The union of more than $\aleph_0$ measurable sets obviously need not be measurable.

What about balls in $\mathbb{R}^n$? - is it possible to find an uncountable family of balls (with nonempty interiors) such that their union is nonmeasurable?

  • $\begingroup$ Of course the union of any family of open balls is open, hence measurable; possibly you should specify "closed balls". $\endgroup$ – David C. Ullrich Jan 15 '18 at 14:50

David Ullrich's answer is incorrect, because $E$ is a null set with regards to the $2-D$ Lebesgue measure.

The claim is at least true for $d=2$, for which I will provide a proof. We need a few results first. The claim is this:

Claim: The arbitrary union of closed discs need not be Borel, but it is measurable.

I'm going to state, and not prove, a result characterizing Lebesgue measurability that we used in my class often.

Theorem: A set $A$ is Lebesgue measurable in $\mathbb{R}^d$ if and only if almost every point in $A$ is a density point, and almost every point in $A^c$ has $0$ density in $A$.

We denote the density of $x$ in $A$ by $d(x,A) = \displaystyle \lim_{r \to 0} \frac{m(B_r(x) \cap A)}{m(B_r(x))}.$

$\textbf{Lemma 1.}\textit{ The Borel sets have the Cardinality of the Continuum.}$

$\textit{Proof.}$ Call the Borel $\sigma-$algebra $B$. First, we know that $\textbf{c}\le |B|$, since there is no countably infinite $\sigma-$algebra. Now, since $\{B_r(q)| q \in \mathbb{Q}^2, r \in \mathbb{Q}\}$ forms a countable basis for the open sets in $\mathbb{R}^2$, we know every open set in $\mathbb{R}$ can be written as a countable union of them. We can create a surjection from $2^\mathbb{N}$ by mapping sequences to open sets based on whether or not they include one of the countably many $B_r(q)$ in the unions that comprise them. This is clearly not an injection, since we don't have disjointness. Anyways, this says $\textbf{c} = |2^\mathbb{N}| \ge |B|$, since the cardinality of the Borel sets is that of the open sets (the Borel sets are formed from countable unions and complements of open sets, not changing the cardinality). Thus the cardinality of the Borel sets is $\textbf{c}$.

$\textbf{Lemma 2.} \textit{ There are more than Continuum many sets which can be written as the uncountable union of discs.}$

$\textit{Proof.}$ For a given set $A \in \mathbb{R}^2$, map each $x\in A$ to the disc of radius $1$ centered at $A$. Then this is an injection from the power set of $\mathbb{R}^2, \mathcal{P}(\mathbb{R}^2)$ to the sets of unions of discs. $

$\textbf{Proposition.}\textit{ If we can prove this problem for discs of radii bounded below, then we are done.}$

$\textit{Proof.}$ This is because if we let $F_n$ be the set of discs of radii $\ge 1/n$ in our union $\displaystyle \bigcup_{i \in I}D_i$, then if we prove $F_n$ is measurable, $\displaystyle \bigcup_{i \in I}D_i = \bigcup_{n \in \mathbb{N}}F_n$ is the countable union of measurable sets.

