Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 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 $k$, and if $x\in U$, we can find $n$ such that $\lVert x\rVert\leqslant n$. As $U^c$ is closed and $x\notin U^c$, $d(x,U^c)$ is positive. So take $k$ such that $d(x,U^c)\geqslant k^{-1}$ and $N:=\max\{n,k\}$. Then $x\in X_N$.

share|cite|improve this answer
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 6:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.