Proof of Growth Bound for a Lipschitz Function with Small Derivative

Note first that $\sigma_{1}=1$, since $\sigma_{1}$ is positive with $\sigma_{1}^{2}=1$ by its defining clause. Hence the Lipschitz image bound reads, in dimension $1$,

$$\lambda^{\ast}\bigl(\psi(S)\bigr)\le 2M\,\lambda^{\ast}(S)$$

for a map $\psi$ Lipschitz with constant $M>0$ on a nonempty domain and $S$ a subset of that domain.

Claim 1. We may assume $A\ne\varnothing$, the case $A=\varnothing$ being trivial. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon<1$.

By outer regularity of the outer measure there is an open set $O'$ with $A\subseteq O'$ and $\lambda(O')\le\lambda^{\ast}(A)+\varepsilon$; put $O=O'\cap I$, which is open, being a finite intersection of open sets (The Open Subsets of a Metric Space Form a Topology), and contains $A$. Both $O$ and $O'$ are Borel, so by agreement on Borel sets and monotonicity, $\lambda(O)=\lambda^{\ast}(O)\le\lambda^{\ast}(O')=\lambda(O')\le\lambda^{\ast}(A)+\varepsilon$.

Let $x\in A$. Since $\phi$ has the derivative $\phi'(x)$ at $x$, there is a real $\delta_{x}>0$ such that

$$\Bigl|\frac{\phi(t)-\phi(x)}{t-x}-\phi'(x)\Bigr|\le\varepsilon\qquad\text{whenever }t\in I,\ 0<|t-x|<\delta_{x},$$

whence $|\phi(t)-\phi(x)|\le(|\phi'(x)|+\varepsilon)|t-x|\le(K+\varepsilon)|t-x|$ for those $t$, and trivially also for $t=x$.

Let $\mathcal{F}$ be the set of pairs $(x,h)$ with $x\in A$, $0<h<\delta_{x}$ and $\bar{B}(x,h)\subseteq O$. It finely covers $A$: given $x\in A$ and a real $\eta>0$, $x$ is an interior point of the open set $O$, so claim 1 of Interior Points in the Metric Topology are Exactly the Centres of Contained Closed Balls supplies a real $\rho>0$ with $\bar{B}(x,\rho)\subseteq O$, and any $h$ with $0<h<\min\{\eta,\delta_{x},\rho\}$ gives a pair of $\mathcal{F}$ whose ball contains $x$ and whose radius is less than $\eta$.

For $(x,h)\in\mathcal{F}$ and $t\in\bar{B}(x,h)$ we have $|t-x|\le h<\delta_{x}$, so $|\phi(t)-\phi(x)|\le(K+\varepsilon)h$. Hence $\phi(\bar{B}(x,h))$ is contained in the closed interval with endpoints $\phi(x)\pm(K+\varepsilon)h$, and therefore, by monotonicity, agreement on Borel sets, and claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness,

$$\lambda^{\ast}\bigl(\phi(\bar{B}(x,h))\bigr)\le 2(K+\varepsilon)h=(K+\varepsilon)\,\lambda\bigl(\bar{B}(x,h)\bigr). \tag{$\ast$}$$

Since $\lambda^{\ast}(A)<\infty$, the Vitali covering theorem supplies pairs $(x_{1},h_{1}),\dots,(x_{N},h_{N})\in\mathcal{F}$ whose closed balls are pairwise disjoint and satisfy $\lambda^{\ast}(A\setminus\bigcup_{i}\bar{B}(x_{i},h_{i}))\le\varepsilon$. Write $A'=A\cap\bigcup_{i}\bar{B}(x_{i},h_{i})$ and $A''=A\setminus\bigcup_{i}\bar{B}(x_{i},h_{i})$, so that $\phi(A)=\phi(A')\cup\phi(A'')$.

