Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a problem in Frank Jones's Lebesgue integration on Euclidean space (p.57),

$$\mathbb{R}^n = N \cup \bigcup_{k=1}^\infty \overline{B}_k$$

where $\lambda(N)=0$, and the closed balls are disjoint.

could any one give some hints?

share|improve this question
    
think "fractally" take same balls, fill in the space between them, then between them etc, and approximate the measure of the compliment of the limit –  yoyo Mar 24 '11 at 15:27
    
You should combine the beginning of the idea in what Jonas wrote below and combine with @yoyo's idea above. Try doing it for the open unit interval first, and then try for the open unit square in $\mathbb R^2$. I think the pattern will emerge quickly at that point. –  Sam Lisi Mar 24 '11 at 18:13
1  
If I were trying this on the unit interval, I would probably try to do the Cantor set construction, taking closed balls instead of open balls. If that worked, then doing it for the entire line would be easy. If that worked, then similar ideas should work on the plane, 3-space, etc... –  Arturo Magidin Mar 24 '11 at 18:35
add comment

5 Answers

up vote 6 down vote accepted

Fix some dimension $d \geq 1$. It suffices to prove that the subspace $X = [0,1)^d \subset \mathbf{R}^d$ is the union of a disjoint family of closed balls and a null set (with respect to Lebesgue measure $\lambda$ on $X$). Let's call any subset of $X$ with the form $\prod_{i=1}^d[x_i,x_i + s)$ where $s>0$ a box. A disjoint union of finitely many boxes (resp. closed balls) will be called a square (resp. round) set. Using a grid construction, it is not difficult to see that every open subset of $X$ is the disjoint union of countably many boxes. Thus follows:

Lemma 1: If $U$ is an open subset of $X$ and $\epsilon > 0$ there is a square set $S \subset U$ with $\lambda(U) - \lambda(S) < \epsilon$.

Let $a \in (0,1)$ be a constant (depending on $d$) such that every box contains a closed ball of $a$ times the measure. Choosing a ball for every box in a square set gives:

Lemma 2: For every square set $S \subset X$ there is a round set $R \subset S$ such that $\lambda(R) = a \lambda(S)$.

Choose $\epsilon > 0$ so that $(1-a) + \epsilon < 1$. We can now construct the desired family of disjoint balls. We will construct recursively for each $n=1,2,\ldots$, a round set $R_n$ such that $\lambda(X - R_n) \leq (1-a)^n + (1-a+\epsilon)^n$.

Since $X$ is square, there is a round set $R_1$ (just a ball actually) with $\lambda(R_1) = a \lambda(X) = a$ whence $\lambda(X - R_1) = 1-a \leq (1-a) + (1-a+\epsilon)$.

Now suppose that $R_n$ given for $n \geq 1$ and that $\lambda(X - R_n) \leq (1-a)^n + (1-a+\epsilon)^n$ holds. Since $X-R_n$ is open, lemma 1 gives us a square set $S$ disjoint from $R_n$ with $\lambda(R_n) - \lambda(S) \leq \epsilon^{n+1}$. Then, by lemma 2, there is a round set $R \subset S$ with $\lambda(R) = a \lambda(S)$. Putting $R_{n+1} = R_n \sqcup R$ we see that:

\begin{align*} \lambda(X-R_{n + 1}) &= \lambda(X - R_n) - \lambda(R)\\ &= [ \lambda(X-R_n) - \lambda(S)] + (1-a) \lambda(S)\\ &\leq [ \lambda(X-R_n) - \lambda(S)] + (1-a) \lambda(X-R_n)\\ &\leq \epsilon^{n+1} + (1-a)[(1-a)^n + (1-a+\epsilon)^n]\\ &= [\epsilon^{n+1} - \epsilon (1-a+\epsilon)^n] + (1-a)^{n+1} + (1-a+\epsilon)^{n+1}\\ &\leq (1-a)^{n+1} + (1-a+ \epsilon)^{n+1} \end{align*}

and the bound is established. It is clear from our construction that $\bigcup_{n=1}^\infty R_n$ is a disjoint union of closed balls and the bound shows that its complement in $X$ is a null set so we are done.

Hopefully this doesn't have too many mistakes and is somewhat readable. I wasn't expecting the analysis portion of the problem to be as finicky as it turned out to be when I started typing this up...

share|improve this answer
    
I have improved the layout of the formulas and corrected a typo. I hope you don't mind. Nice solution. –  Jonas Teuwen Mar 28 '11 at 15:38
add comment

Divide $\mathbf R^n$ into an integer mesh $\mathcal M_0$, that is divide the space into cubes with integer vertices. Now fill the cubes with closed balls of diameter $\frac12$, call this mesh of balls $\mathcal C_0$. Now we can get from $\mathcal M_0$ an infinite sequence of meshes $\mathcal M_k = 2^{-2k} \mathcal M_0$ by bisecting the sides in $2^{2k}$ parts. So for each cube in $\mathcal M_k$ we get $2^{2n}$ cubes in $\mathcal M_{k + 1}$ and each cube in $\mathcal M_k$ has side length $2^{-k-1}$, so they have diameter $\sqrt{n} 2^{-k-1}$.

