Inseparable complete metric space with full Radon probability measure

Question

I am looking for a simple construction of an inseparable complete metric space that carries a Radon probability measure with full support. A similar question was asked before although not in the completely metrizable case.

Answer 1

The support $S$ of a ($\sigma$-)finite Borel measure has the ccc property (e.g. see the proof of the lemma in pre-kidney's answer, but it's a classic fact) and if we're in a metric space that's equivalent to being separable or second countable or Lindelöf etc.), see the equivalence of 6. and 7. in this answer, e.g.

Answer 2

You will not be able to come up with an example. I will use the definition of the support of a Borel probability measure as the smallest closed set of measure one. If a Borel probability measure has a support, that support equals the complement of the union of all open sets of measure zero (a more permissive definition of support). Moreover, a probability measure is supported on every measurable set of measure $1$.

The following can be found in the appendix of Billingsley's (1968) "Convergence of probability measures" (not included in later editions):

Theorem: There exists a Borel probability measure supported on a nonseparable set in a metric space $(M,d)$ if and only if there is a discrete subspace that admits an atomless Borel probability measure.

Since a Borel probability measure on a discrete subspace is defined on all subsets, a metric space supported only on nonseparable subsets exists if and only if there exists a set that admits an atomless probability measure on its power set. It is consistent with the usual axioms of set theory (ZFC) that no such set exists. The relevant keywords are "real-valued measurable cardinal" and "atomlessly measurable cardinal."

Now a compact subset of a discrete metric space must be finite. If the Borel probability measure is Radon and therefore inner regular with respect to compact sets, it must have countable support. So even if a Borel probability measure supported only on nonseparable subsets exists on some metric space, it will never be Radon. Note that every Radon measure does have a support. Indeed, if the union of all open sets of measure zero would have positive measure, so would, by inner regularity, a compact subset. But that compact subset would be covered by finitely many open sets of measure zero.

Answer 3

Lemma 2.1 in this paper of Lawson provides a simple and self-contained proof that every finite Borel measure on a metric space has separable support. In particular this implies a negative answer to the question above.

As pointed out elsewhere in this discussion, some subtleties arise involving the support. To be clear, Lemma 2.1 applies to the set of points whose neighborhoods all have positive measure - which is easily seen to be a closed set, and in particular Borel-measurable. It may, however, have less than full measure - and this case is explored in Proposition 2.3 of the same paper.