nonatomic, sigma-finite measure on polish spaces I'm trying to prove the following conjecture:
Claim. For every uncountable Polish space $(X, \tau)$, there exists a $\sigma$-finite measure $\mu$ on $\mathcal{B}(\tau)$ (the Borel-sigma-algebra generated by $\tau$) that assigns positive measure to every non-empty open set.
I think this is true, but don't quite know how to prove it. In particular, I know that every Polish space contains a homeomorphic to a $G_\delta$ set of the Hilbert cube $[0,1]^\omega$. So one idea is to employ the homeomorphism. But there is no guarantee that the $G_\delta$ set has positive measure under e.g.~the product Lebesgue measure on the Hilbert cube. Another attempt is to try to utilize the Borel isomorphism between $(X, \mathcal{B}(\tau))$ and the unit interval. But it seems in that case one has to pick the isomorphism carefully to ensure that the open sets are mapped to open sets. I think that can be done but am fuzzy on the details. Any hint (including suggestion that the conjecture is wrong) would be greatly appreciated.
 A: Here is a solution that yields a non-atomic Borel measure of full support,  assuming that $X$ has no isolated points.
Lemma. Any uncountable Polish space has a non-atomic Borel probability measure.
Proof.  It is known that any uncountable Polish space is Borel-isomorphic to $[0,1]$, so it suffices to move
Lebesgue's measure over by such an isomorphism. QED
Theorem. Any Polish space $X$ without isolated points admits a non-atomic Borel probability measure of full support.
Proof.
Let $\{U_n\}_n$ be a countable basis for the topology of $X$.   Viewed as a subspace, each $U_n$ is also a Polish space.
Observe that $U_n$ cannot be countable or otherwise by Baire it must have isolated points which would also be isolated
points of $X$.  Therefore $U_n$ is uncountable and hence by the Lemma there is a non-atomic Borel probability measure
$\mu _n$ on $U_n$,  which we view as a measure on $X$ by setting $\mu _n(X\setminus U_n)=0$.
To conclude it is then enough to take
$$
  \mu  = \sum_{n=1}^\infty  2^{-n}\mu _n.
  $$
