TITLE: $\lim_{r\to 0}\frac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)}=|\det f'(a)|$
QUESTION [6 upvotes]: I'm trying to solve this question:

Let $U\subset \mathbb R^m$ be an open set and $f:U\to \mathbb R^m$ a function of class $C^1$. Suppose there is  $a\in U$ such that $f'(a):\mathbb R^m\to \mathbb R^m$ is an isomorphism. Show 
$$\lim_{r\to 0}\frac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)} = |\det f'(a)|$$

My attempt
Using the inverse function theorem, there is $\delta>0$ such that $f_{|B(a,r)}$ is a diffeomorphism (henceforth for simplicity let's still call this restriction $f$). 
Suppose from now on $|r|\lt \delta$. 
Since $f$ is a diffeomorphism and $B(a,r)$ is compact we can use change of variables:
\begin{align}
& \operatorname{vol}f(B(a,r)) \\[10pt]
= {} & \int_{f(B(a;r))}1\cdot dy \\[10pt]
= {} & \int_{B(a;r)}1\cdot (f(a))|\det f'(a)|dx \\[10pt]
= {} & \int1\cdot|\det f'(a)| \, dx \\[10pt]
= {} & |\det f'(a)|\int_{B(a;r)}1 \, dx \\[10pt]
= {} & |\det f'(a)|\operatorname{vol}B(a;r)
\end{align}
So I think I have proved that $\dfrac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)}=|\det f'(a)|$ which is stronger than what the question asks.
What is wrong with my answer and how can I correct that?

REPLY [3 votes]: Actually you have
$$
\text{vol}(f(B(a,r)))=\int_{f(B(a,r))}1\, dy=\int_{B(a,r)}\lvert\det f'(y)\rvert\,dy.
$$
Now you need to use the continuity of $y\mapsto\lvert\det f'(y)\rvert$ on the ball $B(a,r)$ to conclude that
$$\lim_{r\to 0}\frac{\text{vol}(f(B(a,r)))}{\text{vol}(B(a,r))}=\lvert\det f'(a)\rvert.$$
Namely, given $\varepsilon>0$, there exists $r_0>0$ such that
$$
\lvert\det f'(y)-\det f'(a)\rvert<\varepsilon
$$
for $y\in B(a,r)$ and $r<r_0$. Now put everything together.