On the modulus of continuity of functions whose image has positive measure, and metric embeddings into ${\mathbb{R}}^d$ without shrinking

A generalization of the classical Sard theorem in the plane is the following. Let $f$ be a function defined on a subset $A\subset {\mathbb{R}}^2$. If $f$ has modulus of continuity $\omega (r)\lesssim r^2$, then $f(A)\subset \mathbb{R}$ has Lebesgue measure zero. Choquet claimed in \cite{Choquet} that this was a full characterization, i.e. for every $\omega$ for which $\omega (r)\mathrm{/}{r}^2$ converges to $\mathrm{\infty }$ as $r\to 0$, there is a counterexample. We disprove this by showing that the correct characterization, in ${\mathbb{R}}^d$, is ${\int }_0^1\omega (r{)}^{-1\mathrm{/}d}=\mathrm{\infty }$. For the precise statement see Theorem 2. We obtain this as a special case of a more general result. We study which spaces $(X,\rho )$ can be embedded into ${\mathbb{R}}^d$ without decreasing any of the distances in $X$. That is, we ask the question whether there is an $f:X\to {\mathbb{R}}^d$ such that $\mathrm{\parallel }f(x)-f(y)\mathrm{\parallel }\ge \rho (x,y)$ for every $x,y\in X$. We study this problem for some very general distance functions $\rho$ (we do not even assume that it is a metric space, in particular, we do not assume that $\rho$ satisfies the triangle inequality), and find quantitative necessary and sufficient conditions under which such a mapping exists. We will obtain the characterization mentioned above as a special case of our metric embedding results, by choosing $X$ to be an interval in $\mathbb{R}$, and defining $\rho$ by putting $\rho (x,y)=r$ if $\mathrm{\parallel }x-y\mathrm{\parallel }=\omega (r)$.