MathOverflow is a question and answer site for professional mathematicians. 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

It's not difficult to find a function $f\colon \mathbb R \to \mathbb R$ such that its restriction to any (not trivial) interval is surjective. Does anyone know whether such a function is necessarily not (Lebesgue) measurable? I'm pretty sure this is the case, but I cannot prove it.

Here is an example of such an $f$. Let $\sim$ be the equivalence relation defined by $x \sim y$ if and only if $x-y \in \mathbb Q$, with $x$ and $y$ in $\mathbb R$. Let $C$ be $\mathbb R/\sim$ (here we need the axiom of choice), and denote with $\pi$ the projection. Since $\mathbb Q$ is countable, $C$ must have the cardinality of the continuum. Let $\varphi \colon C \to \mathbb R$ a bijection. Then $f:=\varphi \circ \pi$ has the required property.

share|cite|improve this question
up vote 6 down vote accepted

No. Conway's base-13 function is measurable.

See this question.

share|cite|improve this answer
Aw dang, I was about to mention this. – Simon Rose Jul 22 '10 at 21:40
In fact, it is almost everywhere 0. – Willie Wong Jul 22 '10 at 21:41
Wow, amazing! Thank you very much! – Ricky Jul 22 '10 at 21:45
Yeah, it's a cool function. No problem. – Richard Kent Jul 22 '10 at 21:49
That's a great function! It almost ought to appear on the “favourite mathematical jokes” thread… – Peter LeFanu Lumsdaine Jul 22 '10 at 22:16

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.