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

I am reading John Milnor's Topology from a differentiable viewpoint. In Chapter 3 be proves Sard's theorem and claims (page 18) that if $g:R^n\to R^p$ is smooth with set of critical points $C'$ then $g(C')$ is measurable. It is written that this follows from the fact that $g(C')$ can be expressed as a countable union of compact subsets. Can someone explain why $g(C')$ can be expressed in such a way?

share|cite|improve this question
up vote 3 down vote accepted

The set $\mathbb R^n$ is a countable union of compact subsets. Take, for example the compact subsets to be the squares $$D(a) = \{(b_i) \in \mathbb R^n \ | \ a_i \leq b_i \leq a_1 + 1 \ \forall i\}$$ where $a = (a_i)$ ranges over $\mathbb Z^n$ (which is countable).

As the domain is a countable union of compact subsets and $C'$ is closed by intersecting we get that $C'$ is the union of countable compact subsets. You then apply Fubini's Theorem to get that $g(C')$ has measure $0$.

share|cite|improve this answer
Thanks for answering! How do we know that $C'$ is closed? – John Peter Feb 19 '13 at 7:04
@JohnPeter $C'$ is the inverse image of a closed set by a smooth function, smooth functions are continuous. – leo Feb 19 '13 at 17:19

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.