I have come across this problem in Royden. It appears that Royden wants us to extend the Vitali covering lemma to a countable collection of disjoint intervals as opposed to a finite collection. The problem states:

Let $E$ be a set of finite outer measure and $\mathcal{F}$ a collection of closed, bounded interval that covers $E$ in the sense of Vitali. Show that there is a countable disjoint collection $\{ I_k \}_{k = 1}^{\infty}$ of intervals in $\mathcal{F}$ for which $$ m^* \Bigg{[} E \sim \bigcup_{k =1}^{\infty} I_k \Bigg{]} = 0.$$

I am not too sure where to take this proof


1 Answer 1


To all who are reading this: please also read the answer by the other user. An issue with my answer that I didn’t realise was raised in the comment section and fixed in the other answer.


We construct $\{I_k\}_{k=1}^{\infty}$ inductively.

Enumerate the natural numbers with $m$, beginning with $m=1$.

Let $\epsilon=\frac{1}{m}$.

Find inductively, using the Vitali Covering Lemma, a finite disjoint collection of closed intervals, $\{I_{mk}\}_{k=1}^{n_m}$, for which $$m^* (E_m) < \frac{1}{m},$$

where we define $E_1= E \sim \bigcup_{k =1}^{n_1} I_k$ and, for subsequent $m$'s, $E_m=E_{m-1} \sim \bigcup_{k =1}^{n_m} I_k.$

It follows that $m^* (E_m) < \frac{1}{m}$ is arbitrarily small. We let our $\{I_k\}_{k=1}^{\infty}$ to be the collection of all the intervals that we have found. There are $\sum_{m=1}^{\infty}n_m$ of them -- that is a countable collection of intervals.

We can make this collection disjoint, by, at the beginning of each round, taking out the ones that would intersect the intervals already chosen in the previous rounds.


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