# Functions of bounded variation and differentiability

Jordan Bell
March 11, 2016

## 1 Functions of bounded variation

We say that a function $f:A\to\mathbb{R}\cup\{\infty\}$, $A\subset\mathbb{R}$, is increasing if $x\leq y$ implies $F(x)\leq F(y)$, namely if $f$ is order preserving.

Let $a be real. For a function $F:[a,b]\to\mathbb{R}$, define $V_{F}:[a,b]\to[0,\infty]$ by

 $V_{F}(x)=\sup_{N,a=t_{0}

called the variation of $F$. It is apparent that $V_{F}$ is increasing. If $V_{F}$ is bounded, we say that $F$ has bounded variation. $V_{F}$ being bounded is equivalent to $V_{F}(b)<\infty$. If $F$ is increasing then

 $V_{F}(x)=F(x)-F(a),$

so in particular an increasing function has bounded variation.

Define $P_{F}:[a,b]\to[0,\infty]$ by

 $P_{F}(x)=\sup_{N,a=t_{0}

called the positive variation of $F$, and define $N_{F}:[a,b]\to[0,\infty]$ by

 $N_{F}(x)=\sup_{N,a=t_{0}

called the negative variation of $F$. It is apparent that $P_{F}$ and $N_{F}$ are increasing.

We now prove the Jordan decomposition theorem. It shows in particular that if $F$ has bounded variation then $P_{F}$ and $N_{F}$ are bounded.

###### Theorem 1 (Jordan decomposition theorem).

If $F:[a,b]\to\mathbb{R}$ has bounded variation, then for all $x\in[a,b]$,

 $V_{F}(x)=P_{F}(x)+N_{F}(x).$

and

 $F(x)-F(a)=P_{F}(x)-N_{F}(x).$
###### Proof.

For $\epsilon>0$ there is some $L$ and some $a=r_{0} for which

 $\sum_{F(r_{j})\geq F(r_{j-1})}F(r_{j})-F(r_{j-1})>P_{F}(x)-\epsilon,$

and there is some $M$ and some $a=s_{0} for which

 $\sum_{F(s_{j})\leq F(s_{j-1})}-(F(s_{j})-F(s_{j-1}))>N_{F}(x)-\epsilon.$

Let $a=t_{0} with $\{t_{0},\ldots,t_{N}\}=\{r_{0},\ldots,r_{L}\}\cup\{s_{0},\ldots,s_{M}\}$. As $\{r_{0},\ldots,r_{L}\}\subset\{t_{0},\ldots,t_{N}\}$,

 $\sum_{F(t_{j})\geq F(t_{j-1})}F(t_{j})-F(t_{j-1})\geq\sum_{F(r_{j})\geq F(r_{j% -1})}F(r_{j})-F(r_{j-1})$

and as $\{s_{0},\ldots,s_{M}\}\subset\{t_{0},\ldots,t_{N}\}$,

 $\sum_{F(t_{j})\leq F(t_{j-1})}-(F(t_{j})-F(t_{j-1}))\geq\sum_{F(s_{j})\leq F(s% _{j-1})}-(F(s_{j})-F(s_{j-1})).$

Hence

 $V_{F}(x)\geq\sum_{j}|F(t_{j})-F(t_{j-1})|>P_{F}(x)+N_{F}(x)-2\epsilon,$

and as this is true for all $\epsilon>0$ it follows that $V_{F}(x)\geq P_{F}(x)+N_{F}(x)$. And $\sup(f+g)\leq\sup f+\sup g$, so $V_{F}(x)\leq P_{F}(x)+N_{F}(x)$ and therefore $V_{F}(x)=P_{F}(x)+N_{F}(x)$. Now,

 $\displaystyle F(x)-F(a)$ $\displaystyle=\sum_{j}F(t_{j})-F(t_{j-1})$ $\displaystyle=\sum_{F(t_{j})\geq F(t_{j-1})}F(t_{j})-F(t_{j-1})-\sum_{F(t_{j})% \leq F(t_{j-1})}-(F(t_{j})-F(t_{j-1})),$

which implies

 $|F(x)-F(a)-P_{F}(x)+N_{F}(x)|<2\epsilon,$

whence $F(x)-F(a)-P_{F}(x)+N_{F}(x)=0$. ∎

The Jordan decomposition theorem tells us that if $F$ has bounded variation then

 $F(x)=(P_{F}(x)-F(a))-N_{F}(x),$

and as $x\mapsto P_{F}(x)-F(a)$ and $x\mapsto N_{F}(x)$ are increasing, this shows that $F$ is the difference of two increasing functions.

