(Part of) a proof of Sard’s Theorem

I have always wanted to prove Sard’s Theorem. Now I shall stumble my way into proving a deeply unsatisfying special case of it, after a whole day of dead ends and red herrings.

Consider first the special case of a smooth function f:\Bbb{R}\to\Bbb{R}. At first, I thought that the number of critical points of such a function have to be countable. Hence, the number of critical values should also be countable, which would make the measure of critical values 0. However, our resident pathological example of the Cantor set makes things difficult. Turns out that not only can the critical *points* be uncountable, but also of non-zero measure (of course the canonical example of such a smooth function involves a modified Cantor’s set of non-zero measure). In fact, even the much humbler constant function sees its set of critical points having a positive measure of course. However, the set of critical *values* may still have measure 0, and it indeed does.

For f:\Bbb{R}\to\Bbb{R}, consider the restriction of f to [a,b]\subset \Bbb{R}. Note that the measure of critical points of f in [a,b] has to be finite (possibly 0). Note that f'(x) is bounded in [a,b]. Hence, at each critical *point* p in [a,b], given \epsilon>0, there exists a \delta(\epsilon)>0 such that if m(N(p))<\delta(\epsilon), then m(f(N(p)))<\epsilon. This is just another way of saying that we can control the measure of the image.

Note that the reason why I am writing \delta(\epsilon) is that I want to emphasize the behaviour of \frac{\epsilon}{\delta(\epsilon)}. As p is a critical point, at this point \lim\limits_{\epsilon\to 0}\frac{\epsilon}{\delta(\epsilon)}=0. This comes from the very definition of the derivative of a function being 0.

Divide the interval [a,b] into cubes of length <\delta(\epsilon). Retain only those cubes which contain at least one critical point, and discard the rest. Let the final remaining subset of [a,b] be A. Then the measure of f(A)\leq \text{number of cubes}\times\epsilon. The number of cubes is \frac{m(A)}{\delta(\epsilon)}. Hence, m(f(A))\leq m(A)\frac{\epsilon}{\delta(\epsilon)}. Note that f(A) contains all the critical values.

As \epsilon\to 0, we can repeat this whole process verbatim. Everything pretty much remains the same, except for the fact that \frac{\epsilon}{\delta(\epsilon)}\to 0. Hence, m(f(A))\leq m(A)\frac{\epsilon}{\delta(\epsilon)}\to 0. This proves that the set of critical values has measure 0, when f is restricted to [a,b].

Now when we consider f over the whole of \Bbb{R}, we can just subdivide it into \cup [n,n+1], note that the set of critical values for all these intervals has measure 0, and hence conclude that the set of critical values for f over the whole of \Bbb{R} also has measure 0.

Note that for this can be generalized for any f:\Bbb{R}^n\to \Bbb{R}^n.

Also, the case for f:\Bbb{R}^m\to \Bbb{R}^n where m<n is trivial, as the image of \Bbb{R}^m itself should have measure 0.


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Graduate student

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