# Correcting solution of question 6 in Royden fourth edition.

Here is the question and its solution:

Show that the Vitali Covering Lemma extends to the case in which the covering collection consists of nondegenerate intervals that are not necessarily closed

Proof:

Let $$\epsilon > 0$$ be given. Let $$E\subseteq R$$ such that $$m^{*}(E)< \infty$$. Let $$\mathcal{F}$$ contains non degenerate general intervals(not necessarily closed ). Then, $$F$$ contains all kind of intervals such that the length of those intervals is not zero since they are non degenerate intervals. Since $$m^{*}(E)< \infty$$ there is an open set $$\mathcal{O}$$ containing $$E$$ for which $$m(\mathcal{O}) < \infty.$$ because $$\mathcal{F}$$ is a Vitali covering of $$E,$$ we may assume that each interval in $$\mathcal{F}$$ is contained in $$\mathcal{O}.$$

\textbf{Justification:}

Let $$x \in E \subseteq \mathcal{O}$$, there is some $$r > 0$$ such that $$B(x,r) \subseteq \mathcal{O}$$ (as $$\mathcal{O}$$ is an open set, so every point in $$\mathcal{O}$$ is an interior point ).\

Now, given $$\frac{r}{2} > 0$$, there is an $$I = (a,b) \in \mathcal{F}$$ with $$\ell (I) < \frac{r}{2}$$ such that $$x \in I$$ (by definition of covering in the sense of Vitali). Hence $$b-a < \frac{r}{2}$$ or $$b < \frac{r}{2} + a$$, and $$a < x < b$$. From these we get $$a < x < a + \frac{r}{2}$$ or $$|x-a| < \frac{r}{2}$$. Hence, if $$y \in I = (a,b)$$, then

\begin{align} |x-y| &= |(x-a) + (a-y)| \\ &\le |x-a| + |y-a| \\ &< \frac{r}{2} + \frac{r}{2} = r, \\ \end{align} and therefore $$y \in B(x,r)$$, from which it follows $$I \subseteq B(x,r) \subseteq \mathcal{O}$$.\

Now, by countable additivity and monotonicity of measure, we have if $$\{ I_{k}\}_{k=1}^{\infty} \subseteq \mathcal{F}$$ is disjoint,then $$\sum_{k = 1}^{\infty} \ell (I_{k}) \leq m(\mathcal{O}) < \infty.$$ \

Now, since the intervals of $$\mathcal{F}$$ are nondegenerate and not necessarily closed and since by proposition 9 on pg. 17 we have that every nonempty open set (we are speaking about $$\mathcal{O}$$, if the open interval in it is larger than $$\mathcal{F}$$ we can shrink by a similar argument to the justification above, the only difference is that we will keep the interval open ) is the disjoint union of a countable collection of open intervals,we can find an open interval $$(a,b)$$ in $$\mathcal{F}$$ from which we can create closed intervals as follows:\

Assume that we have the nondegenerate ($$a < b$$) open interval $$I = (a,b)$$, then it contains the following closed interval $$[a + \frac{b-a}{3}, b - \frac{b-a}{3}]$$ and we have $$a < a + \frac{b-a}{3} < b - \frac{b-a}{3} < b.$$\

Now, define $$\mathcal{F}_{o} = \{J \in \mathcal{F}: J \textbf{ is closed and bounded and} J \subseteq I\}.$$ \

Now by the justification above we can assume that each $$J$$ in $$\mathcal{F}_{o}$$ is contained in $$\mathcal{O}$$ as $$\mathcal{F}_{o} \subseteq \mathcal{F}.$$ Moreover, by countable additivity and monotonicity of measure, we have if $$\{ J_{k}\}_{k=1}^{\infty} \subseteq \mathcal{F}_{0}$$ is disjoint,then $$\sum_{k = 1}^{\infty} \ell (J_{k}) \leq m(\mathcal{O}) < \infty.$$ \

Moreover, since each $$J_{k}$$ is closed and $$\mathcal{F}_{0}$$ is a Vitali covering of $$E,$$ if $$\{J_{k}\}_{k =1}^{n} \subseteq \mathcal{F}_{0}$$, then $$E \setminus \bigcup_{k =1}^{n} J_{k} \subseteq \bigcup_{J'\in \mathcal{F}_{o}^n} J' \textbf{where }\mathcal{F}_{o}^n = \{ J' \in \mathcal{F}_{0} | J' \cap \bigcup_{k =1}^{n} J_{k} = \emptyset \}$$

\textbf{Justification:}\

Let $$\{J_k\}_{k=1}^n \subseteq \mathcal{F}_{0}$$ and let $$x \in E \setminus \bigcup_{k=1}^n J_{k}$$. Then $$x \notin J_k$$ for every $$k$$, and, as the $$J_k$$ are closed, there exists $$r_k > 0$$ for which $$B(x,r_k) \cap J_k = \emptyset$$. Letting $$r = \frac{1}{2} \min\{r_1,...,r_n\}$$, there exists $$J_r$$ with $$\ell(J_r) < r \le \frac{r_k}{2}$$ for each $$k$$ such that $$x \in J_r$$. But, as we saw above, this means $$J_r \subseteq B(x,r_k)$$ for each $$k$$ and therefore $$J_r \cap J_k = \emptyset$$ for each $$k$$. Hence $$x \in J_r \subseteq \bigcup_{J' \in \mathcal{F}_{o}^n} J'$$. \

If there is a finite disjoint subcollection of $$\mathcal{F}_{o}$$ that covers $$E,$$ the proof is complete. Otherwise, we will define the disjoint countable subcollection $$\{J_{k}\}_{k=1}^{\infty}$$ of $$\mathcal{F}_{o}$$ inductively. Let $$J_{1} \in \mathcal{F}_{o}$$ be arbitrary and suppose that $$J_{1}, ..., J_{n}$$ have been chosen.), which has the following property

$$E \setminus \bigcup_{k =1}^{n} J_{k} \subseteq \bigcup _{k = n +1}^{\infty} 5 \times J_{k} \textbf{ for all n},$$(in this step a problem may arise if $$J_{k}$$ were not closed). Where for a closed bounded interval $$J,$$ $$5 \times J$$ denotes the closed interval that has the same midpoint as $$J$$ and 5 times its length.

To begin this selection, Let $$J_{1} \in \mathcal{F}_{o}$$ be arbitrary and suppose that $$n$$ is a natural number and the finite disjoint subcollection $$J_{1}, ..., J_{n}$$ have been chosen. If $$E \setminus J_{1} = \emptyset$$, then the proof is complete. Otherwise, choose any $$J_{2}\in \mathcal{F}_{o}^1$$ such that $$\ell (J_{2}) > S_{1}/2$$ (continuing in this way i.e. If $$E\setminus (J_{1} \cup J_{2})= \emptyset$$, then the proof is complete and so on )where $$S_{n}$$ is the supremum of the lenths of the intervals in $$\mathcal{F}_{o}^n$$ and its finite since m($$\mathcal{O}$$) ia an upper bound for these lengths and $$S_{n}$$ is greater than $$0$$ because $$E$$ is not covered by $$J_{1}, ..., J_{n}$$. mimic the remaining steps of the proof of vitali covering lemma on pg.110 in our book.

My question is:

1-It turns out that the answer is wrong, specifically the defined set $$\mathcal{F}_{0}$$ and the bold paragraph, but I do not understand why this is wrong, could anyone clarify this for me, please?

2-Also, sharing the correct solution for this problem will be appreciated.