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Let the function $f: [a, b] \rightarrow \mathbb{R}$ be Lipschitz, that is, there is a constant $c \geq 0$ such that for all $u, v \in [a, b]$, $|f( u) - f( v)| \le c |u - v|$. Show that $f$ maps a set of measure zero onto a set of measure zero.

The following is my attempt:

Suppose that $A \subseteq [0,1]$ has measure zero and $\varepsilon > 0$ be given.
By definition of measure, there exists a countable collection of open intervals $\{ I_k \}_{k \geq 1}$ such that $A \subseteq \cup_{k=1}^{\infty}{I_k}$ and $\sum_{k=1}^{\infty}{l(I_k)} < \frac{\varepsilon}{c}. $ Note that $f(A) \subseteq f(\cup_{k=1}^{\infty}{I_k}) = \cup_{k=1}^{\infty}{f(I_k)}.$ For each $k$, denote $I_k = (x_k, y_k).$ Since $f$ is Lipschitz, for each $k$, $|f(x_k) - f(y_k)| \leq c \cdot |x_k - y_k|.$ Therefore, $\sum_{k=1}^{\infty}{|f(x_k) - f(y_k)|} \leq c \cdot \sum_{k=1}^{\infty}{|x_k - y_k|} < \varepsilon.$ Hence, $f(A)$ has measure zero.

Adren
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Idonknow
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    Yes, this is correct. – Andres Mejia Mar 07 '17 at 09:13
  • the first answer here is essentially your own. – Andres Mejia Mar 07 '17 at 09:14
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    @Idonknow I think your answer is not correct. What is $f(I_k)$ and how did you estimate its measure. (It is an interval by IVP but you have to say what the end points are). If $f$ is monotonic you can write down the end points of $f(I_k)$, but you have to fill in some details when $f$ is not monotonic. – Kavi Rama Murthy Aug 14 '18 at 06:06
  • I also agree with @KaviRamaMurthy At first I also thought what you did $m(f(I_k)) = |x_k - y_k|$ holds for every $f$ but this does not hold if $f$ is not monotonic because then it is possible $f(I_k) \supseteq [min(f(x_k), f(y_k)), max(f(x_k), f(y_k))]$ – zia badar Dec 15 '24 at 12:13

1 Answers1

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I am providing my answer as the answer given is question do not explains how measure of $m(f(I_k))$ is computed, from what I understood the answers assumes $m(f(I_k)) = |f(a_k) - f(b_k)|$ where $I_k = [a_k, b_k]$ which is incorrect if $f$ is not monotone.

My answer:

Let $E$ be a measure set of measure zero then its open cover $\bigcup_{i=1}^{\infty}I_k$ has measure $\sum_{i=1}^{\infty}l(I_k) < \epsilon$. Since $f$ is continious $f(I_k)$ is also an interval, since $f$ is lipchitz if $u,v \in [a_k, b_k]$ then $|f(u) -f(v)| \leq c|u-v|$ which implies if $q,r\in f(I_k)$ then $|q-r|\leq c(b_k-a_k)$ as $c(b_k-a_k)$ is the largest possible value on r.h.s in case of $I_k$, which implies $q,r$ has freedom to be at most $c(b_k-a_k)$ distance away from each other which means $m(f(I_k)) \leq c(b_k-a_k)$.

Now $m(f(E)) \leq m(f(\bigcup_{i=1}^{\infty}I_k)) \leq \sum_{i=1}^{\infty}m(f(I_k)) \leq \sum_{i=1}^{\infty}c(b_k-a_k) \lt c\epsilon$

zia badar
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