From Lemma 1 and Lemma 2, we know that $\displaystyle \bigcup_{i \in I}D_i$ need not be Borel. Of course our union is uncountable, else we would trivially have a Borel and thus measurable set. From the proposition, it suffices to consider the radii bounded below by $1$. So, set $X = \displaystyle \bigcup_{i \in I}D_i$, and let $Y = \displaystyle \bigcup_{i \in I}D_i^o$. Then $\overline{Y}\setminus Y \supset X \setminus Y$. We aim to show $\overline{Y}\setminus Y$ is measure zero, and thus so is $X \setminus Y$. So, take $x \in \overline{Y} \setminus Y$. So, a limit point, but not an interior point of any disc. Then for arbitrarily small distances, there is a $D_i$ such that nearly half of $B_r(x)$ is in $D_i$, for sufficiently small $r$. That is, for each $r$ as $r\to 0$ there is a $D_i$ such that $x$ is very close to the boundary of $D_i$, and that $D_i$ fills up almost half of $B_r(x)$. Since the $D_i$ are all of a certain size, they do not get degenerately small. Since circles are smooth $1-$manifolds, locally they look like $\mathbb{R}$. Hence as $r$ gets very small and $D_i$ gets very close to $x$, the boundary of $D_i$ is like a straight line. Hence one sees we can fit a small ball of say, radius $r/10$ (this need not be optimal) in the part of $B_r(x)$ that's in $D_i$. But then $d(x, \overline{Y}\setminus Y) < \displaystyle \frac{\pi r^2 - \pi(r/10)^2}{\pi r^2} = \frac{99}{100} <1$. And thus $x$ is not a density point. But since $x$ was arbitrary, this says $\overline{Y}\setminus Y$ has no density points, and thus is measure zero. Now we're done.

Note: The same result holds for squares and other polyhedra. We can fit balls inside of squares, and other polyhedra, and there was nothing particularly important about circles in our argument above.

  • $\begingroup$ Is the map in the proof of Lemma 2 really an injection? Say $A$ is a disc of radius greater than $1$ and $A'$ is $A$ minus the center of $A$; it looks to me like $A$ and $A'$ get mapped to the same union of disks. (The lemma is certainly true, for example, heh, if $E\subset\Bbb R$ there is a union of disks that intersects $\Bbb R$ in precisely $E$...) $\endgroup$ – David C. Ullrich Jan 15 '18 at 16:04
  • $\begingroup$ @DavidBowman May I ask where I can find a reference for the theorem you mentioned about the characterization of measurable sets using density points? Does it work for other Radon measure as well? $\endgroup$ – BigbearZzz May 2 '19 at 12:57

I thought the answer was yes in $\Bbb R^d$ for $d>1$; no for $d=1$.

David Bowman pointed out an error in the proof for $d=2$ below - I'm leaving it here as an exercise. See if you can spot the error:

Wrong example in $\Bbb R^2$: Say $E\subset\Bbb R$ is non-measurable. Regard $\Bbb R$ as a subset of $\Bbb C$ in the usual way. For every $x\in E$ let $$B_x=\overline{D(x+i,1)},$$so in particular $B_x\cap\Bbb R=\{x\}$.

Then $\bigcup_{x\in E}B_x$ cannot be measurable, because $$E=\Bbb R\cap \bigcup_{x\in E}B_x$$is not measurable.

That's wrong - if you give up here's why:

There's no contradiction in showing $E$ is measurable here! Since Lebesgue measure is complete any subset of $\Bbb R$ is measurable, as a subset of $\Bbb R^2$.

On the other hand, suppose $E\subset\Bbb R$ is the union of a family of (non-degenerate!) closed intervals. Say $C$ is a connected component of $E$. If $x\in C$ then we have $x\in[a,b]\subset E$ with $b>a$. Hence $[a,b]\subset C$, so in particular $C$ contains a rational.

So $E$ is a countable union of (open, closed or half-open) intervals, hence $E$ is measurable. Borel, in fact.

  • $\begingroup$ The arbitrary union of closed discs is measurable, see my proof below. $\endgroup$ – David Bowman Jan 15 '18 at 15:25
  • $\begingroup$ @DavidBowman Well, one of us is wrong. The example I gave is very simple - do you see an error? If the error is on my end finding it should be easy. (If you want to look for an error in your post you might look for a step that fails for the family of disks I constructed...) $\endgroup$ – David C. Ullrich Jan 15 '18 at 15:30
  • $\begingroup$ I mentioned your error in my post. The $2-d$ Lebesgue measure is complete, so any subset of $\mathbb{R}$ is a null set in $\mathbb{R}^2$. $\endgroup$ – David Bowman Jan 15 '18 at 15:33
  • $\begingroup$ @DavidBowman Ah, right. Thanks. $\endgroup$ – David C. Ullrich Jan 15 '18 at 15:38

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