Lipschitz function $f\colon A \to \mathbb{R}$, $A \subseteq \mathbb{R}$ measurable.

Question

Suppose $A \subseteq \mathbb{R}$ is measurable and $f\colon A \to \mathbb{R}$ is Lipschitz on the set $A$, i.e there is some $K\ge 0$ such that $\lvert f(x)-f(y)\rvert \le K \lvert x-y\rvert$ for $x,y \in A$.

I'm trying to prove that $$ m^\ast(f(E)) \le K\,m^\ast(E)\textrm{ for every set }E \subseteq A. $$

I've tried covering the set $E$ by intervals, yet the function $f$ may not have been defined outside the set $A$. Also, approaching the set $E$ from inside can encounter problem when $E$ is non-measurable.

Answer 1

We do not know if the function can be extended to $\mathbb{R}$ by continuous extension theorem, so we handle it using the method we use to prove Growth Lemma.

For $\forall E \subseteq A$, we can cover $E$ with $\{I_k\}$, s.t. $$ E \subseteq \cup_k I_k\quad \textrm{and}\quad \sum_k m(I_k) \le m^\ast (E) + \varepsilon $$ Since $A$ is measurable, we have $\{I^\prime_k\}$ with $I_k^\prime = \ A \cap I_k$ measurable for every $k$. $$ E \subseteq \cup_k I_k^\prime\quad \textrm{and}\quad \sum_k m(I^\prime_k) \le m^\ast (E) + \varepsilon $$ For $\forall x, y \in I_k^\prime$,

$\because x, y \in A$ and being Lipschitz on the set $A$.

$\therefore |f(x) - f(y)| \le K |x - y|$.

$\Rightarrow \textrm{diam}(f(I_k^\prime)) = \textrm{sup}\{|f(x) - f(y)|:x,y \in I_k^\prime\} \le K m(I_k^\prime)$

Since $f$ maps $A$ to $\mathbb{R}$, and interval $Q_k$ with $m(Q_k) = {\rm diam}(f(I_k^\prime))$ can cover $f(I_k^\prime)$.

$\Rightarrow m^\ast (f(I_k^\prime)) \le m(Q_k) = {\rm diam}(f(I_k^\prime))$

Hence, $m^\ast (f(I_k^\prime)) \le K m(I_k^\prime)$. Then we can conclude that \begin{align*} m^\ast (f(E)) & \le m^\ast(\cup_k f(I_k^\prime))\\ & \le \sum_k m^\ast (f(I_k^\prime))\\ & \le K \sum_k m(I_k^\prime)\\ & \le K \sum_k m(I_k)\\ &\le K (m^\ast (E) + \varepsilon) \end{align*}

Since $\varepsilon$ is arbitrary, the result follows.

Answer 2

I've tried covering the set $E$ by intervals, yet the function $f$ may not have been defined outside the set $A$.

Good that you noticed that problem.

But that problem can be dealt with, we can extend $f$ to a Lipschitz continuous $F \colon \mathbb{R} \to \mathbb{R}$ with the same Lipschitz constant $K$:

First, since Lipschitz functions are uniformly continuous, $f$ has a unique continuous extension $\overline{f} \colon \overline{A} \to \mathbb{R}$. This extension is also Lipschitz continuous with Lipschitz constant $K$: for $x,y \in \overline{A}$, choose sequences $(x_n),\, (y_n)$ in $A$ with $x_n \to x,\, y_n \to y$. Then

$$\lvert \overline{f}(x) - \overline{f}(y)\rvert = \lim_{n\to\infty} \lvert f(x_n) - f(y_n)\rvert \leqslant \lim_{n\to\infty} K\lvert x_n - y_n\rvert = K\lvert x-y\rvert.$$

Then we can extend $\overline{f}$ to all of $\mathbb{R}$. $\mathbb{R}\setminus \overline{A}$ is a disjoint union of countably many open intervals. For a bounded interval $(a,b)$ in the complement of $\overline{A}$, we have $a,b \in \overline{A}$ and can define the extension by linear interpolation,

$$F(x) := \frac{x-a}{b-a} \overline{f}(b) + \frac{b-x}{b-a} \overline{f}(a).$$

If $\overline{A}$ is bounded above, and $a = \max \overline{A}$, define $F(x) = \overline{f}(a)$ for $x > a$. If $\overline{A}$ is bounded below and $b = \min \overline{A}$, define $F(x) = \overline{f}(b)$ for $x < b$. (And if $A = \varnothing$, there's nothing to do, but we can set $F(x) = 0$ for all $x$.)

Verify that the extension $F$ thus defined is a globally Lipschitz function with $\lvert F(x) - F(y)\rvert \leqslant K\lvert x-y\rvert$ for all $x,y\in \mathbb{R}$.

Then covering $E$ by intervals and looking at the corresponding cover of $f(E) = F(E)$ works.