Question about a proof in Rudin R&CA: $ \mu(\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \eta\}) < \varepsilon$ 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 $\varepsilon&gt;0$ there exists a $\eta&gt;0$ s.t.  $$
 \mu(\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \eta\}) &lt;
 \varepsilon. $$ 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&amp;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 $\mu(E_1)&lt;+\infty$ 
$$
\lim_{n\to+\infty} \mu(E_n)=0.
$$
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 $\mu(E_1)&lt;+\infty$ 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.
The set $A(\mathbb R^k)$ is contained in an hyperplane.

The set $A(B_1(0))$ is bounded.

So the intuitive idea is that the set of points which are at a distance less than $\eta$ away from $A(B_1(0))$ is a very flat set, with thickness less than $2\eta$. Its Lebesgue measure is thus small.

More precisely:


Up to an isometry (which does not change the Lebesgue measure of the sets you are considering), you can suppose that $A(B_1(0))$ is a bounded subset of $Span(e_1,\dots,e_{k-1})$, with $e_j:=(0,\dots,0,1,0,\dots,0)$, say $$A(B_1(0))\subset[-M,M]^{k-1}\times \{0\}$$
for some $M&gt;0$.
Then observe that for any $b&gt;0$ there exist an $\eta$ such that $$\{x\in\mathbb R^k\,|\,d(A(B_1(0)),x)\leq \eta)\}\subset[-(M+1),(M+1)]^{k-1}\times [-b,b]\,.$$
The measure of the set on the right hand side is $2b(2(M+1))^{k-1}$ it is enough to take $b$ small enough to get the result. ($b\leq\varepsilon/(2(2(M+1))^{k-1})$)
The set $A(\mathbb R^k)$ is contained in an hyperplane.

The set $A(B_1(0))$ is bounded.

So the intuitive idea is that the set of points which are at a distance less than $\eta$ away from $A(B_1(0))$ is a very flat set, with thickness less than $2\eta$. Its Lebesgue measure is thus small.

More precisely:


Up to an isometry (which does not change the Lebesgue measure of the sets you are considering), you can suppose that $A(B_1(0))$ is a bounded subset of $Span(e_1,\dots,e_{k-1})$, with $e_j:=(0,\dots,0,1,0,\dots,0)$, say $$A(B_1(0))\subset[-M,M]^{k-1}\times \{0\}$$
for some $M&gt;0$.
Then observe that for any $b&gt;0$ there exist an $\eta$ such that $$\{x\in\mathbb R^k\,|\,d(A(B_1(0)),x)\leq \eta)\}\subset[-(M+1),(M+1)]^{k-1}\times [-b,b]\,.$$
The measure of the set on the right hand side is $2b(2(M+1))^{k-1}$ it is enough to take $b$ small enough to get the result. ($b\leq\varepsilon/(2(2(M+1))^{k-1})$)