Given smooth manifolds $M,N$ ($m$- and $n$- manifolds respectively) Sard's theorem says that for $f:M \to N$ in $C^k$ ; $k \geq 1$, the image of the set of critical points of $f$ in $M$ --points in $M$ where the Jacobian has rank $< m$ --has measure zero in N:


Now, I know what it means for a set to have measure zero in the case of $M, N$ being Euclidean spaces, but I am not clear on how we define measures on manifolds; I know we may pull back different types of objects from $\mathbb R^n$ into a manifold, like $k$-forms, etc., but I don't know if/how one pullsback a measure from $\mathbb R^n$, since I am not even clear on what kind of object a measure is. Is it just a $0$-form, i.e., just a function? If so, can we define a measure globally on a manifold, or do we just pullback separately for each chart?

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    $\begingroup$ Well, to define measure zero you don't really need a measure. You can just say that a set has measure zero if it has measure zero in all charts. This answer of Willie Wong seems to address your more specific question. $\endgroup$ – t.b. Nov 5 '11 at 20:35
  • $\begingroup$ Thanks, but I am not even sure of how to define a measure on a single given chart, because I don't even know how a measure on a manifold nor on a given chart are defined. $\endgroup$ – J.K Nov 5 '11 at 20:37
  • $\begingroup$ In the case of an oriented Riemannian manifold, we usually integrate against a volume form, which is a global generator for the top exterior power of the cotangent bundle. But we may define measures in more general settings. Have you studied any measure theory? $\endgroup$ – Zhen Lin Nov 5 '11 at 20:37
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    $\begingroup$ @J.K: You said you know how to define measure zero on subsets of $\mathbb{R}^n$. Now call a set in $M$ "of measure zero" if its image in every chart has measure zero in $\mathbb{R}^n$, that's not entirely satisfactory but sufficient for most purposes. Please have a look at Willie's answer I linked to but I'm sure somebody will provide an answer to your actual question. $\endgroup$ – t.b. Nov 5 '11 at 20:41
  • $\begingroup$ @t.b: thanks; I will read Willie's answer, but, just for now, let me follow up with Zhen Lin: yes I know some measure theory; generalized measure spaces and sigma algebras. I guess I would/could then define a sigma algebra on my manifold, maybe the sigma-algebra generated by the open sets in the manifold (with its standard metric topology given by the interinsic metric). But then what? Do I use some sort of pullback metric by the chart map? $\endgroup$ – J.K Nov 5 '11 at 21:01

Having measure zero is something that can be defined without actually defining a measure on the manifold. A subset $A$ of $N$ can be defined to have measure zero if for every chart $\phi:U\to\mathbb R^n$ of $N$, $\phi(U\cap A)$ has measure zero in $\mathbb R^n$.

You would want this to be something that is invariant under diffeomorphism. That is the case because smooth maps between subsets of $\mathbb R^n$ preserve the property of having Lebesgue measure $0$.

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