Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

If $f$ is a continuous function on $X$ and E a Lebesgue measurable set, can we conclude that $f^{-1}(E)$ is measurable?

share|cite|improve this question
I presume $f : X \to Y$ is a continuous function between topological spaces equipped with their Borel $\sigma$-algebras? – Qiaochu Yuan Sep 28 '12 at 2:35
I know that "The inverse image of any Borel set is measurable". But I want to see if the general case (any measurable) holds. – Anita Sep 28 '12 at 2:37
No. See here for the standard counterexample:… – Chris Janjigian Sep 28 '12 at 2:39
What do you mean by "the general case" here? What is the codomain of $f$ and what $\sigma$-algebra are you using on it? – Qiaochu Yuan Sep 28 '12 at 2:45
Actually the example in the link provided by Chris is used to proof that "measurable" is really more general than "Borel": there is a measurable set $B$ such that its inverse image by a continuous map is not measurable so is not Borel. The same function in that post serves to show that measurability is not a topological property. – leo Sep 28 '12 at 4:57

Your Answer


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

Browse other questions tagged or ask your own question.