Let $f: M \rightarrow N$ be a smooth map between two submanifolds of $\mathbb{R}^{m}$, $\mathbb{R}^{n}$ respectively. Sard's famous theorem asserts that the set of critical values $C$ of $f$ has measure zero.
My question is: Does every null set in $\mathbb{R}^n$ arise as the set of critical values of some smooth map $f$ as above?
As a start: For $n=1$ and $C \subset \mathbb{R}$ countable, I think one can construct such a map. Namely, let $M=\coprod_{c \in C} \mathbb{R}$ be the disjoint union of $|C|$ copies of $\mathbb{R}$ and $f: M \rightarrow \mathbb{R}$ be defined by $(x,c) \mapsto x^2+c$. $M$ is a one-dimensional real manifold (it is $2$nd countable, Hausdorff and carries a natural smooth structure coming from the one on $\mathbb{R}$). Hence, by Whitney, it can be embedded as a submanifold of some $\mathbb{R}^m$. Moreover, the set of critical values of $f$ is exactly $C$.
But this approach does not seem to work in general. If $C \subset \mathbb{R}$ is an uncountable null set, for example, then $M=\coprod_{c \in C} \mathbb{R}$ is not $2$nd countable, hence is not a smooth manifold in the usual sense and Whitney's theorem does not apply.
Any help towards an answer to my question is much appreciated! In particular, references are also welcome.
The answer is no.
Sard gave a refined version of his theorem in 1965. It states if
$$f : U \to \mathbb R^m$$
with $U \subset \mathbb R^n$ and $f$ is $C^k$ for $k \geq \max(n-m+1, 1)$, and if we let
$$ A_r = \{ p \in U : Df_p \text{ has rank } \leq r \}$$
Then $f(A_r)$ has Hausdorff dimension $\leq r$.
So the the case you're interested in, the critical values not only has measure zero, but the Hausdorff dimension is at most $\min(m-1,n-1)$.
You can of course realize this number fairly easily.