If $f:\mathbb{R}^n\to\mathbb{R}^m$ is of class $C^1$ and $\operatorname{rank} Df(x\_o)=k$, then clearly $\operatorname{rank} Df\geq k$ in a neighborhood of $x\_o$. It is not particularly difficult to prove the following counterpart of this result for Lipschitz mappings (very nice exercise). Recall that by the Rademacher theorem Lipschitz mappings are differentiable a.e.



> 
> **Theorem.** If $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lipschitz, differentiable at $x\_o$ and $\operatorname{rank} Df(x\_o)=k$, then in
>  any neighborhood of $x\_o$ the set of points satisfying
>  $\operatorname{rank} Df\geq k$ has positive measure.
> 
> 
> 


This result is so natural that I am sure it has been observed before.



> 
> **Question.** Do you know a reference for this result?
> 
> 
> 


I might use this result in my research, and I would prefer to quote it rather than prove it.



Rademacher theorem
While I don't have a reference, let me add another perspective. The following result is proved e.g. in the book *Functions of Bounded Variation and Free Discontinuity Problems* by Ambrosio, Fusco and Pallara (see the proof of Theorem 2.16).



> 
> **Theorem.** Let $n\ge k$. If $f\_j\to f\_\infty$ is a converging sequence of Lipschitz maps from $\mathbb R^n$ to $\mathbb R^k$ (pointwise or equivalently locally uniformly), then
> $$ \det(\partial\_{1}f\_j,\dots,\partial\_kf\_j)\rightharpoonup\det(\partial\_{1}f\_\infty,\dots,\partial\_kf\_\infty) $$
> in the weak\*-$L^\infty$ topology.
> 
> 
> 


Your observation can be seen as a nice corollary of the above theorem.
Indeed, assuming wlog that $m=k$ and $L(x):=\sum\_{i=1}^k\partial\_i f(x\_0)x\_i$ is an invertible linear map, you can apply the theorem to $f\_j(\cdot):=r\_j^{-1}f(x\_0+r\_j \cdot)$, for any sequence of radii $r\_j\downarrow 0$, getting that
$$ \mathcal{L}^n(\{x\in B\_1:\det(\partial\_{1}f\_j(x),\dots,\partial\_kf\_j(x))\neq 0\}>0 $$
eventually (since $f\_\infty=L$), which trivially implies your claim.


