If $M$ is a smooth manifold with boundary, $f \in C^\infty(M)$, $b$ a regular value of $f$, then is $f^{-1}((-\infty,b])$ a regular domain in $M$? 
The above is taken from John Lee's Introduction to Smooth Manifolds (p. 121). In Proposition 5.47, one supposes that $M$ is a smooth manifold. Does Proposition 5.47 also hold if $M$ is changed to a smooth manifold with boundary?

Proposition 5.47. Suppose $M$ is a smooth manifold and $f \in C^\infty(M)$.
  
  
*
  
*For each regular value $b$ of $f$, the sublevel set $f^{-1}((-\infty,b])$ is a regular domain in $M$.
  
*If $a$ and $b$ are regular values of $f$ with $a < b$, then $f^{-1}([a,b])$ is a regular domain in $M$.

 A: Definitely not. 
First recall the definition of a regular domain: it's a properly embedded (hence closed) codimension-$0$ smooth submanifold with boundary in $M$.
The basic problem is that wherever the boundary of a sublevel set intersects the boundary of $M$, you're likely to get a corner or worse. A simple counterexample is to take $M$ to be the closed upper half-plane $\mathbb R \times [0,\infty)$, and take $f(x,y) = x$. Then $f^{-1}((-\infty,0])$ is the quadrant $\{(x,y): x \le 0,\ y\ge 0\}$, which is a smooth manifold with corners but not a regular domain.
But it can be much worse than that -- for example, the "corner points" can have an accumulation point, which prevents the sublevel set from even being a smooth manifold with corners. For example, with $M$ as above, 
define $f\colon M\to\mathbb R$ by $f(x,y) = u(x)-y$, where
$u\colon\mathbb R\to \mathbb R$ is given by
\begin{equation*}
u(x) = 
\begin{cases}
e^{-1/x^2} \sin \frac{1}{x}, & x\ne 0,\\
0, & x=0.
\end{cases}
\end{equation*}
 Then there are infinitely many points on the $x$-axis where the boundary of the sublevel set $f^{-1}((-\infty,0])$ is not smooth, and they accumulate at the origin.