The following says that a function of bounded variation is continuous at a point if and only if its variation is continuous at that point.11 1 V. I. Bogachev, Measure Theory, volume 1, p. 333, Proposition 5.2.2.

###### Theorem 2.

If $F:[a,b]\to\mathbb{R}$ has bounded variation, then $F$ is continuous at $x$ if and only if $V_{F}$ is continuous at $x$.

###### Theorem 3.

If $F:[a,b]\to\mathbb{R}$ has bounded variation then there are at most countably many $x\in[a,b]$ at which $F$ is not continuous.

###### Proof.

According to the Jordan decomposition theorem, $V_{F}=P_{F}+N_{F}$, so it suffices to prove that if $f:[a,b]\to\mathbb{R}$ is increasing then there are at most countably many $x\in[a,b]$ at which $f$ is not continuous. Let $f(a^{-})=f(a)$ and for $a let

 $f(x^{-})=\lim_{y\to x,y

and let $f(b^{+})=f(b)$ and for $a\leq x let

 $f(x^{+})=\lim_{y\to x,y>x}f(y);$

this makes sense because $f$ is increasing, and also because $f$ is increasing we have $f(x^{-})\leq f(x)\leq f(x^{+})$. Let $E$ be the set of those $x\in[a,b]$ at which $f$ is not continuous. If $x\in E$, then $f(x^{-}) and hence there is some $r_{x}\in(f(x^{-}),f(x^{+}))\cap\mathbb{Q}$. If $x,y\in E$, $x, then as $x we have $f(x^{+})\leq f(y^{-})$, and as $x,y\in E$, $f(x^{-}) and $f(y^{-}), so $r_{x}. Therefore $x\mapsto r_{x}$ is one-to-one $E\to\mathbb{Q}$, showing that $E$ is countable. ∎

## 2 Coverings

The following is the rising sum lemma, due to F. Riesz.22 2 Elias M. Stein and Rami Shakarchi, Real Analysis, p. 118, Lemma 3.2. (We don’t use the rising sun lemma elsewhere in these notes, and instead use the Vitali covering theorem, stated next.)

###### Lemma 4 (Rising sun lemma).

Let $G:[a,b]\to\mathbb{R}$ be continuous and let $E$ be the set of those $x\in(a,b)$ for which there is some $x satisfying $G(y)>G(x)$. $G$ is open, and if $G$ is nonempty then $G$ is the union of countably many disjoint $(a_{k},b_{k})\subset[a,b]$. If $a_{k}>a$ then $G(b_{k})=G(a_{k})$, and if $a_{k}=a$ then $G(b_{k})\geq G(a_{k})$.

###### Proof.

If $x_{0}\in E$, there is some $x_{0} with $G(y_{0})>G(x_{0})$. Writing $\epsilon=G(y_{0})-G(x_{0})$, as $G$ is continuous there is some $\delta>0$, $(x_{0}-\delta,x_{0}+\delta)\subset[a,b]$, such that if $|x-x_{0}|<\delta$ then $|G(x)-G(x_{0})|<\epsilon$, so

 $\displaystyle G(y_{0})-G(x)$ $\displaystyle=\epsilon+G(x_{0})-G(x)$ $\displaystyle\geq\epsilon-|G(x)-G(x_{0})|$ $\displaystyle>0.$

Thus if $x\in(x_{0}-\delta,x_{0}+\delta)$ then $G(y_{0})>G(x)$, which shows that $E$ is open.