From these meshes we get meshes of closed balls $\mathcal C_k$ just like we have obtained $\mathcal C_0$.

Now define $\Omega_k = \{Q \in \mathcal C_n : n = 1,\ldots, k - 1\}$, these are all the previous balls. So the "cover" of the space is now

$$\mathcal F = \bigcup_k \{Q \in \mathcal C_k : Q \cap \Omega_k = \emptyset\}.$$

Note that $\mathcal F^c = \bigcap_k \{Q \in \mathcal C_k : Q \cap \Omega_k \neq \emptyset\}$.

Try #2. The idea is to fill up the mesh with balls where there is horizontally and vertically the space of one diameters between them, then we split up the mesh-sides again in $4$ pieces, fill those up again, then make sure you only select the disjoint ones.

share|improve this answer
1  
I do not think this works for $n=1$. And if $n>1$, are the balls in $\mathcal{C}_0$ disjoint? For $n=2$ thay each ball will touch four other balls. –  Julián Aguirre Mar 24 '11 at 17:54
    
@Julián: Well, not in $\mathcal C_0$ but in $\mathcal F$ they are, no? Oh I see, I include $\mathcal C_0$. Let me see if I can fix that. –  Jonas Teuwen Mar 24 '11 at 18:09
    
I hope it is better this time. –  Jonas Teuwen Mar 24 '11 at 18:33
    
@Jonas: I don't quite get what you're doing. Do you explain anywhere why the remainder is a null set? Moreover, I don't understand what you mean by $Q\cap\Omega_k$: $Q$ is a ball (the variable name is a bit unlucky here; I'd expect a cube) and $\Omega_k$ is a set of balls, right? –  Hendrik Vogt Mar 25 '11 at 11:36
    
@Hendrik: Oops, it doesn't look very good. I'll see if I'll edit or delete it. My idea was to do something like the cantor set but instead of removing the middle $\frac13$ piece I put a ball there and continue like this. –  Jonas Teuwen Mar 27 '11 at 19:10
show 1 more comment

it's overkill but you can use Vitali-Lebesgue covering theorem (they cover subsets of finite measure, so e.g. decompose $\mathbb{R}^n$ to cubes by hyperplanes and then cover the interior of each of those cubes)

share|improve this answer
add comment

Here is an idea without all of the details.

Fix some sphere packing with positive density $d$ so that all spheres have radius $1$.

A positive proportion $c$ of the complement of any collection of spheres of radius at least $1$ is of distance at least $\epsilon$ away from the spheres.

Rescale the sphere packing to be spheres of radius $\epsilon$ to cover at least $dc$ of the complement.

Iterate, so at step $n$ you use spheres of radius $\epsilon^n$ to cover at least $dc$ times what you haven't covered before.

share|improve this answer
add comment

This is an idea which I cannot see if it ends up working or not. Maybe someone can?

Consider the set $\mathcal S$ of all families of closed balls in $\mathbb R^n$ which are pairwise disjoint. Ordered by inclusion, this poset satisfies the hypothesis in Zorn's lemma, so there exist maximal elements $S\in\mathcal S$. One could hope for $S$ to be a candidate...

Now: is $\mathbb R^n\setminus\bigcup_{B\in S}B$ a null set? I don't see how to prove this...

Notice, though, that every decomposition like those gylns wants does give a maximal element in $\mathcal S$.

Later: This idea does not work: Mike has explicitely constructed a counterexample in the comments below.

Is there a way to fix this? I mean: can one select a smaller set $\mathcal S$ such that the union of its maximal elements have null complement?

share|improve this answer
    
@Mariano: I don't see how to prove that that thing is a null set, but isn't AC a little bit overkill? –  Jonas Teuwen Mar 24 '11 at 20:48
4  
There are maximal collections of closed balls whose complement is not a null set. –  Douglas Zare Mar 24 '11 at 20:55
    
@Jonas: I like my vector spaces to have bases, my products of compact spaces to be compact, and my commutative rings to have lots of maximal ideals. AC is true, as far as I am concerned, and a basic fact of nature :) I see the point of worrying about how much choices is needed for this or that, but for the most part, I don't mind lots of choice. –  Mariano Suárez-Alvarez Mar 24 '11 at 22:35
    
@Douglas, I suspected so. Can you construct an explicit example? –  Mariano Suárez-Alvarez Mar 24 '11 at 22:36
1  
@Mariano: Let $(q_i)_{i \in \mathbb{Z}^+}$ be an enumeration of the points in $\mathbb{R}^d$ with all coordinates rational. Let $B_1$ be a closed ball containing $q_1$ of (Lebesgue) measure $1/2$. For $n=2,3,\ldots$, proceed as follows: if $q_n \in B_i$ for some $i < n$ then do nothing; otherwise, choose a closed ball $B_n$ which contains $q_n$, is disjoint from the $B_i$ with $i < n$, and has measure smaller than $2^{-n}$. –  Mike F Mar 25 '11 at 1:44
show 3 more comments

Your Answer

 
discard

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.