# Is there a second countable topological space, which can not be equipped with a finite borel measure of full support?

If I have a second countable topological space X, can i Always find a finite borel measure, such that every non-empty open set has positive measure?

without second countability, the discrete topology on $$\mathbb R$$ is a counter example.

• Yes (if you mean "every nonempty open subset"), and actually on any separable topological space. Indeed, assuming $X$ nonempty, consider a dense countable subset with a fully supported discrete probability measure: this defines a measure on all subsets, which is positive on nonempty open subsets. – YCor Jan 26 at 21:30
• Thanks. Of course i meant non-empty. – Paul Pfeiffer Jan 26 at 21:46
• perhaps one can ask a similar question, further requiring the measure to have no atoms (assuming no isolated points)? – erz Jan 27 at 6:13
• @erz again no: $X$ can be countable with no isolated point, such as $\mathbf{Q}$ with the topology of inclusion into $\mathbf{R}$, so all measures are atomic. So one should at least assume that $X$ has no nonempty countable open subset. – YCor Jan 28 at 5:28
• In this article by Gardner and Gruenhage (ams.org/journals/proc/1981-081-04/S0002-9939-1981-0601743-5/…), it is notably proved that it is consistent that every Borel probability measure on a metrizable space of cardinal $\aleph_1$ is atomic (of course this is incompatible with the continuum hypothesis). We can indeed find in $\mathbf{R}$ subsets of cardinal $\aleph_1$ in which every nonempty open subset has cardinal $\aleph_1$ (pick $A\subset\mathbf{R}$ of cardinal $\aleph_1$ and consider $A+\mathbf{Q}$). – YCor Jan 28 at 7:12

A simple solution: if $$X$$ is second countable, let $$D=\{d_n : n =1,2,3,\ldots\}$$ be a dense subset of $$X$$ and define $$\mu(A)= \sum_{n:d_n \in A}\frac{1}{2^n}$$ for all subsets of $$X$$.

Then clearly $$\mu(X)=1$$ and $$\mu(O)>0$$ for all $$O$$ non-empty and open.

If you want an atomless measure, we need at least that $$X$$ is crowded, and then we must maybe assume some more on $$X$$, e.g. to avoid cases like $$\mathbb{Q}$$ which is second countable and crowded but all of whose measures are atomic by countability of the space.

• So, you used: 2nd countable implies separable to prove a more general result. – Gerald Edgar Jan 29 at 12:02
• @GeraldEdgar I could have sufficed with separable indeed, but the OP asked for second countable. – Henno Brandsma Jan 29 at 17:24

See Corollary 2.8 in this paper:

If $$X$$ is perfect, compact and metrizable, then there is a non-atomic regular Borel measure of full support on $$X$$.

• Reference: Hebert, Lacey, On supports of regular Borel measures, Pacific JM 27(1), 1968. – YCor Jan 29 at 4:32
• Compact metrizable seems awfully strong. What if $X$ is only Polish and perfect? It seems like the following would work: take a countable base $U_n$, find a Cantor set $C_n$ inside each $U_n$, and let $\mu_n$ be Cantor measure on $C_n$. Then let $\mu = \sum 2^{-n} \mu_n$. – Nate Eldredge Jan 29 at 5:07
• Having lots of Cantor sets inside the space seems enough. As @NateEldredge suggested, one in each open set will get the support property. – Henno Brandsma Jan 29 at 17:26
• The "compact" assumption can be weakened to countable union of compact subsets, as any $\sigma$-finite measure can be rescaled to a finite measure with the same support. – Paul Pfeiffer Feb 4 at 2:40