Let $f:U\to \mathbb{R}^{m}$ be a $C^{1}$ function defined on an open subset $U\subset \mathbb{R}^{m}$. Prove that if $f'(a)$ is not an isomorphism then $$\lim_{r\to 0} \frac{\mathrm{vol}~ f(\bar{B}_{r}(a))}{\mathrm{vol}~\bar{B}_{r}(a)}=0.$$
It seems to be necessary to use Sard theorem, but I was not able to do it. I appreciate any help.
Edit: Since $f'(a)$ is a linear transformation which is not an isomorphism, $f'(\bar{B}_{r}(a))$ is contained in $\mathrm{Im}~ f'(a)$, which is an hipersurface of $\mathbb{R}^{m}$ such it dimension is smaller than $m$. Therefore, $f'(\bar{B}_{r}(a))$ has empty interior and thus, $\mathrm{vol}~ f'(\bar{B}_{r}(a))= 0$. Does this imply the thesis?
We will demonstrate considering $B(a,r) = \{ x\in \mathbb{R}^{m};$ $|x-a|_1$ $< r\}$ (this isn't a problem because in $\mathbb{R}^m$ all the norms are equivalent, realize that now $\text{Vol}(B_r(a)) = 2^m r^m $) moreover, suppose without loss of generality that $f(a) = 0$. And consider $\delta >0$ as a real number such that $\overline{B_\delta} (a) \subset U$.
Given $\varepsilon>0$, let $0<r<\delta$ be a real number sufficiently close to zero satisfying
$$\left|Df(a)(y-a) - f(y) - f(a) \right| < \varepsilon |y-a| \leq \varepsilon r$$
for all $y$ $\in$ $B_r:=\overline{B_r}(a)$.
By hypothesis $Df(a)$ isn't an isomorphism, then the set $\{Df(a)(y-a); y \in B_r \}$ lies in a $(m-1)$-dimentional subspace $V\subset \mathbb{R}^n $, consequently the set $\{f(y) - f(a); y\in B_r\}$ lies within $\varepsilon r$ of the $(m-1)$-plane $V+f(a) = V$.
On the other hand, using the compactness of $\overline{B_\delta}(a)$ and the mean value inequality, there is $M$ $\in$ $\mathbb{R}$, such that
$$|f(y)|=|f(y) - f(a)| < M |y-a|, \hspace{0.2cm}\forall\ y\in B_\delta(a),$$
$$\Rightarrow |f(y)|=|f(y) - f(a)| < M |y-a| \leq Mr, \hspace{0.2cm} \forall\ y\in B_r(a),\ \mbox{and }\forall\ 0<r<\delta.$$
Thus, if we change the canonical basis of $\mathbb{R}^m$, $\{e_1,...,e_m\}$, to an orthonormal basis $\{v_1,...,v_{m-1}, w\}$, where $\{v_1,...,v_{m-1}\}$ is a basis of the vector space $V$, and $w$ is normal unit vector to $V$.
Then we can consider $\mathbb{R}^m = \text{span}(v_1) \times ...\times\text{span}(v_{m-1})\times \text{span}(w)$, implying
$$f(B_r) \subset \left[-Mr,Mr\right]\times ... \left[-Mr,Mr\right]\times [-r\varepsilon , r\varepsilon],$$
so
$$\text{Vol}(f(B_r)) \leq 2^m \varepsilon M^{m-1} r^m = \varepsilon M^{m-1} \text{Vol}(B_r)$$ $$\Rightarrow \frac{ \text{Vol}(f(B_r))}{\text{Vol}(B_r)} \leq \varepsilon M^{m-1}.$$
Since $M$ is a fixed constant the result is proved, i.e;
$$\lim_{r\to 0 } \frac{ \text{Vol}\left(f\left(\overline{B_r}(a) \right)\right)}{\text{Vol}(\overline{B_r}(a) )} =0.$$
Remark: If $f(B_r) \subset \left[-Mr,Mr\right]\times ... \left[-Mr,Mr\right]\times [-r\varepsilon , r\varepsilon]$ doesn't make sense to you, the correct way to write this would be
$$ f(B_r) \subset \{t\cdot v_1\ ; t\in[-Mr,Mr]\}\times...\times \{t\cdot v_{m-1}\ ; t\in[-Mr,Mr]\}\times \{t\cdot w\ ; t\in[-\varepsilon r,\varepsilon r]\},$$ since $\{v_1,...,v_{m-1},w\}$ is a ortonortonal basis is immediate that
$$\text{Vol}\left(\{t\cdot v_1\ ; t\in[-Mr,Mr]\}\times...\times \{t\cdot v_{m-1}\ ; t\in[-Mr,Mr]\}\times \{t\cdot w\ ; t\in[-\varepsilon r,\varepsilon r]\}\right) = 2^m M^{m-1}\varepsilon.$$