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Here is the proof in multiple dimensions Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?

Here is my trial in 1-dimensional i.e. $f \in Lip[0,1],$ but I am unable to complete:

Let $f \in Lip[0,1]$, Let $c > 0$ be a Lipschitz constant for $f$ on $[0,1],$ that is, $$|f(u) - f(v)| \leq c|u - v|, \quad \forall u,v \in [0,1].$$

Now, we want to show that this $f$ belongs to $\mathcal{F}$ which denotes the collection of real-valued functions defined on $[0,1]$ that map every set of measure zero to a set of measure 0.\

Let $A$ be a set of measure $0.$ then for every $\epsilon > 0$ by the definition of outer measure on pg.31 $A$ can be covered by a countable collection of nonempty open bounded intervals $ \{(a_n,b_n)\}_{n = 1}^{\infty }$ (i.e $A \subset \bigcup_{n=1}^\infty (a_n,b_n)$) with $\sum_{n} (b_n-a_n) <\epsilon$. \

Now, since $f \in Lip[0,1],$ then $f((a_n,b_n))$ is contained in an interval of length $c\epsilon $

My questions are:

1-What is the argument that $f$ should take? is it an interval or a set ?

2-Should the $\epsilon$ depend on $n$?

3-Could anyone help me complete the proof?

Intuition
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1 Answers1

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For a given interval $(a,b)$, use the Lipschitz-continuity for the center $\frac{a+b}{2}$ of the interval, i.e. you have that $|f(\frac{a+b}{2})-f(x)| \leq c|\frac{a+b}{2}-x|$ for every $x \in (a,b)$. Now, we know that the distance from $x$ to the center to the interval is at most $\frac{1}{2}$ times the size of the interval, i.e. $c|\frac{a+b}{2}-x| \leq \frac{c}{2}|a-b| = \frac{c}{2}(b-a)$. This means that $|f(\frac{a+b}{2})-f(x)| \leq \frac{c}{2}(b-a)$, or in other words, $f(x) \in (f(\frac{a+b}{2})-\frac{c}{2}(b-a),f(\frac{a+b}{2})+\frac{c}{2}(b-a))$. Since this holds for every $x \in (a,b)$, it follows that $f((a,b)) \subseteq (f(\frac{a+b}{2})-\frac{c}{2}(b-a),f(\frac{a+b}{2})+\frac{c}{2}(b-a))$, an interval of length $c(b-a)$, which is $c$ times the length of the original interval. This means that if $I$ is an interval of length $l$, $f(I)$ is contained in an interval of length $\leq cl$. Can you go on from here?

Edit: I'll complete the answer since you asked for it, but the rest follows directly. You've already stated that for any set $A$, we have that $\lambda(A) \Leftrightarrow \forall\varepsilon>0 \exists (I_n): A \subseteq \cup_{n\in\mathbb{N}} I_n \land \sum_{n\in\mathbb{N}} \lambda(I_n) < \varepsilon$. If we replace $\varepsilon$ with $\frac{\varepsilon}{c}$, we obtain that $\sum_{n\in\mathbb{N}} \lambda(I_n) < \frac{\varepsilon}{c}$ and since we just proved that $\lambda(f(I_n)) \leq c\lambda(I_n)$ for every interval, it follows that $\sum_{n\in\mathbb{N}} \lambda(f(I_n)) < c\frac{\varepsilon}{c} = \varepsilon$. Also, since for any interval $I$, $f(I)$ is an interval as well (why?), and since $\cup_{n\in\mathbb{N}} f(I_n)$ is a cover of $f(A)$, it follows (again from the definition of an outer measure) that $f(A)$ has Lebesgue-measure zero.

Nicolas
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