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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?

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    I think you need to change $a$ for $x$ in the integrals – 2017-01-27
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    I am guessing you'll need to use continuity of the derivative and determinant to get $\int\limits_{B_r(a)} dx\ |\det f'(a) - \det f'(x)| \to 0$ as $r \to 0.$ – 2017-01-27
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    @WillM. thank you for your comments. – 2017-01-28

1 Answers 1

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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

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    Thank you very much for your answer. I'm working on to solve some technical issues related to the absolute values. Generally is the volume defined as being always positive? If $\text{vol}B(a,r)$ is negative I couldn't put it inside the absolute value of the difference of the determinants you've just written above. – 2017-01-27
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    The volume is always nonnegative. From you assumption that $f'(a)$ is an isomorphism you know that the determinant is nonzero on a neighborhood of $a$ (in particular it takes always the same sign, or if you prefer $\lvert \text{det}f'(y)\rvert$ is positive on a neighborhood of $a$). – 2017-01-27
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    One last question. If $f'(a)$ is not an isomorphism, is it direct that the limit $\lim_{r\to 0}\frac{\text{vol}f(B(a;r))}{\text{vol}B(a;r)}=0$? – 2017-01-28
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    Not really but it is similar: you can write $\lvert\text{det} f'(y)\rvert<\varepsilon$ and so you can estimate from above easily. From below since the volume is nonnegative we do have $\lim_{r\to0}\ge0$. The matter is instead that since $f'(a)$ is not an isomorphism one cannot use the change of variables formula... but the volume is still at most the integral on the right-hand side. – 2017-01-28