We know for the Borel $\sigma$-algebra that each singleton set is measurable. I was working on the problem of proving that each infinite $\sigma$-algebra has uncountably many members. My solution went something like this: if $\sigma$ were countable, intersect all of the measurable sets which contain $x$. I derive a contradiction from this. Of course, $\sigma$-algebras aren't necessarily closed under arbitrary intersection, so it might not make sense to talk about "smallest measurable set" containing a point. My question is what conditions can we put on the measure space so that it does make sense to talk of the smallest measurable set containing a point?