Vitali Covers of a Set

Definition: Let $E$ be a set. A collection $\mathcal I$ of intervals is a Vitali Cover or Vitali Covering of $E$ if for all $\epsilon > 0$ we have that for every $x \in E$ there exists an interval $I \in \mathcal I$ such that $x \in I$ and $l(I) < \epsilon$. Furthermore, we say that $\mathcal I$ Covers $E$ in the sense of Vitali.

From the definition above, a collection of intervals $\mathcal I$ is a Vitali cover of $E$ if for every element $x$ in $E$ we can find an interval $I$ in the collection $\mathcal I$ containing $E$ of arbitrarily small length.

For example, consider the set $\mathbb{N}$ of natural numbers and let:

(1)

\begin{align} \quad \mathcal I_{\mathbb{N}} = \left \{ I_{n, k} = \left ( n - \frac{1}{2k}, n + \frac{1}{2k} \right ) : n, k \in \mathbb{N} \right \} \end{align}

We claim that $\mathcal I_{\mathbb{N}}$ is a Vitali cover of $\mathbb{N}$. To prove this, let $\epsilon > 0$ be given and let $n \in \mathbb{N}$.

For each $n \in \mathbb{N}$ we note that $l(I_{n, k}) = \frac{1}{k}$. So choose $k = k(\epsilon) \in \mathbb{N}$ such that $\displaystyle{\frac{1}{k(\epsilon)} < \epsilon}$. Then $n \in I_{n, k(\epsilon)}$ and:

(2)

\begin{align} \quad l(I_{n,k(\epsilon)}) = \frac{1}{k(\epsilon)} < \epsilon \end{align}

Therefore $\mathcal I_{\mathbb{N}}$ is indeed a Vitali cover of $\mathbb{N}$.

Note that for any set $E \subseteq \mathbb{R}$ that the following collection if a Vitali cover of $E$:

(3)

\begin{align} \quad \mathcal I_E = \left \{ I_{x, k} = \left ( x - \frac{1}{2k}, x + \frac{1}{2k} \right ) : x \in E, \: k \in \mathbb{N} \right \} \end{align}