Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider Lebesgue measure on $\mathbb{R}$ and let $f:\mathbb{R}\to\mathbb{R}$ be differentiable. Does $f$ necessarily preserve measure zero sets? Does $f$ necessarily preserve measurable sets?

Note that if $f$ is $C^1$ then $f$ preserves measure zero sets since $C^1$ functions are locally Lipschitz. Therefore $C^1$ functions also preserve measurable sets since a measurable set is the union of an $F_\sigma$ set and an measure zero set, and continuous functions preserve $F_\sigma$ sets. More generally if $f$ is absolutely continuous on each interval then $f$ preserves both measure zero sets and measurable sets.

However, I'm not sure about the differentiable case. I would guess that the answer to both questions is no. I'm interested in a counter example or proof in each case.

share|cite|improve this question
This is a special case of… (Shine posted this as a comment on an answer, and I wanted to make it more prominent). – Jonas Meyer Jul 16 '14 at 16:37
up vote 5 down vote accepted

If $f$ preserves null-sets, it also preserves Lebesgue-measurability (but not necessarily Borel measurability), because by regularity, every Lebesgue measurable set $M$ can be written as

$$ M = N \cup \bigcup K_n $$

with $K_n$ compact and $N$ a null-set. By continuity, $f$ preserves compact sets.

Now a theorem in Rudin, Real and Complex Analysis (Lemma 7.25) shows in particular that every everywhere differentiable function maps null-sets to null-sets, so that your claim is true.

share|cite|improve this answer
Ah sorry about that, you are right. – Chris Janjigian Jul 16 '14 at 16:29
Never mind, I should not have posted such a harsh reply to your comment :) – PhoemueX Jul 16 '14 at 20:08

The following claim implies your result.

Claim: Let $E \subseteq \mathbb{R}$ be arbitrary. Suppose $|f'(x)| \leq M$ for all $x \in E$. Then $\mu^{\star}(f[E]) \leq M\mu^{\star}(E)$.

Proof: Fix $\epsilon > 0$. Get an open set $U \supseteq E$ with $\mu(U) < \mu^{\star}(E) + \epsilon$. For each $x \in E$, let $\delta_x > 0$ be such that $(x - \delta_x, x + \delta_x) \subseteq U$ and $|f(y) - f(x)| \leq (M + \epsilon)|y - x|$ for every $y \in (x - \delta_x, x+ \delta_x)$. Consider the family of closed intervals: $V = \{[f(x), f(y)], [f(y), f(x)] : x \in E, |y - x| < \delta_x\}$ - This is a Vitali covering of $E$ which means that every point in $E$ is covered by arbitrarily small intervals in $V$. Using Vitali covering theorem, get a countable subfamily $C$ of pairwise disjoint intervals which covers all but a null part of $f[E]$. The (disjoint) union of the intervals in $C$ has measure at most $(M + \epsilon) \mu(U)$ which is less than $(M + \epsilon)(\mu^{\star}(E) + \epsilon)$. Now, let $\epsilon$ go to zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.