### (Part of) a proof of Sard’s Theorem

#### by ayushkhaitan3437

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 . 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 . 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 , and it indeed does.

For , consider the restriction of to . Note that the measure of critical points of in has to be finite (possibly ). Note that is bounded in . Hence, at each critical *point* in , given , there exists a such that if , then . This is just another way of saying that we can control the measure of the image.

Note that the reason why I am writing is that I want to emphasize the behaviour of . As is a critical point, at this point . This comes from the very definition of the derivative of a function being .

Divide the interval into cubes of length . Retain only those cubes which contain at least one critical point, and discard the rest. Let the final remaining subset of be . Then the measure of . The number of cubes is . Hence, . Note that contains all the critical values.

As , we can repeat this whole process verbatim. Everything pretty much remains the same, except for the fact that . Hence, . This proves that the set of critical values has measure , when is restricted to .

Now when we consider over the whole of , we can just subdivide it into , note that the set of critical values for all these intervals has measure , and hence conclude that the set of critical values for over the whole of also has measure .

Note that for this can be generalized for any .

Also, the case for where is trivial, as the image of itself should have measure .