In this proof below, I can't show why our choice of $\eta$ gives $m(E)<\epsilon r^k$. It seems like $E$ is just obtained by stretching $E_\eta$ by a factor of $r$, but I don't know how to show this fact. I would greatly appreciate any help.
The assumptions are given as below.
Let $D=A(B(0,1))$. Then $E_{\eta}=\{x:d(x,D) <\eta \}$. We know that $T(B(0,r))\subset \{x:rD)<\eta r\}$. [ By linearity of $A$, $rD$ is same as $A(B(0,r))$]. Now $x \in E_{\eta}$ iff $rx \in \{x:rD)<\eta r\}$. [This follows from the fact that $\|ry\|=r\|y\|$]. Hence $m(T(B(0,r)) \leq m(rE_{\eta})=r^{k}m(E_{\eta})<r^{k}\epsilon$.
Notation used: for any subset $H$ of $\mathbb R^{k}$ and any $r>0$ the set $rH$ is defined as $\{rx:x \in H\}$. If $H$ is measurable then $m(rH)=r^{k}m(H)$].