Let $X$ be a separable metric space and $B(X)$ the Borel Sigma algebra. In addition, we have $m$ as a measure on the measurable space $(X, B(X))$.
If $Z = \{w$ open : $m(w)=0\}$. Why is Z NOT empty (ok, because the measure is 0, right?) and why does Z admit a maximal element (with respect to set inclusion of course)?
Let $U$ be the union of all open sets of measure $0$. (The empty set is open and has measure $0$ so the collection $Z$ is not empty). Since $X$ is separable we can write $U$ as a countable union of sets of measure $0$ and hence $U$ itself has measure $0$. It is obviously the largest open set of measure $0$. [ Any seprable metric space is second countable; any union of open sets can be expressed as the union of a countable subfamily].