\classheader{2013-01-25}

Today we'll talk about Sard's theorem.

All manifolds today will be second-countable and Hausdorff.

\begin{definition}
Given $\Cinfty$ manifolds $M$ and $N$, and a $\Cinfty$ map $f:M\to N$, we define
\[\Crit(f)=\{x\in M\mid f'(x)\text{ is not surjective}\}.\]
Thus, $f(\Crit(f))\subset N$. The complement $\complement{f(\Crit(f))}$ is called the set of regular values of $f$.
\end{definition}

\begin{theorem}[Sard's theorem]
The set $f(\Crit(f))$ has measure zero.
\end{theorem}

\subsection*{Preliminaries}

Let $M$ be a $\Cinfty$ manifold, and $E\subseteq M$. We say that the measure of $E$ is zero when, for all diffeomorphisms
\begin{center}
\begin{tikzcd}
\Omega \ar{r}{\varphi} \ar[hook]{d}[swap]{\text{open}} & U\ar[hook]{d}{\text{open}} \\
\R^n & M
\end{tikzcd}
\end{center}


we have $m^*(\varphi^{-1}(E))=0$, where $m^*$ denotes outer Lebesgue measure.

\begin{lemma}
Given an open $\Omega\subseteq\R^n$, a subset $E\subseteq \Omega$, and a $C^1$ map $f:\Omega\to\R^k$, then
\begin{itemize}
\item[(a)] if $k=n$ and $m^*(E)=0$, then $m^*(f(E))=0$, and
\item[(b)] if $k>n$, then $m^*(f(\Omega))=0$.
\end{itemize}
\end{lemma}
\begin{proof}
For (a), it suffices to prove that $m^*(f(E\cap K))=0$ where $K$ is compact. Given $x,y\in\Omega$ such that the line between them is contained in $\Omega$, i.e. such that $tx+(1-t)y\in\Omega$ for all $0\leq t\leq 1$, then
\[\|f(y)-f(x)\|\leq\sup\{\|f'(z)\|\mid z=tx+(1-t)y,\;0\leq t\leq 1\}\cdot\|y-x\|.\]
Given a compact $K\subset \Omega$, there is a $\delta>0$ such that
\[K_\delta:=\{x+v\mid x\in K,\,\|v\|\leq\delta\}\subseteq\Omega.\]
If $m^*(E\cap K)=0$, then for all $\epsilon>0$, there is some covering
\[E\cap K\subseteq\bigcup_{i=1}^\infty C_i\]
where the $C_i$ are translates of $[-a_i,a_i]^n$ with $\diam(C_i)<\delta$ and $\sum m^*(C_i)<\epsilon$.

We have $\diam(f(C_i))\leq S\diam(C_i)$ where $S=\sup\{\|f'(z)\|\mid z\in K_\delta\}$.

There is a fixed ratio $S'$ between the volume of a ball and the volume of a cube.

Thus, $f(C_i)\subseteq$ an open ball of volume $\leq SS'\vol(C_i)$.

To prove (b), note that the claim reduces to considering mapping a cube $C$ in $n$ variables into an open ball $B$ in $k$ variables. Because we have
\[\frac{\vol(B)}{\vol(C)}<\text{constant}\cdot(\diam(C))^{k-n},\]
by covering a compact set $K$ with cubes of sufficiently small diameter we are done.
\end{proof}

\begin{corollary}
Given an open $\Omega\subseteq \R^n$ and open $U\subseteq M$, and a diffeomorphism $\varphi:\Omega\to U$, then if $E\subseteq \Omega$ has $m^*(E)=0$, then $\varphi(E)$ has measure zero.
\end{corollary}

\begin{remark-N}
To prove Sard's theorem, it suffices to consider the case when the domain is an open subset of $\R^n$ and the codomain is $\R^k$. This is because, given our map $f:M\to N$,  we can cover $N$ by countably many open sets $N=\bigcup_{m=1}^\infty U_m$, so that $M=\bigcup_{m=1}^\infty f^{-1}(U_m)$, and we can cover each $f^{-1}(U_m)$ by subsets diffeomorphic to open subsets of $\R^n$. Then note that the union of the $\Crit(f^{-1}(U_m)\to U_m)$ is $\Crit(f)$.
\end{remark-N}

\begin{remark-N}
Part (b) of the lemma implies Sard's theorem in the case that $\dim(\text{domain})<\dim(\text{codomain})$.
\end{remark-N}

\begin{proof}[Proof of Sard's theorem]
$\text{}$

\subsubsection*{Step 1}
We take $\Omega\subseteq\R^n$ an open set and a $\Cinfty$ map $f:\Omega\to\R$.

