I am reading Measure Theory and Fine Properties of Functions by Evans and Gariepy. In the proof Vitali covering theorem, it states that

Let $\mathcal{F}$ be any collection of nondegenerate closed balls in $\mathbb{R}^n$ with $$\sup\{\text{diam } B\; |\; B\in \mathcal{F}\}< \infty$$ Then there exists a countable family $\mathcal G$ of disjoint balls in $\mathcal{F}$ such that $$\bigcup_{B\in \mathcal{F}} B \subset \bigcup_{B\in \mathcal{G}} 5B.$$

However in the proof. It only showed that there exists a subcollection $\mathcal{G}$ which has the property $$\bigcup_{B\in \mathcal{F}} B \subset \bigcup_{B\in \mathcal{G}} 5B,$$ how to deduce that $G$ can be chosen as a countable collection? Is this something trivial?

The proof given in the book is also on wikipedia you can find here. And there is a remark says that in the context of a general metric space (instead of $\mathbb{R}^n$) the resulting subcollection may not be countably infinite.


2 Answers 2


Note $\mathcal G$ is a collection of disjoint balls. As every ball in $\mathbb R^n$ contains a rational point, there cannot be more than countably many of them.


Here is another cute proof, which may apply in other context as well, e.g. to cases where one is working with other types of sets which are not open and may not contain rational, etc.

Fix a huge (but finite) box in $Q \subset R^n$. How many disjoint balls of measure more than 1 can you fit in Q? Only finitely many. How many disjoint balls of measure more than 1/2 can you fit in? Only finitely many... Inductively, for any natural number k, there are only finitely many balls of measure more than 1/k in Q?

How many balls of positive measure can you fit in Q? At most countably many because every ball of positive measure is a ball of measure more than 1/k for some k.

Now $R^n$ is covered in turn by a countable union of such Q. For details (or a visual explanation) see my video on YT: https://www.youtube.com/watch?v=YWWj97i39_Y


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