J-measurable sets and functions of class $C^1$ If $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is $C^1$ class and $det f^{\prime}(0)=0$ show that, when $r\rightarrow 0$
$$\dfrac{Vol(f(B[0,r]))}{ Vol(B[0,r])} \rightarrow 0$$
where $Vol(X)$ is n-dimensional volume to set $J$-measurable $X$.
Any suggestions are welcome, thanks
 A: The determinant of the Jacobian (what the OP denotes by $f'$) is precisely the coefficient by which an infinitesimal element of volume is modified when $f$ is applied to it. Since the Jacobian is assumed to be zero at the point, it is going to be smaller than $\epsilon>0$ in a suitable neighborhood for any $\epsilon$, by the $C^1$ assumption on $f$. Then the volume of the image will be at most $\epsilon$ times the volume of the original ball, i.e. the ratio $\dfrac{Vol(f(B[0,r]))}{ Vol(B[0,r])}$ is bounded above by $\epsilon$.
A: By the change of variable formula you have
$$ \int_{f(B(0,r))}1 dx = \int_{B(0,r)} 1 \cdot \text{Jac }f(x) dx $$ 
Since the Jacobian of $f$ at zero is zero and $f$ is $C^1$ it follows that for $\varepsilon >0$ given, there exists $r_\varepsilon >0$ such that for any $r <r_0$ we have $|\text {Jac} f(x)| <\varepsilon$ on $B(0,r)$. From the above inequality we get 
$$ | f(B(0,r))| \leq |B(0,r)| \cdot \varepsilon,\ \forall r<r_\varepsilon.$$
This expresses the fact that $\frac{|f(B(0,r))|}{|B(0,r)|}$ goes to $0$ as $r \to 0$.