The case of $n=0$ is evident. We proceed by induction on $n>0$. Note that
\[X_1:=\Crit(f)=\{x\in\Omega\mid \partial_i f(x)=0\text{ for all }1\leq i\leq n\},\]
and we can form a decreasing sequence of closed sets
\[X_k:=\{x\in\Omega\mid \text{all partial derivatives of order $r$ vanish, for all }1\leq r\leq k\}.\]
It will suffice to show that
\begin{itemize}
\item[(a)] $f(X_m-X_{m+1})$ has measure zero for all $m=1,2,3,\ldots$, and
\item[(b)] $f(X_\infty)$ has measure zero, where $X_\infty=X_1\cap X_2\cap X_3\cap\cdots$.
\end{itemize}
For each multi-index $(i_1,\ldots,i_{m+1})$, consider the set
\[Y=\{x\in\Omega\mid \partial_{i_1}\partial_{i_2}\cdots\partial_{i_{m+1}}f(x)\neq 0\text{ and }\underbrace{\partial_{i_2}\cdots\partial_{i_{m+1}}f(x)}_{:=\,\varphi(x)}=0\}.\]
We have a finite collection of such $Y$'s, and $X_m-X_{m+1}$ is contained in the union of these $Y$. It suffices to show that $f(Y\cap X_1)$ has measure zero for each such $Y$. Rewrite $Y$ as
\[Y=\{x\in\Omega\mid \varphi(x)=0\text{ and }\tfrac{\partial \varphi}{\partial x_{i_1}}(x)\neq 0\}.\]
$Y$ is a $\Cinfty$ submanifold of dimension $n-1$ of $\Omega$ by \underline{\hphantom{$\;\;\;\;\;\;\;\;\;$}}. We have
\begin{align*}
X_1\cap Y&=\{x\in Y\mid f'(x):T_x\Omega\to\R\text{ is zero}\}\\
&=\{x\in Y\mid f'(x)|_{T_xY}:T_xY\to\R\text{ is zero}\}\\
&=\Crit(f|_Y)
\end{align*}
It follows that $f(Y\cap X_1)\subseteq f(\Crit(f|_Y))$ which has measure zero by induction hypothesis.


Now we turn to part (b). We have
\[X_\infty=\{p\in\Omega\mid \text{the Taylor expansiom of }f-f(p)\text{ is zero at }p\}.\]
Fix a compact set $K\subset \Omega$, and points $a,x\in K$ such that the line segment joining them is contained inside $K$. Let $a\in X$

Cover $X_\infty \cap K$ by cubes of diameter $\delta$, such that the sum of the volume of the cubes is $\leq D$. The key fact is that there is some constant $L$ such that
\[\diam(f(C))\leq L\cdot\diam(C)^{n+1},\]
which follows from the inequality $|f(x)-f(a)|\leq C\|x-a\|^{n+1}$. We have
\begin{align*}
m^*({\textstyle\bigcup} f(C)) &\leq L\sum \diam(C)^{n+1}\\
&\leq \diam(C)\cdot L\sum\diam(C)^n\\
&\leq \diam(C)\cdot LL'D
\end{align*}
and $\diam(C)$ can be shrunk to zero, so we are done.

\subsubsection*{Step 2}

We have proven Sard's theorem for a map $f:\Omega\to\R$ where $\Omega\subseteq\R^n$ is open. Now we need to prove the claim for maps $f:\Omega\to\R^k$ for all $k\geq 1$. We proceed by induction on $k$, because we have just dealt with the case of $k=1$.

{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad I missed the details here / possible mistakes\quad $\blacktriangledown$
\end{center}}

Consider the projection
\begin{center}
\begin{tikzcd}
\Omega \ar{r}{f} \ar{rd}[swap]{\pi\circ f} & \R^k \ar{d}{\pi}\\
& \R^{k-1}
\end{tikzcd}
\end{center}
where $\pi(x_1,\ldots,x_k)=(x_1,\ldots,x_{k-1})$. We clearly have that $\Crit(\pi\circ f)\subseteq\Crit(f)$. Note that if $E\subseteq \R^{k-1}$ has $m^*(E)=0$, then we have $m^*(E\times \R)=0$. Thus, it suffices to show that
\[f(\{x\in \Omega\mid f'(x)\text{ is not onto but }(\pi\circ f)'(x)\text{ is onto}\})\]
is of measure zero. If $(\pi\circ f)'(x)$ is onto, we can choose coordinates on $\Omega$ such that
\[(\pi\circ f)(x_1,\ldots,x_n)=(x_1,\ldots,x_{k-1}),\]
i.e. we can assume that $f(x_1,\ldots,x_n)=(x_1,\ldots,x_{k-1},\varphi(x_1,\ldots,x_n))$. We have submersions $\pi$ and $\pi'$ such that
\begin{center}
\begin{tikzcd}[column sep=0.1in]
\Omega \ar{rr}{f} \ar{rd}[swap]{\pi'} & & \R^k\ar{ld}{\pi}\\
& \R^{k-1}
\end{tikzcd}
\end{center}
Then $x\in \Crit(f)$ if and only if $x\in\Crit(f|_{(\pi')^{-1}\pi'(x)\to \pi^{-1}\pi'(x)})$.

We've essentially frozen $(x_1,\ldots,x_{k-1})$. Then $\frac{\partial \varphi}{\partial x_i}=0$ for all $i=k,\ldots,n$.
\begin{align*}
E&=f(\Crit(f))\cap (\{a\}\times\R)\\
&=\{\tfrac{\partial \varphi}{\partial x_i}(a,x_k,\ldots,x_n)=0\text{ for all }i\}
\end{align*}
where $a=(a_1,\ldots,a_{k-1})$.

The set $E\cap \pi^{-1}(a)$ is measure zero for all $a$.

We'd like to use Fubini's theorem to finish, but the problem is that Fubini's theorem requires that the function be measurable to start with, which we can't guarantee. Thus, we'll have to use the same sort of estimates we've been using to produce a correct proof. You can find the details in Milnor's book.
\end{proof}


{\color{myred}{\rule{\textwidth}{0.02in}}}