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Why is this true?

I think I can find a countable union of compact sets $\cup_{k=1}^\infty X_k$ such that $\cup X_k \subseteq U$ and the lebesgue measure of $U \setminus \cup X_k$ is zero.

(for any $k\in \mathbb{N}$, we can find a closed set $Y_k \subset U$ such that $\lambda(U\setminus Y_k)<\frac{1}{k}$. (Take $X_k=B(k)\cap Y_k$ where $B(k)$ is the ball of radius $k$.)

But that doesn't solve the problem.

Davide Giraudo
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MA3590
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    Visualize it in $\mathbb{R}^2$: at step $n$ draw a square grid of mesh size $1/n$. Add those squares that are completely contained in $U$. Alternatively: every open set is the union of countably many balls. Show that each ball of radius $r$ is a countable union of compact sets (closed balls with same center and radius $r-1/n$). – t.b. Apr 24 '12 at 11:18

1 Answers1

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Let $$X_k:=\left\{x\in \mathbb R^n,\left\lVert x\right\rVert\leqslant k\}\cap \{x,d\left(x,U^c\right)\geqslant k^{-1}\right\}.$$ Then $X_k$ is closed (as an intersection of such sets) and bounded in $\mathbb R^n$ hence compact. We have $U=\bigcup_{k\geqslant 1}X_k$. Indeed, by definition $X_k\subset U$ for all $k$, and if $x\in U$, we can find $n$ such that $\lVert x\rVert\leqslant n$. As $U^c$ is closed and $x\notin U^c$, $d\left(x,U^c\right)$ is positive. So take $k$ such that $d\left(x,U^c\right)\geqslant k^{-1}$ and $N:=\max\left\{n,k\right\}$. Then $x$ belongs to $X_N$.

Davide Giraudo
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  • In the second set notion, do you intend to take $x$ from $U$ or $\mathbb{R}^{n}$? – T. Eskin Apr 24 '12 at 11:23
  • As you want, since it will yield the same definition. – Davide Giraudo Apr 24 '12 at 11:24
  • You're right. One more question though. If $x\in U$, why would there exist $k\geq 1$ so that $d(x,U^{c})\leq \frac{1}{k}$? Say $U=B(0,100)\subset \mathbb{R}^{n}$ and we choose $x=0$. Then isn't $d(x,U^{c})=100$? – T. Eskin Apr 24 '12 at 11:31
  • Alright, now it works. +1 for nice and simple solution. – T. Eskin Apr 24 '12 at 11:36
  • @Davide Can you please explain the proof in detail? – Ester Dec 15 '12 at 06:13
  • Davide, is it true that the same argument, but with $X_k:={x\in U,\lVert x\rVert < k}\cap {x,d(x,U^c)> k^{-1}}.$ would give an increasing sequence of open, compactly contained in each other sets $X_1\subset\subset X_2\subset\subset....X_k\subset\subset ...U$ such that $U=\bigcup\limits_{k=1}^{\infty}{X_k}$ ? At least, I can not see a problem. – Svetoslav May 16 '16 at 15:30
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    @DavideGiraudo I think, the proof is incorrect: the first set ${x\in U,\lVert x\rVert\leqslant k}$ is not closed. Consider $U = (0,1)$ in $\mathbb{R}$. Then $ {x\in (0, 1),\lVert x\rVert\leqslant 1} = (0,1)$. It should be ${x\in \mathbb{R}^n,\lVert x\rVert\leqslant k}$. –  Sep 22 '17 at 13:53
  • @Goodnighyn You are right, thanks you I edited. – Davide Giraudo Sep 22 '17 at 14:36