Consider $f:R^n\to R^m$, with continuous Jacobian $J(f)(x)$ of rank $m$ (we assume $n>m$) for all $x\in R^n$.
Now let $E\subset R^m$ be a set of Lebesgue measure 0. Is it always true that $f^{-1}(E)$ has also measure 0?
Does this result have a name?
edit: I believe Theorem 1 in https://math.stackexchange.com/a/3216190/484640 provides an answer.
By constant rank theorem (as long as $J(f)$ is $C^1$), $f$ locally looks like projection, and the preimage of a null set under projection is indeed null, so it is true.
More precisely, if $f(x)=y$, then there are neighborhoods $U\ni x,V\ni y$ and $C^1$ diffeomorphisms $\alpha:U\rightarrow\mathbb{R}^n,\beta:V\rightarrow\mathbb{R}^m$ such that $\beta\circ f\circ\alpha^{-1}(x_1,...,x_m,x_{m+1},...,x_n)=(x_1,...,x_m)$. $C^1$ diffeomorphisms preserve the property of having measure zero, so we can work with the projection $\beta\circ f\circ\alpha^{-1}$. This shows that every $y\in\text{Im}f\ \cap E$ has a neighborhood $V_y$ such that $f^{-1}(E\cap V)$ is null. Let $y$ vary and cover $\text{Im}f\ \cap E$ by countably many such $V_y$'s.