Suppose now that $E$ is nonempty, and for $x\in E$ let

 $A_{x}=\inf\{t\in\mathbb{R}:(t,x)\subset E\},\quad B_{x}=\sup\{t\in\mathbb{R}:(% x,t)\subset E\}.$

As $E$ is open, there is some $\delta_{x}>0$ such that $(x-\delta_{x},x+\delta_{x})\subset E$, so $A_{x}\leq x-\delta_{x} and $B_{x}\geq x+\delta_{x}>x$. Furthermore, as $E$ is open it follows that $A_{x}\not\in E$ and $B_{x}\not\in E$. For $x,y\in E$, either $(A_{x},B_{x})\cap(A_{y},B_{y})=\emptyset$ or $(A_{x},B_{x})=(A_{y},B_{y})$, and as $(A_{x},B_{x})$ contains at least one rational number,

 $E=\bigcup_{x\in E\cap\mathbb{Q}}(A_{x},B_{x}).$

As $E\cap\mathbb{Q}$ is countable, there are pairwise disjoint $(a_{k},b_{k})\subset[a,b]$, $a_{k}\not\in E,b_{k}\not\in E$, $k\in I$, such that

 $E=\bigcup_{k\in I}(a_{k},b_{k}).$

For $k\in I$, suppose by contradiction that $G(b_{k}). Let

 $C_{k}=\left\{c\in(a_{k},b_{k}):G(c)=\frac{G(a_{k})+G(b_{k})}{2}\right\},$

which is nonempty by the intermediate value theorem. Let $c_{k}=\sup C_{k}$, and because $G$ is continuous, $c_{k}\in C_{k}$. $c_{k}=b_{k}$ would imply $G(b_{k})=\frac{G(a_{k})+G(b_{k})}{2}$, contradicting $G(b_{k}); hence $c_{k}\in(a_{k},b_{k})\subset E$. Then because $c_{k}\in E$, there is some $c_{k} satisfying $G(d)>G(c_{k})$. If $d>b_{k}$ then as $b_{k}\in(a,b)\setminus E$ it holds that $G(d)\leq G(b_{k}), a contradiction, and if $d=b_{k}$ then $G(d)=G(b_{k}), a contradiction; hence $d. As $G(d)>G(c_{k})>G(b_{k})$, by the intermediate value theorem there is some $c\in(d,b_{k})$ such that $G(c)=G(c_{k})$. But then we have $c\in C_{k}$ and $c>c_{k}$, contradicting $c_{k}=\sup C_{k}$. Therefore,

 $G(b_{k})\geq G(a_{k}).$

If $a_{k}\neq a$ then $a_{k}\in(a,b)\setminus E$, which means that there is no $a_{k} satisfying $G(y)>G(a_{k})$. Hence $G(b_{k})\leq G(a_{k})$, which shows that for $a_{k}\neq a$, we have $G(b_{k})=G(a_{k})$. ∎

Let $\lambda$ be Lebesgue measure on the Borel $\sigma$-algebra of $\mathbb{R}$ and let $\lambda^{*}$ be Lebesgue outer measure on $\mathbb{R}$.

A Vitali covering of a set $E\subset\mathbb{R}$ is a collection $\mathcal{V}$ of closed intervals such that for $\epsilon>0$ and for $x\in E$ there is some $I\in\mathcal{V}$ with $x\in I$ and $0<\lambda(I)<\epsilon$. The following is the Vitali covering theorem.

###### Theorem 5 (Vitali covering theorem).

Let $U$ be an open set in $\mathbb{R}$ with $\lambda(U)<\infty$, let $E\subset U$, and let $\mathcal{V}$ be a Vitali covering of $E$ each interval of which is contained in $U$. Then for any $\epsilon>0$ there are pairwise disjoint $I_{1},\ldots,I_{n}\in\mathcal{V}$ such that

 $\lambda^{*}\left(E\setminus\bigcup_{j=1}^{n}I_{j}\right)<\epsilon.$

## 3 Differentiability

Let $F:[a,b]\to\mathbb{R}$ be a function. The Dini derivatives of $F$ are the following. $D^{-}F(x):(a,b]\to\mathbb{R}\cup\{\infty\}$ is defined by

