I have a question concerning the Proof of the Vitali Covering Theorem in Evans (Thm. 1.26, p. 35). So the Theorem states that we can cover any open, bounded set $\Omega \subset \mathbb{R}^n$ by countable many closed, disjoint balls with radii smaller than any constant $\delta > 0$ up to a Lebesgue nullset.
In the proof, it is stated that given some $\theta \in (1- \frac{1}{5^n}, 1)$, we find some countable family $\tilde{G}$ such that \begin{equation} \mathcal{L}^n(\Omega\setminus\cup_{B \in \tilde{G}} B) \leq (1- \frac{1}{5^n}) \mathcal{L}^n(\Omega). \end{equation} This, I do understand. But then it says that we can find a finite subfamily $G \subset \tilde{G}$ such that \begin{equation} \mathcal{L}^n(\Omega \setminus \cup_{B \in G} B) \leq \theta \mathcal{L}^n(\Omega). \end{equation}
Why does this hold? I had the feeling it has something to do with the inner regularity of the Lebesgue measure, but don't see how it works out.
The key of the question is Borel sets are Lebesgue-measurable.
$$\mathcal{L}^n(\Omega \backslash \cup_{B\in \tilde{G}} B)\leq (1-\frac{1}{5^n})\mathcal{L}^n(\Omega) < \theta \mathcal{L}^n(\Omega)$$
Note closed balls in $B$ are countable(Vitali’s Covering Theorem), so we have:
$$\Omega \backslash \cup_{B\in \tilde{G}} B = \Omega - \cup_{k=1}^{\infty} B_k=lim_{m\rightarrow \infty}(\Omega - \cup_{k=1}^{m} B_k)$$
Notice $(\Omega - \cup_{k=1}^{m} B_k)$ is decreasing by m, $\Omega$ and $\{B_k\}_{k=1}^{\infty}$ are all Lebesgue-measurable(closed sets and open sets are Borel sets), wolg we assume $\mathcal{L}^n(\Omega) < \infty$, we have:
$$\begin{align} \mathcal{L}^n(\Omega \backslash \cup_{B\in \tilde{G}} B) & = lim_{m\rightarrow \infty} \mathcal{L}^n(\Omega - \cup_{k=1}^m B_k)\\\\ & < \theta \mathcal{L}^n(\Omega) \end{align}$$
And by definition of limit, $\exists m_0\in N_{+}$ s.t. $$\mathcal{L}^n(\Omega - \cup_{k=1}^{m_0}B_k)\leq \theta \mathcal{L}^n(\Omega)$$
If $\mathcal{L}^n(\Omega) = \infty$, we apply the above to $\Omega_m=\{x\in \Omega | m<|x|< m+1\}$