Suppose we have a linear function $A\colon \mathbb R^k \to \mathbb R^k$ which is not injective (hence not surjective). Clearly, we have $\mu(A(\mathbb R^k))=0$ where $\mu(\cdot)$ is the Lebesgue measure on $\mathbb R^k$.
Then, for every $\varepsilon>0$ there exists a $\eta>0$ s.t. $ \mu(\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \eta\}) < \varepsilon. $ where $B_1(\mathbf 0)$ is the unit ball in $\mathbb R^k$.
Is this fact so obvious? I find it in the proof of a theorem on Rudin, R&CA (Chapter 7, thm. 7.24, pag 152) and I'm quite puzzled. I tried to justify this passage in the following manner: consider the sets $ E_n :=\left\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \frac{1}{n}\right\} $ Then $ \bigcap_{n\in \mathbb N}E_n=\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) =0\}=\overline{A(B_1(\mathbf0))} $ (I'm not so sure about the last equality...). Then, by monotonicity of the measure, we get $ \mu\left(\bigcap_{n \in \mathbb N}E_n\right)=0 $ which implies, if $\mu(E_1)<+\infty$ $ \lim_{n\to+\infty} \mu(E_n)=0. $ From this, I think that Rudin's claim follows (just writing the $\varepsilon-\delta$ definition of limit). But I'm not sure: indeed, I can't show that $\mu(E_1)<+\infty$ and, secondly, I think there is something I'm missing, it can't be too difficult (otherwise Rudin would have explained it!).
Thanks in advance.