 $D^{-}F(x)=\limsup_{h\to 0,h<0}\frac{F(x+h)-F(x)}{h},$

$D_{-}F(x):(a,b]\to\mathbb{R}\cup\{-\infty\}$ is defined by

 $D_{-}F(x)=\liminf_{h\to 0,h<0}\frac{F(x+h)-F(x)}{h},$

$D^{+}F(x):[a,b)\to\mathbb{R}\cup\{\infty\}$ is defined by

 $D^{+}F(x)=\limsup_{h\to 0,h>0}\frac{F(x+h)-F(x)}{h},$

$D_{+}F(x):[a,b)\to\mathbb{R}\cup\{-\infty\}$ is defined by

 $D_{+}F(x)=\liminf_{h\to 0,h>0}\frac{F(x+h)-F(x)}{h}.$

For $x\in[a,b]$, the upper derivative of $F$ at $x$ is

 $\overline{D}F(x)=\limsup_{h\to 0,h\neq 0}\frac{F(x+h)-F(x)}{h},$

and the lower derivative of $F$ at $x$ is

 $\underline{D}F(x)=\liminf_{h\to 0,h\neq 0}\frac{F(x+h)-F(x)}{h}.$

Let

 $\mathscr{L}=\{x\in(a,b]:D^{-}F(x)=D_{-}F(x)\},$
 $\mathscr{R}=\{x\in[a,b):D^{+}F(x)=D_{+}F(x)\}.$

For $x\in\mathscr{L}$, the left-derivative of $F$ at $x$ is

 $F^{\prime}_{-}(x)=D^{-}F(x)=D_{-}F(x),$

and for $x\in\mathscr{R}$, the right-derivative of $F$ at $x$ is

 $F^{\prime}_{+}(x)=D^{+}F(x)=D_{+}F(x).$

For $x\in(a,b)$, for $F$ to be differentiable at $x$ means that

 $-\infty

We prove that the set of points at which $F$ is left-differentiable and right-differentiable but $F^{\prime}_{-}(x)\neq F^{\prime}_{+}(x)$ is countable.33 3 V. I. Bogachev, Measure Theory, volume 1, p. 332, Lemma 5.1.3.

###### Lemma 6.

$\{x\in\mathscr{L}\cap\mathscr{R}:F^{\prime}_{-}(x)\neq F^{\prime}_{+}(x)\}$ is countable.

###### Proof.

Let $\mathbb{Q}=\{r_{k}:k\geq 1\}$, $r_{k}\neq r_{j}$ for $k\neq j$, and let

 $E=\{x\in\mathscr{L}\cap\mathscr{R}:F^{\prime}_{-}(x)

For $x\in E$, as $F^{\prime}_{-}(x) there is a minimal $k$ with $F^{\prime}_{-}(x). As $r_{k}>F^{\prime}_{-}(x)$, there is a minimal $m$ such that $r_{m} and for all $t\in(r_{m},x)$, $\frac{F(t)-F(x)}{t-x} and hence $F(t)-F(x)>r_{k}(t-x)$. Likewise, as $r_{k}, there is a minimal $n$ such that $r_{n}>x$ and for all $t\in(x,r_{n})$, $\frac{F(t)-F(x)}{t-x}>r_{k}$ and hence $F(t)-F(x)>r_{k}(t-x)$. Hence

 $F(t)-F(x)>r_{k}(t-x),\qquad t\in(r_{m},r_{n}),t\neq x.$ (1)

Now for distinct $x,y\in E$ suppose by contradiction that $(k(x),m(x),n(x))=(k(y),m(y),n(y))$. As $x,y\in(r_{m},r_{n})$, using (1) with $t=y$ and $t=x$ we get

 $F(y)-F(x)>r_{k}(y-x),\qquad F(x)-F(y)>r_{k}(x-y),$

yielding $r_{k}(x-y), a contradiction. Therefore $x\mapsto(k(x),m(x),n(x))$ is one-to-one $E\to\mathbb{N}^{3}$, for $\mathbb{N}$ the positive integers, which shows that $E$ is countable.

We similarly prove that

 $\{x\in\mathscr{L}\cap\mathscr{R}:F^{\prime}_{-}(x)>F^{\prime}_{+}(x)\}$

is countable. ∎

## 4 Differentiability of increasing functions

We now use the Vitali covering lemma to prove that the Dini derivatives of an increasing function are finite almost everywhere.44 4 Russell A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, p. 55, Lemma 4.8.

