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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.

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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
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1 Answer

up vote 4 down vote accepted

Let $$X_k:=\{x\in U,\lVert x\rVert\leqslant k\}\cap \{x,d(x,U^c)\geqslant k^{-1}\}.$$ 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 $k$, an if $x\in U$, we can find $n$ such that $\lVert x\rVert\leqslant n$. As $U^c$ is closed and $x\notin U^c$, $d(X,U^c)$ is positive. So take $k$ such that $d(x,U^c)\geqslant k^{-1}$ and $N:=\max\{n,k\}$. Then $x\in X_N$.

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In the second set notion, do you intend to take $x$ from $U$ or $\mathbb{R}^{n}$? –  Thomas E. 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 $d(x,U^{c})=100$? –  Thomas E. Apr 24 '12 at 11:31
    
Alright, now it works. +1 for nice and simple solution. –  Thomas E. Apr 24 '12 at 11:36
    
@Davide Can you please explain the proof in detail? –  Ester Dec 15 '12 at 6:13
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