These definitions are necessary:
There is the theorem:If
$(a)$$V$ is open in $R^{k}$.
$(b)$$T : V \to R^{k}$ is continuous, and
$(c)$$T$ is differentiable at some point $x \in V$, then $$ \lim_{r \to 0} \frac{m(T(B(x,r)))}{m(B(x,r))} = \Delta(T'(x)). $$
There is the proof: Assume, without loss of generality, that $x = 0$ and $T(x) = 0$. Put $A = T'(0)$.
The following elementary fact about linear operators one finite-dimensional vector spaces will be used: A linear operator $A$ on $R^{k}$ is one-to-one if and only if the range of $A$ is all of $R^{k}$. In that case, the inverse $A^{-1}$ of $A$ is also linear.
Accordingly, we split the proof into two cases.My question is about case 2: $A$ is not one-to-one. In this case $A$ maps $R^{k}$ into a subspace of lower dimension, i.e., into a set of measure $0$. Given $\epsilon \gt 0$, there is therefore an $\eta \gt 0$ such that $m(E_{\eta}) \lt \epsilon$ if $E_{\eta}$ is the set of all points in $R^{k}$ whose distance from $A(B(0,1))$ is less that $\eta$. Since $A = T'(0)$, there is a $\delta \gt 0 $ such that $|x| \lt \delta$ implies $$|T(x) - Ax| \leq \eta |x| $$
If $r \lt \delta$, then $T(B(0,r))$ lies therefore in the set $E$ that consists of the points whose distance from $A(B(0,r))$ is less that $\eta r$. Our choice of $\eta$ shows that $m(E) \lt \epsilon r^{k}$.
I don't understand how do we conclude that there is a $\delta \gt 0$ such that $|x| \lt \delta$ implies that $|T(x) - Ax| \leq \eta |x|$.
I also don't understand why is $m(E)$ less Than$\epsilon r^{k}$ by our choice of $\eta$.
Any help would be appreciated.