###### Lemma 7.

Let $F:[a,b]\to\mathbb{R}$ be an increasing function and let

 $A^{-}=\{x\in(a,b]:D^{-}F(x)=\infty\},\;A_{-}=\{x\in(a,b]:D_{-}F(x)=-\infty\},$
 $A^{+}=\{x\in[a,b):D^{+}F(x)=\infty\},\;A_{+}=\{x\in[a,b):D_{+}F(x)=-\infty\}.$

Then

 $\lambda^{*}(A^{-})=0,\lambda^{*}(A_{-})=0,\lambda^{*}(A^{+})=0,\lambda^{*}(A_{% -})=0.$
###### Proof.

Because $F$ is increasing, for any $h\neq 0$, $\frac{F(x+h)-F(x)}{h}\geq 0$, and therefore $A_{-}=\emptyset$ and $A_{+}=\emptyset$. Suppose by contradiction that

 $\lambda^{*}(A^{-})=\alpha>0.$

As $\alpha>0$, there is some $r>0$ satisfying

 $\frac{r\alpha}{2}>F(b)-F(a).$

For $x^{-}$, because $D^{-}F(x)=\infty$ there is an increasing sequence $t_{x,k}\in[a,b]$ that tends to $x$ such that for each $k\geq 1$,

 $\frac{F(x)-F(t_{x,k})}{x-t_{x,k}}\geq r.$ (2)

Let

 $\mathcal{V}=\{[t_{x,k},x]:x\in A^{-},k\geq 1\},$

which is a Vitali covering of $A^{-}$, and so by the Vitali covering theorem there are pairwise disjoint $[t_{x_{j},k_{j}},x_{j}]\in\mathcal{V}$, $1\leq j\leq n$, such that

 $\lambda^{*}\left(A^{-}\setminus\bigcup_{j=1}^{n}[t_{x_{j},k_{j}},x_{j}]\right)% <\frac{\alpha}{2}$

and then

 $\lambda^{*}(A^{-})\leq\lambda^{*}\left(A^{-}\setminus\bigcup_{j=1}^{n}[t_{x_{j% },k_{j}},x_{j}]\right)+\lambda\left(\bigcup_{j=1}^{n}[t_{x_{j},k_{j}},x_{j}]% \right),$

hence

 $\displaystyle\sum_{j=1}^{n}\lambda([t_{x_{j},k_{j}},x_{j}])$ $\displaystyle=\lambda\left(\bigcup_{j=1}^{n}[t_{x_{j},k_{j}},x_{j}]\right)$ $\displaystyle\geq\lambda^{*}(A^{-})-\lambda^{*}\left(A^{-}\setminus\bigcup_{j=% 1}^{n}[t_{x_{j},k_{j}},x_{j}]\right)$ $\displaystyle>\alpha-\frac{\alpha}{2}.$

That is,

 $\sum_{j=1}^{n}(x_{j}-t_{x_{j},k_{j}})>\frac{\alpha}{2}.$

Now, by (2), $F(x_{j})-F(t_{x_{j},k_{j}})\geq r(x_{j}-t_{x_{j},k_{j}})$, so

 $\sum_{j=1}^{n}(F(x_{j})-F(t_{x_{j},k_{j}}))\geq\sum_{j=1}^{n}r(x_{j}-t_{x_{j},% k_{j}})>\frac{r\alpha}{2}>F(b)-F(a).$

But because the intervals $[t_{x_{j},k_{j}},x_{j}]$ are pairwise disjoint and $F$ is increasing, $\sum_{j=1}^{n}(F(x_{j})-F(t_{x_{j},k_{j}}))\leq F(b)-F(a)$, contradicting the above inequality. Therefore $\lambda^{*}(A^{-})=0$.

