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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?

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  • $\begingroup$ Please clarify the assumptions. What if $X$ is Polish itself? Are you asking for an example of a $T_0$-space with the above property? $\endgroup$ – Dieter Kadelka Feb 12 at 23:11
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    $\begingroup$ Is it better now? I am asking if such a measure can exist, or if it is excluded. It is excluded if $X$ is good enough, but all I have is $T_0$. $\endgroup$ – Bedovlat Feb 12 at 23:13
  • $\begingroup$ The point is that, if the underlying space is not Hausdorff, a non-zero Borel measure can actually have empty support, which is sad. There is an example at the Wiki page referred to above in my question. Now whether or not that can happen in a standard Borel structure is the question. Thank you. $\endgroup$ – Bedovlat Feb 13 at 2:22
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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".

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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.

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