construct a counterexample in measure theory

Question

Problem: Construct a $\sigma$-algebra $\mathscr{F}$ of subsets of $R$ such that no open interval is measurable with respect to $\mathscr{F}$, although any singleton $\{x\}$ is ($x\in R$). I tried to construct an example like the complement of any open interval is not in $\mathscr{F}$, but I did not make it. Can someone give a hint?

Answer 1

Consider the collection $\mathscr{F}$ where $A \in \mathscr{F}$ if $A$ is either empty, at most countable, or has a complement which is at most countable. This is a $\sigma$-algebra on $\mathbb{R}$ where no interval is contained in it (unless you consider $\mathbb{R}$ itself which needs to be in there anyway).