Suppose by contradiction that

 $\lambda^{*}(A^{+})=\alpha>0.$

As $\alpha>0$, there is some $r>0$ satisfying

 $\frac{r\alpha}{2}>F(b)-F(a).$

For $x\in A^{+}$, because $D^{+}F(x)=\infty$ there is a decreasing sequence $t_{x,k}\in[a,b]$ that tends to $x$ such that for each $k\geq 1$,

 $\frac{F(t_{x,k})-F(x)}{t_{x,k}-x}\geq r.$ (3)

Let

 $\mathcal{V}=\{[x,t_{x,k}]:x\in A^{+},k\geq 1\},$

which is a Vitali covering of $A^{+}$, and so by the Vitali covering theorem there are pairwise disjoint $[x_{j},t_{x_{j},k_{j}}]\in\mathcal{V}$, $1\leq j\leq n$, such that

 $\lambda^{*}\left(A^{+}\setminus\bigcup_{j=1}^{n}[x_{j},t_{x_{j},k_{j}}]\right)% <\frac{\alpha}{2}$

and then

 $\lambda^{*}(A^{+})\leq\lambda^{*}\left(A^{+}\setminus\bigcup_{j=1}^{n}[x_{j},t% _{x_{j},k_{j}}]\right)+\lambda\left(\bigcup_{j=1}^{n}[x_{j},t_{x_{j},k_{j}}]% \right),$

hence

 $\displaystyle\sum_{j=1}^{n}\lambda([x_{j},t_{x_{j},k_{j}}])$ $\displaystyle=\lambda\left(\bigcup_{j=1}^{n}[x_{j},t_{x_{j},k_{j}}]\right)$ $\displaystyle\geq\lambda^{*}(A^{+})-\lambda^{*}\left(A^{+}\setminus\bigcup_{j=% 1}^{n}[x_{j},t_{x_{j},k_{j}}]\right)$ $\displaystyle>\alpha-\frac{\alpha}{2}.$

That is,

 $\sum_{j=1}^{n}(t_{x_{j},k_{j}}-x_{j})>\frac{\alpha}{2}.$

Now, by (3), $F(t_{x_{j},k_{j}})-F(x_{j})\geq r(t_{x_{j},k_{j}}-x_{j})$, so

 $\sum_{j=1}^{n}(F(t_{x_{j},k_{j}})-F(x_{j}))\geq\sum_{j=1}^{n}r(t_{x_{j},k_{j}}% -x_{j})>\frac{r\alpha}{2}>F(b)-F(a).$

But because the intervals $[x_{j},t_{x_{j},k_{j}}]$ are pairwise disjoint and $F$ is increasing, $\sum_{j=1}^{n}(F(t_{x_{j},k_{j}})-F(x_{j}))\leq F(b)-F(a)$, contradicting the above inequality. Therefore $\lambda^{*}(A^{+})=0$. ∎

We now prove that an increasing function is differentiable almost everywhere.55 5 Russell A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, p. 55, Theorem 4.9.

###### Theorem 8.

Let $F:[a,b]\to\mathbb{R}$ be increasing and let

 $E=\{x\in(a,b):-\infty

Then $\lambda^{*}([a,b]\setminus E)=0$.

###### Proof.

Let

 $A=\{x\in(a,b):D_{+}F(x)

and suppose by contradiction that $\lambda^{*}(A)>0$. Since

 $A=\bigcup_{p,q\in\mathbb{Q},p

which is a union of countably many sets, there are some $p,q\in\mathbb{Q}$, $p, such that $\lambda^{*}(B)=\beta>0$,

 $B=\{x\in(a,b):D_{+}F(x)

Let $\epsilon>0$. There is an open set $U\subset(a,b)$ with $B\subset U$ and $\lambda(U)<\lambda^{*}(B)+\epsilon=\beta+\epsilon$. For $x\in B$, because $D_{+}F(x) and because $x$ belongs to the open set $U$, there is a sequence $t_{x,k}\in(x,x+1/k)$, $[x,t_{x,k}]\subset U$, such that for each $k\geq 1$,

 $\frac{F(t_{x,k})-F(x)}{t_{x,k}-x}

Then

 $\mathcal{V}=\{[x,t_{x,k}]:x\in B,k\geq 1\}$

is a Vitali covering of $B$, so by the Vitali covering theorem there are pairwise disjoint $[x_{j},t_{x_{j},k_{j}}]\in\mathcal{V}$, $1\leq j\leq m$, such that

 $\lambda^{*}\left(B\setminus\bigcup_{j=1}^{m}[x_{j},t_{x_{j},k_{j}}]\right)<\epsilon,$

and then, as the intervals $[x_{j},t_{x_{j},k_{j}}]$ are pairwise disjoint and are all contained in $U$,

