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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.

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Your argument is OK, and $\mu(E_1) < \infty$ follows because $E_1$ is a bounded set. –  Lukas Geyer Nov 28 '12 at 15:49
    
Thank you for your comment. Would you be so kind to explain me why $E_1$ is bounded? I can's see it. Thank you very much. –  Romeo Nov 28 '12 at 17:05
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$B_1(0)$ is bounded, so its image under the linear map $A(B_1(0))$ is bounded, too, as is its $1$-neighborhood $E_1$. –  Lukas Geyer Nov 28 '12 at 17:07
    
Yes, now I get it. Thanks a lot, if you want you can write your comments as an answer, I'll accept it. Thank you very much. –  Romeo Nov 28 '12 at 22:39

1 Answer 1

up vote -1 down vote accepted

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 $M>0$.
  • Then observe that for any $b>0$ 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. ($b\leq\varepsilon/(2(2(M+1))^{k-1})$)
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