Let $\Omega \subset \mathbb R^n$ an open set and $f:\Omega \to f(\Omega)$ a $C^1$ diffeomorphism. For each $a = (a_1,...,a_n) \in \Omega$ and $r>o$, define the following set $$ I[a,r] = [a_1 - r/2, a_1 + r/2]\times...\times[a_n - r/2, a_n + r/2] $$ Proof that $$ \lim_{r \to 0^+} \frac{m(f(I[a,r]))}{r^n} = |\, \det f'(a)\, | $$ Where $m$ is the Lebesgue-measure.
My attempt:
Since $\Omega$ is an open subset, we can find $r_0$ sufficiently small such that $I[a,r] \subset \Omega$, for all $0<r\leq r_0$. Besides, $I[a,b]$ is Lebesgue-measurable, so, using the C.V.T, we have $$m(f(I[a,r])) = \int_{I[a,r]} |\, \det f'(x)\, | dm(x) = \int \chi_{I[a,r]} |\, \det f'(x)\, | dm(x) $$. $$ \therefore \quad \frac{m(f(I[a,r]))}{r^n} = \frac{1}{r^n}\int \chi_{I[a,r]} |\, \det f'(x)\, | dm(x) = \int \frac{\chi_{I[a,r]}}{r^n} |\, \det f'(x)\, | dm(x) $$
I was thinking in using a convergence theorem, although I'm not sure if I can do it for the family $\left ( \frac{\chi_{I[a,r]}}{r^n} \right )_{0<r\leq r_0}$ and how it would save my problem. Thank you for any help!!
Hint:
$$\frac{m(f(I[a,r]))}{r^n} - |\, \det f'(a)\, |= \frac{1}{r^n}\int_{I[a,r]} (|\, \det f'(x)\, | - |\, \det f'(a)\, |)\, dm(x).$$
Use the continuity of $\det f'(x)$ at $a.$