Let $(\Omega,O)$ be a topological space with countable basis $\mathcal F$. Furthermore, let $\mu$ be a Borel-measure (resp. a measure on the measure space $(\Omega,\mathcal B(\Omega))$.
Prove that if for a family of open sets $(U_i)_{i\in I}$ with $\mu(U_i)=0 \ \ \forall i\in I$ it follows that $\mu(\bigcup_{i\in I}U_i)=0$
The problem here is that $I$ may be an uncountable index set, so I cannot use the $\sigma-$sub-additivity. We have:
$$U_i = \bigcup_{\substack{F\in\mathcal F \\ F\subset U_i}}F$$ Hence $$\bigcup_{i\in I}U_i=\bigcup_{i\in I}\bigcup_{\substack{i\in F\in\mathcal F \\ F\subset U_i}}F=\bigcup_{\substack{F\in\mathcal F \\ F\subset\bigcup_{i\in I} U_i}}F$$ since $(U_i)_{i\in I}\subset O \implies \bigcup_{i\in I}U_i\subset O$. Can I now conclude that $$\mu(\bigcup_{i\in I}U_i)=\mu(\bigcup_{\substack{F\in\mathcal F \\ F\subset\bigcup_{i\in I} U_i}}F)=\sum_{\substack{F\in\mathcal F \\ F\subset\bigcup_{i\in I} U_i}}\mu(F)=0 \ \ ?$$ Thanks in advance!
In a space with a countable basis any union $\cup_{i\in I} U_i$ can be written as $\cup_{i\in } U_{i_n}$ for some $i_1,i_2,...$. Proof: if $x \in \cup_{i\in I} U_i$ then there exists $i$ such that $x \in U_i$. There is a member of the basis $V$ such that $x \in V \subset U_i$. The collection of all $V$'s that arise this way is countable and each of then is contained in some $U_i$. This way we get a countable number of $U_i$'s whose union includes every $x \in \cup_{i\in I} U_i$.