How exactly do we clean up this heuristic proof of Sard's theorem, from Schwarz' `Differential Topology for Physicists' using the Jacobian $J(f)(x)$:

"A heuristic justification for Sard's theorem is the following: if $f$ is a smooth one-to-one map on $S \subset E$, the volume of $S' = f(S)$ equals $\int_S |J(f)(x)|dx$. If $f$ is smooth, but not necessarily one-to-one, the volume of $S'$ is at most this integral. Applying this to the case where $S$ is the set of singular points of $f$, we conclude that the volume of $S'$ cannot be greater than zero, since $J(f)(x) = 0$ for all $x \in S$. This is not a complete proof because, strictly speaking, the formula $\mathrm{vol}S' = \int_S |J(f)(x)| dx$ only applies if $S$ satisfies certain conditions (for example, if $S$ is open). But it is not hard to make the proof watertight." ?

Does it link up with, or even maybe motivate, the proof from Milnor of Sard,

If $F:M^m \to N^n$ is a smooth map between smooth manifolds of dimension $m$ and $n$ respectively, then $F(C)$ has measure zero in $N$ (where $C = \{x \in M : {rank}\; dF_x < n\}$)

motivated here & given here, which amounts to writing

$C = (C\smallsetminus C_1)\cup (C_1 \smallsetminus C_2) \cup \cdots \cup (C_{k-1} \smallsetminus C_k) \cup C_k$ (where $C_i$ denotes the set of points in $M$ for which all partial derivatives of order $\le i$ vanish at $x$) and then show that the image under $F$ of each of the sets in the union on the r.h.s. has measure zero in $N$.

and invoking the constant rank theorem, Fubini's theorem & Taylor's theorem at various steps?

  • $\begingroup$ I haven't thought about it too much, but my immediate response is ... NO. $\endgroup$ – Ted Shifrin Jun 15 '16 at 1:59
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    $\begingroup$ Schwarz's argument seems to be a heuristic for the case $m=n$. This is much easier than the general $m>n$ case which Milnor is talking about. I think it shouldn't be too hard to make a Schwarz-like argument work when $m=n$. $\endgroup$ – David E Speyer Jun 21 '16 at 14:08

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