 $\displaystyle\sum_{j=1}^{m}(F(t_{x_{j},k_{j}})-F(x_{j}))$ $\displaystyle<\sum_{j=1}^{m}p(t_{x_{j},k_{j}}-x_{j})$ $\displaystyle=p\sum_{j=1}^{m}\lambda([x_{j},t_{x_{j},k_{j}}]]$ $\displaystyle=p\lambda\left(\bigcup_{j=1}^{m}[x_{j},t_{x_{j},k_{j}}]\right)$ $\displaystyle\leq p\lambda(U)$ $\displaystyle

Let $C=B\cap\bigcup_{j=1}^{n}(x_{j},t_{x_{j},k_{j}})$, for which

 $\beta=\lambda^{*}(B)\leq\lambda^{*}(C)+\lambda^{*}\left(B\setminus\bigcup_{j=1% }^{n}[x_{j},t_{x_{j},k_{j}}]\right)<\lambda^{*}(C)+\epsilon,$

so

 $\lambda^{*}(C)>\beta-\epsilon.$

For $y\in C$ there is some $i$ for which $y\in(x_{i},t_{x_{i},k_{i}})$, and because $D^{+}F(y)>q$ there is a sequence $u_{y,l}\in(y,y+1/l)$, $[y,u_{y,l}]\subset(x_{i},t_{x_{i},k_{i}})$, such that for each $l\geq 1$,

 $\frac{F(u_{y,l})-F(y)}{u_{y,l}-y}>q.$

Then

 $\mathcal{W}=\{[y,u_{y,l}]:y\in B,l\geq 1\}$

is a Vitali covering of $C$, so by the Vitali covering theorem there are pairwise disjoint $[y_{j},u_{y_{j},l_{j}}]\in\mathcal{W}$, $1\leq j\leq n$, such that

 $\lambda^{*}\left(C\setminus\bigcup_{j=1}^{n}[y_{j},u_{y_{j},l_{j}}]\right)<\epsilon,$

so

 $\lambda^{*}(C)\leq\lambda^{*}\left(C\setminus\bigcup_{j=1}^{n}[y_{j},u_{y_{j},% l_{j}}]\right)+\lambda\left(\bigcup_{j=1}^{n}[y_{j},u_{y_{j},l_{j}}]\right)<% \epsilon+\sum_{j=1}^{n}\lambda([y_{j},u_{y_{j},l_{j}}]),$

and then

 $\sum_{j=1}^{n}(F(u_{y_{j},l_{j}})-F(y_{j}))>\sum_{j=1}^{n}q(u_{y_{j},l_{j}}-y_% {j})>q(\lambda^{*}(C)-\epsilon)>q(\beta-2\epsilon).$

Now for $1\leq i\leq m$ let $\pi_{i}=\{1\leq j\leq n:[y_{j},u_{y_{j},l_{j}}]\subset(x_{i},t_{x_{i},k_{i}})\}$. Because $F$ is increasing, if $j\in\pi_{i}$ then $F(u_{y_{j},l_{j}})-F(y_{j})\leq F(t_{x_{i},k_{i}})-F(x_{i})$, and because each $[y_{j},u_{y_{j},l_{j}}]$ is contained in some $(x_{i},t_{x_{i},k_{i}})$,

 $\displaystyle q(\beta-2\epsilon)$ $\displaystyle<\sum_{j=1}^{n}(F(u_{y_{j},l_{j}})-F(y_{j}))$ $\displaystyle=\sum_{i=1}^{m}\sum_{j\in\pi_{i}}(F(u_{y_{j},l_{j}})-F(y_{j}))$ $\displaystyle\leq\sum_{i=1}^{m}(F(t_{x_{i},k_{i}})-F(x_{i}));$

the last inequality also uses that the intervals $[y_{j},u_{y_{j},l_{j}}]$ are pairwise disjoint. But we have found $\sum_{i=1}^{m}(F(t_{x_{i},k_{i}})-F(x_{i})), so $q(\beta-2\epsilon). As this is true for all $\epsilon>0$, it holds that $q\beta\leq p\beta$, and as $\beta>0$ we get $q\leq p$, contradicting that $p. Therefore $\lambda^{*}(A)=0$. ∎