The Vitali Covering Lemma Part 1

# The Vitali Covering Lemma Part 1

Recall from the Vitali Covers of a Set page that if $E$ is a set of real numbers then a collection of intervals $\mathcal I$ is said to be a Vitali cover of $E$ if for all $\epsilon > 0$ we have that for all $x \in E$ there exists an interval $I \in \mathcal I$ with $x \in I$ and $l(I) < \epsilon$.

We now prove a very important result called the Vitali covering lemma.

The proof of the result below is rather lengthy, so we split it into two parts.

• The Vitali Covering Lemma Part 1
 Lemma (The Vitali Covering Lemma): Let $E$ be a set of real numbers such that $m^*(E) < \infty$ and let $\mathcal I$ be a Vitali cover of $E$. Then for all $\epsilon > 0$ there exists a finite collection $\{ I_1, I_2, ..., I_N \}$ of mutually disjoint intervals from $\mathcal I$ such that $\displaystyle{m^* \left ( E \setminus \bigcup_{n=1}^{N} I_n \right ) < \epsilon}$.
• Proof: We first make an important observation.
• In the proof of this result, we may assume that $\mathcal I$ is a collection of closed intervals. This is because if $\mathcal I$ is a Vitali cover of $E$ then the set of the closures of each element in $\mathcal I$ is a Vitali cover of $E$. Let $\mathcal{\overline{I}}$ denote this set:
(1)
\begin{align} \quad \mathcal{\overline{I}} = \{ \overline{I} : I \in \mathcal I \} \end{align}
• To show that $\mathcal{\overline{I}}$ is a Vitali cover of $E$, let $\epsilon > 0$ be given and let $x \in E$. Since $\mathcal I$ is a Vitali cover of $E$ there exists an interval $I \in \mathcal I$ such that $x \in E$ and $l(I) < \epsilon$. Then $\overline{I} \in \mathcal{\overline{I}}$, $x \in \overline{I}$, and $l(\overline{I}) = l(I) < \epsilon$. Thus $\mathcal{\overline{I}}$ is indeed a Vitali cover of $E$.
• If the conclusion of the Vitali Covering Lemma holds for the Vitali cover $\mathcal I$, that is, for all $\epsilon > 0$ there exists a finite collection $\{ \overline{I_1}, \overline{I_2}, ..., \overline{I_N} \}$ of mutually disjoint intervals from $\mathcal{\overline{I}}$ such that $\displaystyle{m^* \left ( E \setminus \bigcup_{n=1}^{N} \overline{I_n} \right ) < \epsilon}$ then for all $\epsilon > 0$ there exists a finite collection $\{ I_1, I_2, ..., I_N \}$ of mutually disjoint intervals from $\mathcal I$ such that:
(2)
\begin{align} \quad m^* \left ( E \setminus \bigcup_{n=1}^{N} I_n \right ) = m^* \left ( E \setminus \bigcup_{n=1}^{N} \overline{I_n} \right ) < \epsilon \end{align}
• Now since $m^*(E) < \infty$ there exists an open set $O$ such that $E \subseteq O$ and:
(3)
\begin{align} \quad m^*(O) \leq m^*(E) + 1 < \infty \end{align}
• We may assume that every interval $I \in \mathcal I$ is contained in $O$. This is because if $\mathcal I$ is a Vitali cover of $E$ then the set of intervals in $\mathcal I$ that are also contained in $O$ is a Vitali cover of $E$. Let $\mathcal I_O$ denote this set:
(4)
\begin{align} \quad \mathcal I_O = \{ I : I \in \mathcal I, I \subseteq O \} \subseteq \mathcal I \end{align}
• To show that $\mathcal I_O$ is a Vitali cover of $E$, let $\epsilon^* > 0$ be given. Let $x \in E$. Since $O$ is an open set, there exists an open interval centered at $x$ contained in $O$. Say $x \in (x - \alpha, x + \alpha) \subseteq O$. Then $l((x - \alpha, x + \alpha)) = 2\alpha$.
• Since $\mathcal I$ is a Vitali cover of $E$, for this $x$ there exists an interval $I \in \mathcal I$ such that $x \in I$ and $l(I) < \min \{ \epsilon^*, \alpha \}$. Then we must have that this $I$ is such that $l(I) < \epsilon^*$ and $I \subseteq (x - \alpha, x + \alpha) \subseteq O$, i.e., $I \in \mathcal I_O$.
