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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.

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    $\begingroup$ 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. $\endgroup$ – YCor Jan 26 at 21:30
  • $\begingroup$ Thanks. Of course i meant non-empty. $\endgroup$ – Paul Pfeiffer Jan 26 at 21:46
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    $\begingroup$ perhaps one can ask a similar question, further requiring the measure to have no atoms (assuming no isolated points)? $\endgroup$ – erz Jan 27 at 6:13
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    $\begingroup$ @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. $\endgroup$ – YCor Jan 28 at 5:28
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    $\begingroup$ 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 $A+\mathbf{Q}$). $\endgroup$ – YCor Jan 28 at 7:12
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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.

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

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  • $\begingroup$ Reference: Hebert, Lacey, On supports of regular Borel measures, Pacific JM 27(1), 1968. $\endgroup$ – YCor Jan 29 at 4:32
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    $\begingroup$ 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 $\mu = \sum 2^{-n} \mu_n$. $\endgroup$ – Nate Eldredge Jan 29 at 5:07
  • $\begingroup$ Having lots of Cantor sets inside the space seems enough. As @NateEldredge suggested, one in each open set will get the support property. $\endgroup$ – Henno Brandsma Jan 29 at 17:26
  • $\begingroup$ 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. $\endgroup$ – Paul Pfeiffer Feb 4 at 2:40

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