The measure of an uncountable union of a family of open sets. Given that $m$ is a regular measure and $\mathcal{O} = \{O_{\alpha}\}$ be an uncountable family of open sets.
Then certainly $\bigcup_{\alpha} O_{\alpha}$ is measurable since it is open.
I want to to prove that
$$ m(\bigcup_{\alpha}O_{\alpha}) = \sup \{ \sum_{i=1}^n m(O_{\alpha_i}) \mid n \in \mathbb{N} \text{ and } O_{\alpha_i} \in \mathcal{O} \}
$$
One direction of inequality is trivial, however I could not prove the other direction. Any hint please? Or maybe this equality is just false?
Edit:
I have to add the condition that $\{O_{\alpha}\}$ is a collection of disjoint sets, otherwise some comments and answers pointed out this equality is false.
 A: 
A regular measure on a topological space is a measure for which every measurable set can be approximated from above by open measurable sets and from below by compact measurable sets.
Wikipedia

So let $\mu$ be a regular measure.  We assume the sigma-algebra is the Borel sets (the OP doesn't say that, but does assume that open implies measurable).
Let $A = \mu(\bigcup_{\alpha} O_\alpha) \in (0,+\infty]$.  Let $B \in \mathbb R$ such that $B < A$.  By regularity, there is a compact measurable set $K \subset \bigcup_\alpha O_\alpha$ with $\mu(K) \ge B$.  By compactness, the open cover $O_\alpha$ of $K$ has a finite subcover $K \subseteq O_{\alpha_1}\cup\cdots\cup O_{\alpha_n}$.  Then
$$
B \le \mu(K) \le \mu(O_{\alpha_1})+\dots+\mu(O_{\alpha_n})
$$
Therefore
$$
\sup \left\{ \sum_{i=1}^n m(O_{\alpha_i}) \mid n \in \mathbb{N} \text{ and } O_{\alpha_i} \in \mathcal{O} \right\} \ge B
$$
This is true for all $B < A$, so
$$
\sup \left\{ \sum_{i=1}^n m(O_{\alpha_i}) \mid n \in \mathbb{N} \text{ and } O_{\alpha_i} \in \mathcal{O} \right\} \ge A = \mu\big(\bigcup_\alpha O_\alpha\big);
$$
A: That is not true. Just imagine what will happen when all $O_\alpha$'s are equal.
A: If $\mathscr{O}$ is an uncountable family of pairwise disjoint open sets in $\Bbb R^n$, let $$\mathscr{O}_0=\{O\in\mathscr{O}:O\ne\varnothing\}\,;$$ then $\mathscr{O}_0$ is countable, since each member of $\mathscr{O}_0$ must contain a different element of $\Bbb Q^n$. That is, $\mathscr{O}$ has only countably many non-empty elements, and you might as well be working with the countable family $\mathscr{O}_0$ in the first place.
Of course all of this applies equally well to any separable metric space.
