# (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 is trivial, as the image of $\Bbb{R}^m$ itself should have measure $0$.