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It states:

Let $g:A \to R^n$ be continuously differentiable, where $A \subset R^n$ is open, and let $B=${${x \in A: \det g'(x)=0}$}. Thne $g(B)$ has measure $0$.

Okay.... obviously this theorem is right... but why don't constant functions violate this? After all, the derivative of a constant function is $0$ EVERYWHERE.... so that can't possibly be measure zero!

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Well, technically, it's hardly "obvious" that Sard's Theorem is true -- but yes, if you take the theorem on faith, then sure. –  Jesse Madnick Jul 4 '12 at 15:01

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up vote 4 down vote accepted

It is $g(B)$, not $B$, that has measure $0$. If $g$ is constant, then g(B) contains a single point (the constant).

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x) thanks... its obvious now. –  Squirtle Jul 4 '12 at 6:03

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