2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.