If a function has a jacobian with linearly independent columns on a set B, is the following union equal to B? Suppose we have a function $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ with the condition that the rows of the Jacobian matrix of $f$ are linearly independent on $B \subseteq \mathbb{R}^m$.
Let $B_p$ = $\{ t \in B : \forall s \in \mathbb{R}^m  \; \; \lvert s - t \rvert < \frac{1}{p} \implies \frac{\lvert s - t \rvert}{p} \leq \lvert f(s) - f(t) \rvert \}$
I believe it's the case that $\lambda(\bigcup\limits_{p=1}^{\infty} B_{p}) = 0 \implies \lambda(B) = 0$, where $\lambda$ is the Lebesgue measure on $\mathbb{R}^m$, but I've been unable to prove it.

Noting that the precondition necessitates that $ n \leq m$, the case where $m = 1$ is pretty simple, as the precondition reduces to $f$ having non-zero derivative on $B$.
I'm not entirely sure how to proceed. An argument by induction on $m$ is tempting, but I'm struggling to see exactly how to utilise the linear independence of the rows in this proof.

Consider the map $\Bbb R^2\to\Bbb R$, $(x,y)\mapsto x$. The Jacobian of this map is at every point simply $\begin{pmatrix}1&0\end{pmatrix}$, which is one non-zero column hence the columns are all linearly independent.
However for any $(x,y)\in\Bbb R^2$ and $p\in\Bbb R_{>0}$ you have that $\|(x,y)-(x,y+\frac1{2p})\|<\frac1p$ but
$$\|f(x,y)-f(x,y+\frac1{2p})\|=0\not≥ \frac{\|(x,y)-(x,y+\frac1{2p})\|}{p}$$
hence for any set $B$ the set $B_p$ is empty.