The balls $\bar{B}(x_{i},h_{i})$ are pairwise disjoint members of $\mathcal{B}(\mathbb{R})$ contained in $O$, so $\sum_{i}\lambda(\bar{B}(x_{i},h_{i}))\le\lambda(O)$ by claims 1 and 2 of Basic Properties of a Measure. Using countable subadditivity of $\lambda^{\ast}$, applied to the sequence whose first $N$ terms are the sets $\phi(\bar{B}(x_{i},h_{i}))$ and whose remaining terms are empty, and then $(\ast)$,

$$\lambda^{\ast}\bigl(\phi(A')\bigr)\le\sum_{i=1}^{N}\lambda^{\ast}\bigl(\phi(\bar{B}(x_{i},h_{i}))\bigr)\le(K+\varepsilon)\sum_{i=1}^{N}\lambda\bigl(\bar{B}(x_{i},h_{i})\bigr)\le(K+\varepsilon)\bigl(\lambda^{\ast}(A)+\varepsilon\bigr).$$

By the displayed Lipschitz image bound applied to $\phi$ and $S=A''$, together with $\lambda^{\ast}(A'')\le\varepsilon$, we get $\lambda^{\ast}(\phi(A''))\le 2L\varepsilon$. Adding, and using countable subadditivity once more for the two-set union $\phi(A)=\phi(A')\cup\phi(A'')$,

$$\lambda^{\ast}\bigl(\phi(A)\bigr)\le(K+\varepsilon)\bigl(\lambda^{\ast}(A)+\varepsilon\bigr)+2L\varepsilon\le K\lambda^{\ast}(A)+\varepsilon\,c,\qquad c=\lambda^{\ast}(A)+K+1+2L,$$

using $\varepsilon<1$. The real number $c$ is positive and does not depend on $\varepsilon$; taking $\varepsilon=1/2$ in the display shows in particular that $\lambda^{\ast}(\phi(A))$ is finite. If $\lambda^{\ast}(\phi(A))>K\lambda^{\ast}(A)$ we could choose, by The Archimedean Property of the Real Numbers, a real $\varepsilon$ with $0<\varepsilon<1$ and $\varepsilon c<\lambda^{\ast}(\phi(A))-K\lambda^{\ast}(A)$, contradicting the display. Hence $\lambda^{\ast}(\phi(A))\le K\lambda^{\ast}(A)$.

Claim 2. Let $x,y\in I$. If $x=y$ the assertion is trivial, so assume $x<y$ after exchanging them if necessary; note $|y-x|=y-x$. Since $I$ is an interval containing $x$ and $y$, the closed interval $[x,y]$ is contained in $I$.

Put $A=[x,y]\setminus N$. Then $\lambda^{\ast}(A)\le\lambda([x,y])=y-x<\infty$ by monotonicity and claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness, and by hypothesis $\phi$ has at every point of $A$ a derivative bounded in absolute value by $K$. Claim 1 therefore gives $\lambda^{\ast}(\phi(A))\le K(y-x)$. The set $[x,y]\cap N$ is null, being a subset of the null set $N$, by the null-set claim; hence $\phi([x,y]\cap N)$ is null by the null-image claim. As $\phi([x,y])=\phi(A)\cup\phi([x,y]\cap N)$, countable subadditivity gives

$$\lambda^{\ast}\bigl(\phi([x,y])\bigr)\le K(y-x).$$

The function $\phi$ is continuous on $[x,y]$, being Lipschitz and hence uniformly continuous by A Lipschitz Map is Uniformly Continuous and Continuous. By Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals, $\phi$ attains on $[x,y]$ every value between $\phi(x)$ and $\phi(y)$; hence $\phi([x,y])$ contains the closed interval $J$ with endpoints $\phi(x)$ and $\phi(y)$, whose measure is $|\phi(y)-\phi(x)|$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness. By monotonicity and agreement on Borel sets,

$$|\phi(y)-\phi(x)|=\lambda^{\ast}(J)\le\lambda^{\ast}\bigl(\phi([x,y])\bigr)\le K(y-x)=K|y-x| .$$