen nonempty intervals, choose $I_n$ disjoint from those earlier nonempty intervals and with length at least half the supremum of the lengths of all such candidates; once no disjoint candidate remains, set $I_n=\mathrm{\varnothing }$ and all later terms equal to $\mathrm{\varnothing }$.

Let $M:=\{n\ge 1:I_n\mathrm{\ne }\mathrm{\varnothing }\}$. The nonempty intervals $(I_n{)}_{n\in M}$ are pairwise disjoint and lie in the finite-open set $U$, so ${\sum }_{n\in M}\mathrm{\mid }{I}_n\mathrm{\mid }\le \lambda (U)<\mathrm{\infty }\mathrm{.}$ If $M$ is finite, write $M=\{1,\dots ,N\}$ and let $R:=E\setminus {\bigcup }_{n=1}^N{I}_n$. If $x\in R$, then the finite union ${\bigcup }_{n=1}^N{I}_n$ is closed and does not contain $x$, so $d:=\mathrm{dist}(x,{\bigcup }_{n=1}^N{I}_n)>0$. Because ${\mathcal{V}}_U$ is a fine cover of $E$, choose $J\in {\mathcal{V}}_U$ with $x\in J$ and $\mathrm{\mid }J\mathrm{\mid }<d$. Then $J$ is disjoint from $I_1,\dots ,I_N$, contradicting the terminal clause in step 1.1. Therefore $R=\mathrm{\varnothing }$. In this finite-termination case the theorem already holds with the finite family $I_1,\dots ,I_N$, and the finite $\epsilon$-version is immediate as well.

It remains to treat the case where $M$ is infinite. Then no $I_n$ is empty, because empties persist forever after the first one, so ${\sum }_{n\ge 1}\mathrm{\mid }{I}_n\mathrm{\mid }<\mathrm{\infty }$ and therefore $\mathrm{\mid }{I}_n\mathrm{\mid }\to 0$. Let $R:=E\setminus {\bigcup }_{n\ge 1}{I}_n$. Fix $x\in R$ and $m\ge 1$. Since $x$ does not lie in the finite union $I_1\cup \cdots \cup I_{m-1}$, the distance $d_m:=\mathrm{dist}\text{\hspace{0.17em}⁣}(x,{\bigcup }_{n=1}^{m-1}{I}_n)$ is positive. Also $s_m:={\sup }_{n\ge m}\mathrm{\mid }{I}_n\mathrm{\mid }>0,$ because there are infinitely many nonempty intervals in the tail. Choose $J\in {\mathcal{V}}_U$ with $x\in J,\mathrm{\mid }J\mathrm{\mid }<\min (d_m,s_m)\mathrm{.}$ Then $J$ is disjoint from $I_1,\dots ,I_{m-1}$. If $J$ were disjoint from every $I_n$ with $n\ge m$, then $J$ would remain a candidate at every later stage. The half-maximal choice would then give $\mathrm{\mid }J\mathrm{\mid }\le 2\mathrm{\mid }{I}_n\mathrm{\mid }$ for every $n\ge m$, contradicting $\mathrm{\mid }{I}_n\mathrm{\mid }\to 0$. So $J$ meets some $I_n$ with $n\ge m$; let $n$ be the least such index. Then $J$ is disjoint from $I_1,\dots ,I_{n-1}$, so it is an admissible candidate at stage $n$, and the half-maximal choice gives $\mathrm{\mid }J\mathrm{\mid }\le 2\mathrm{\mid }{I}_n\mathrm{\mid }$. Intersecting intervals with comparable lengths satisfy $J\subseteq 5I_n$. Therefore $x\in {\bigcup }_{n\ge m}5I_n$. Since $m$ was arbitrary, $R\subseteq {\bigcap }_{m\ge 1}{\bigcup }_{n\ge m}5I_n\mathrm{.}$

By countable subadditivity of outer measure and the interval formula for Lebesgue outer measure, ${\lambda }^{\ast }(R)\le {\inf }_{m\ge 1}{\lambda }^{\ast }\text{\hspace{0.17em}⁣}\left({\bigcup }_{n\ge m}5I_n\right)\le {\inf }_{m\ge 1}5{\sum }_{n\ge m}\mathrm{\mid }{I}_n\mathrm{\mid }=0.$ This proves the countable disjoint conclusion in the infinite case. For the finite version, choose $N$ so large that $5{\sum }_{n>N}\mathrm{\mid }{I}_n\mathrm{\mid }<\epsilon$ and repeat the same estimate with $R_N:=E\setminus {\bigcup }_{n=1}^N{I}_n$.

Steps 1.1 through 4.1 prove both forms of the theorem.

Assume first that $E$ is null, and let $\epsilon >0$. Choose an open set $U\supseteq E$ with $\lambda (U)<\epsilon$. Restrict $\mathcal{V}$ to intervals lying in $U$; the restricted family is still a fine cover of $E$. Applying The Vitali covering theorem for fine covers on the real line gives a countable disjoint subfamily $(J_n)$ of $\mathcal{V}$ with $\lambda \text{\hspace{0.17em}⁣}(E\setminus {\bigcup }_n{J}_n)=0.$ Because the $J_n$ lie in $U$ and are disjoint, ${\sum }_{n\ge 1}\mathrm{\mid }{J}_n\mathrm{\mid }\le \lambda (U)<\epsilon \mathrm{.}$ Thus $(J_n)$ is the required cheap countable subfamily.