• So for all $\epsilon^* > 0$ we have that for all $x \in E$ there is an interval $I \in \mathcal I_O$ such that $x \in I$ and $l(I) < \epsilon$, so $\mathcal I_O$ is a Vitali cover of $E$.
• So we may assume that if $I \in \mathcal I$ then $I \subseteq O$.
• Now let $I_1 \in \mathcal I$ be any interval in $\mathcal I$. There are two cases to consider.
• Case 1.1: If $E \subseteq I_1$ then we're done since $\{ I_1 \}$ is a finite collection of (mutually disjoint) intervals from $\mathcal I$ such that:
(5)
\begin{align} \quad m^*(E \setminus I_1) = m^*(\emptyset) = 0 < \epsilon \end{align}
• Case 2.1: If $E \not \subseteq I_1$ then there exists an element $x \in E$ such that $x \not \in I_1$. Let $k_1$ be the supremum of the lengths of all intervals in $I$ that are disjoint from $I_1$.
(6)
\begin{align} \quad k_1 = \sup \{ l(I) : I \in \mathcal I, I \cap I_1 = \emptyset \} \end{align}
• We note that $k_1 < \infty$. This is because each $I$ is assumed to be such that $I \subseteq O$ and $m^*(I) \leq m^*(O) < \infty$. Consider the number $\displaystyle{\frac{1}{2}k_1}$. By one of the properties of the supremum, we must have that there exists an interval $I_2 \in \mathcal I$ with $I_1 \cap I_2 = \emptyset$ such that:
(7)
• (If not, then this would contradiction $k_1$ being the supremum defined above). There are two cases to consider once again.
• Case 1.2: If $E \subseteq (I_1 \cup I_2)$ then we're done since $\{ I_1, I_2 \}$ is a finite collection of mutually disjoint intervals from $\mathcal I$ such that:
(8)
\begin{align} \quad m^*(E \setminus (I_1 \cup I_2)) = m^*(\emptyset) = 0 < \epsilon \end{align}
• Case 2.2: If $E \not \subseteq (I_1 \cup I_2)$ then let $k_2$ be the supremum of the lengths of all intervals in $\mathcal I$ that are mutually disjoint from $I_1$ and $I_2$, that is:
(9)
\begin{align} \quad k_2 = \sup \left \{ l(I) : I \in \mathcal I, \: I \cap ( I_1 \cup I_2) = \emptyset \right \} \end{align}
• By one of the properties of the supremum, we must have that there exists an interval $I_3 \in \mathcal I$ with $(I_1 \cup I_2) \cap I_3 = \emptyset$ such that:
(10)
• We proceed in this manner. Suppose that $\{ I_1, I_2, ..., I_n \}$ have been chosen as above with corresponding $k_1, k_2, ..., k_{n-1}$. There are two cases to consider.
• Case n.1: If $E \subseteq \bigcup_{j=1}^{n} I_j$ then we are done since $\{ I_1, I_2, ..., I_n \}$ is a finite collection of mutually disjoint intervals from $\mathcal I$ such that:
(11)
\begin{align} \quad m^* \left ( E \setminus \bigcup_{j=1}^{n} I_j \right ) = m^*(\emptyset) = 0 < \epsilon \end{align}
• Case n.2: If $E \not \subseteq \bigcup_{j=1}^{n} I_j$ then let $k_n$ be the supremum of the lengths of all intervals in $\mathcal I$ that are mutually disjoint from $I_1$, $I_2$, …, $I_n$, that is:
(12)
\begin{align} \quad k_n = \sup \left \{ l(I) : I \in \mathcal I, \: I \cap \left ( \bigcup_{j=1}^{n} I_j \right ) = \emptyset \right \} \end{align}
• By one of the properties of the supremum, we must have that there exists an interval $I_{n+1} \in \mathcal I$ with $(I_1 \cup I_2 \cup ... \cup I_n) \cap I_{n+1} = \emptyset$ such that:
(13)
• If at some $N \in \mathbb{N}$, $\{ I_1, I_2, ..., I_N \}$ we have that $E \subseteq \bigcup_{n=1}^{N} I_n$ then we're done. If not, then we will have a sequence of mutually disjoint intervals $(I_n)_{n=1}^{\infty}$ and a sequence $(k_n)_{n=1}^{\infty}$ of positive numbers. Furthermore, since $(I_n)_{n=1}^{\infty}$ is a collection of mutually disjoint intervals contained in $O$ and $m^*(O) < \infty$ we have that:
• Therefore $\displaystyle{\sum_{n=1}^{\infty} l(I_n)}$ is a convergent series of positive real numbers. So for $\epsilon_1 = \frac{\epsilon}{5} > 0$ there exists an $N \in \mathbb{N}$ such that: