The Vitali Covering Lemma Part 2

# The Vitali Covering Lemma Part 2

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.

We continue the proof of the Vitali Covering Lemma. The first part is linked below for reference.

**The Vitali Covering Lemma Part 2**

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 (Continued):**The Vitali Covering Lemma Part 1

- We claim that $\{ I_1, I_2, ..., I_N \}$ is the set of mutually disjoint intervals in $\mathcal I$ that provide us with the conclusion of the lemma.

- Let $x \in E \setminus \bigcup_{n=1}^{N} I_n$. Since $\mathcal I$ is a Vitali cover of $E$ there exists an interval $I \in \mathcal I$ such that $x \in I$ and $\displaystyle{I \cap \bigcup_{n=1}^{N} I_n = \emptyset}$ (this can be done by taking the length of $I$ to be equal to half of the distance from $x$ to $\displaystyle{\bigcup_{n=1}^{N} I_n}$).

- Furthermore, $I$ must intersect at least one of the intervals in the sequence $(I_n)_{n=1}^{\infty}$, say $I_j$. If not, then for every $j \in \mathbb{N}$ we have that $I \cap \left ( \bigcup_{n=1}^{j} I_n \right ) = \emptyset$. But then:

\begin{align} \quad l(I) \leq k_j \overset{\dagger} < 2 l(I_{j+1}) \end{align}

- (See $\dagger$ from part 1).

- But $2 l(I_{j+1}) \to 0$ from the convergence of $\displaystyle{\sum_{n=1}^{\infty} l(I_n)}$. Therefore $l(I) = 0$ which is a contradiction since $I$ is an interval and must have positive length. Therefore the assumption that $I$ does not intersect at least one of the intervals in the sequence $(I_n)_{n=1}^{\infty}$ is false. So there exists a $j \in \mathbb{N}$ such that $I$ intersects $I_j$.

- Let $j_0$ be the smallest natural number greatest than or equal to $N+1$ such that $I$ intersects $I_j$, that is:

\begin{align} \quad j_0 = \min \{ j : j \geq N+1, \: I \cap I_j \neq \emptyset \} \end{align}

- Since $j_0$ is the smallest number such that $I \cap I_j \neq \emptyset$ we have that $I$ does not intersect the sets $I_1, I_2, ..., I_{j_0 - 1}$. Furthermore, $l(I) \leq k_{j_0-1} < 2l(I_{j_0})$. Let $c_{j_0}$ be the center of the interval $I_{j_0}$. Then:

\begin{align} \quad d(x, c_0) \leq l(I) \leq l(I) + \frac{1}{2} l(I_{j_0}) \leq 2l(I_{j_0}) + \frac{1}{2} l(I_{j_0}) = \frac{5}{2} l(I_{j_0}) \end{align}

- Let $I_{j_0}'$ be the closed interval with center $c_{j_0}$ and whose length is $5l(I_{j_0})$ . Then $x \in I_{j_0}'$ and:

\begin{align} \quad E \setminus \bigcup_{n=1}^{N} I_n \subseteq \bigcup_{n=N+1}^{\infty} 5I_n \end{align}

- Therefore:

\begin{align} \quad m^* \left ( E \setminus \bigcup_{n=1}^{N} I_n \right ) \leq m^* \left ( \bigcup_{n=N+1}^{\infty} 5I_n \right ) = 5 \cdot \frac{\epsilon}{5} = \epsilon \quad \blacksquare \end{align}