$f^{-1}f(\Sigma)$ has measure zero for analytic map from manifolds $M$ to $N$ Differential Topology Hirsch Chapter 3 Section 1 Problem 4: (a) Let $M$ be a connected manifold and $f: M \rightarrow N$ an analytic map. Let $\Sigma \subset M$ be the set of critical points. If $\Sigma \neq M$ then $f^{-1}f(\Sigma)$ has measure zero.
(b) If $f$ is merely $C^{\infty}$ The conclusion (a) can be false.
For part (a) A function is analytic iff its Taylor series about $x_0$ converges to the function in some neighborhood for every $x_0$ in the domain. The definition of analytic function is the same as $C^{\infty}$ function except for the notion of convergence in a neighborhood of nonzero radius in the former. Therefore, we can apply the Morse-Sard Theorem and conclude that the set of regular values is dense. A critical point cannot be a regular value so $f^{-1}f(\Sigma)$, which is the set of critical points must have measure 0.
For part (b) I didn't use anything about the function being analytic in part (a) so I'm not sure how I would approach this part.
Thank you!
 A: I am trying to do this question myself, and I have not been able to do it yet . Now for the counterexample part of a smooth function $f: M\rightarrow N$ such that $f^{-1}(f(\Sigma))$ doesn't have null measure, I think an idea would be to consider $M=N=\mathbb{R}$ and take a bump function. We would have that $\Sigma\neq \mathbb{R}$ since there are points where the derivative is non-zero and so the differential is surjective, and we would have that $f^{-1}(f(\Sigma))$ wouldn't have measure zero.  So I don't think your argument works, but I haven't been able to prove this myself. I am guessing it will use some local property of real analytic functions , but I am not very aware of these so I will try to look things up.
Here's the attempt at proving the statement , the idea is similar to what Hirsch did to prove the Morse-Sard theorem, and notice we can assume that $\dim M\geq \dim N$:
Notice that we can do this locally and assume we have a function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ analytic.
Let $\Sigma_f$ denote the set of critical points of $f$.We will see that $m(\Sigma_f)=0$.  We have that $f_i(x_1,...,x_m)=\sum a^i_{k_1}...a{k_m}(z)(x_1-z_1)^{k_1}...(x_m-z_m)^{k_m}$. We can divide $\Sigma_f$ into two sets, let $\Sigma_1$ denote the set of points $x\in \Sigma$ such that $\nabla f_i(x)\neq 0$ for some differential operator of order $\geq 2$, and $\Sigma_2$ the set of $x\in \Sigma$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ for some $i\in \{1,..,n\}$ and $j\in \{1,..,m\}$. We notice that $\Sigma\subset \Sigma_1 \cup \Sigma_2$ since if we have that $\frac{\partial f _i}{\partial x_j}(z)=0$ and $\nabla f_i(z)=0$ for every $i,j$ and differential operator of order $\geq 2$ then we will have that $a^i_{k_1}...a_{k_m}(z)=0$ which would contradict the fact that $f\neq 0$ since $M$ is connected.
Now Let's see that $\Sigma_1- \Sigma_2$ has null measure. We can consider the set of points $x\in M$ such that $\nabla f(x)= 0$ and $\frac{\partial f_i}{\partial x_j}\neq 0$, and this set will form a submanifold of dimension $m-1$ which we will call $M_{\nabla },i,j$. We have that $\Sigma_1-\Sigma_2 \subset \cup_{\nabla,i,j}X_{\nabla,i,j}$ where this union is over a countable set , and since $M$ is a $n-1$\submanifold it will have null measure so that $\Sigma_1-\Sigma_2$ has null measure.
If $x\in \Sigma_2$ we will have that there exists $f_i$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ and so there exists an open set $U$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ for $x\in U$.Now using the Implicit function theorem for analytic functions we have that there exists a function $h:A\times B\rightarrow U$ such that $f_i(h(x,t))=t$, for $(x,t)\in A\times B$. And so $f(h(x,t))=(u_t(x),t)$ and so being a critical point of $f$ is equivalent to being a critical point for $u_t:A\subset \mathbb{R}^{n-1}\rightarrow \mathbb{R}^n$, and so by induction hypothesis we have that $m(\Sigma_{u_t})=0$ and then using the fubini theorem we get that $m(\Sigma_f\cap U)=0$. Since we can cover $\Sigma_f$ by a coutable number of these open sets we get the desired result.
Now we already know that $\Sigma_f\subset f^{-1}(f(\Sigma))$ has zero measure , we just need to see that $f^{-1}(f(\Sigma))-\Sigma_f$ also has null measure. Let $x\in f^{-1}(f(\Sigma_f))-\Sigma_f$ then we will have that $d_x f$ is surjective. And so there will exist an open set $x\in U=A\times B\subset M$ and parametrizations such that $\bar f=\psi^\circ f\circ \phi^{-1} : U\rightarrow \mathbb{R}^n$ looks like the canonical projection. Since $K:=\hat f(\phi(\Sigma_f))\subset A$ has null measure by Sard's theorem, we will have that $\hat f^{-1}(K)=K\times B$ will have null measure by Fubini's theorem, and so $m(\hat f^{-1}(f(\Sigma_f)))=0$. Now since $\phi^{-1}$ takes null measure sets into null measure sets we have that $m(\phi^{-1}(\hat f^{-1}(f(\Sigma_f))))=m( f^{-1}(f(\Sigma))\cap U)=0$.
