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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?

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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 $0$.

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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
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