Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 31 / 4 / i / Solution

For a Borel probability measure $\mu$ on a Polish space $E$, its support of a measure isEquivalently, for a compatible metric, every open ball about $x$ has positive measure. The complement is open: any zero-measure open neighborhood of a point is a zero-measure neighborhood of all its points. Hence the support is closed.

A Polish space has a countable base. Its support complement is therefore a countable union of zero-measure basic open sets, so $\mu (\mathrm{supp}\mu )=1$. If $F$ is any closed full-measure set, a point outside $F$ has the zero-measure open neighborhood $E\setminus F$ and is outside the support. Thus the support is also the smallest closed set of full measure. Countability is what justifies the full-measure assertion; an arbitrary uncountable union of null sets would not suffice.