Positive Borel measure with empty support on a standard measurable space

Question

Let $X$ be a $T_0$ topological space (not $T_1$), and let $\Sigma_X$ be the Borel $\sigma$-algebra. Assume that $(X,\Sigma_X)$ is a standard measurable space, i.e., measurably isomorphic to the Borel $\sigma$-algebra $(Y,\Sigma_Y)$ of a complete separable metric space $Y$.

Question (amended to honour Dieter Kadelka): Is it true that there does NOT exist a positive measure $\mu$ on $(X,\Sigma_X)$ with empty support, $\mathrm{supp}\mu=\emptyset$?

Note that the measurable isomorphism between $X$ and $Y$ is a priori merely Borel measurable, and can potentially map open sets to sets without interior and vice versa.

Thank you.

Discussion: The question arises from the discrepancy in definitions of the support of a Borel measure on a topological space. I think that the standard definition is this, in which case, unless $X$ is a very good space, you don't have to expect $\mu(X\setminus\mathrm{supp}\mu)=0$. However, in Propositiom 8.6.8 in Dixmier's "$C^*$-algebras", the author defines the support of a measure as the smallest closed subset with negligible complement (the standard analysis definition). Now I wonder if the two definitions coincide in this context. Note that the space in question is only $T_0$ in general, possesses a dense locally compact open subset (Proposition 4.4.5 in the book), and the Borel structure is measurable isomorphic to that of a complete separable metric space (Proposition 4.6.1 in the same book).

Discussion 2: Well, there is more to the spectrum of a separable postliminal $C^*$-algebra - it is locally qusicompact, second countable etc. But the question remains as it is: does the mere Borel isomorphism to a Polish space rule out empty support?

Answer 1

An example of a $T_0$-space which is not $T_1$ is $\mathbb{R}$ with the right order topology $\tau$ (Steen/Seebach: Counterexamples in Topology, 50). The topology $\tau$ is coarser than the usual topology $\tau_1$ on $\mathbb{R}$, and both have the same Borel-$\sigma$-algebra. Since any open neighbourhood of $x \in \mathbb{R}$ w.r.t. $\tau$ is also an open neighbourhood w.r.t $\tau_1$, the support of any $\mu \not= 0$ is not empty (w.r.t. $\tau$). So I think your question should be "Is there a $T_0$ space $X$ and a Borel-measure $\mu$ with the properties in the question".

Answer 2

There exists a counterexample under the Continuum Hypothesis, which implies that the unit interval $[0,1]$ admits a well-order $\preceq$ such that for every $x\in[0,1]$ the initial interval ${\downarrow}x=\{y\in [0,1]:y\preceq x\}$ is at most countable. On $[0,1]$ consider the Hausdorff topology $\tau$ generated by the subbase consisting of the sets $[0,a)$, $(a,1]$ and ${\downarrow}a$ where $a\in [0,1]$. It is easy to see that the Borel $\sigma$-algebra generated by the topology $\tau$ coincides with the Borel $\sigma$-algebra generated by the standard topology on $[0,1]$. We claim that the Lebesgue measure on $[0,1]$ has empty support (in the topology $\tau$). This follows from the fact that each point $x\in [0,1]$ has the open neighborhood ${\downarrow}x$ of Lebesgue measure zero.