Existence of doubling non-Polish metric measure spaces Let $(X,d,\mu)$ be a metric measure space (i.e. $(X,d)$ is a metric space and $\mu$ is a Borel measure on $X$). Let's say that $X$ is doubling if there exists a constant $C \geq 1$ such that $0 < \mu(B(x,2r)) \leq C\mu(B(x,r)) < \infty$ for all $x \in X$ and $r > 0$ (note the finiteness and positivity conditions which don't always appear in this definition).
The metric space $(X,d)$ is Polish if it is separable and completely metrisable (i.e. there exists a complete metric, not necessarily $d$, which determines the same topology).
Are there any examples of metric measure spaces $(X,d,\mu)$ which are doubling but not Polish?
One can prove that if $\mu(B(x,r)) \in (0,\infty)$ for all $x \in X$ and $r > 0$, then $(X,d)$ is separable, so we only need to concern ourselves with complete metrisability. (I haven't seen this in the literature anywhere, so I have no idea if this is already known, but I have a proof of my own.)
My approach to this problem was to take a non-Polish metric space which is geometrically doubling (every ball can be covered by uniformly finitely many balls of half the radius) and invoke the theorem which guarantees the existence of a doubling measure (Theorem 13.3 in Heinonen's Lectures on Analysis on Metric Spaces). But alas, a key hypothesis of this theorem is that $(X,d)$ is complete (hence Polish), so this doesn't work.
 A: Expanding comment to answer: here's an example that is perhaps somewhat trivial but does achieve what you ask.
Let $(X_0, d, \mu_0)$ be your favorite doubling metric measure space, and suppose $X \subset X_0$ is a subset which is Borel but not $G_\delta$ and has full measure.  Consider $(X,d)$ as a metric space in its own right, and equip it with the Borel measure $\mu$ which is the restriction of $\mu_0$ to $X$.  (Note that every Borel subset of $(X,d)$ is also a Borel subset of $(X_0, d)$ so this makes sense.)  It is a standard theorem that since $X$ is not $G_\delta$ in $X_0$, the metric space $(X,d)$ is not completely metrizable.  And for each $x \in X$ and $r \ge 0$, we have $\mu(B_X(x,r)) = \mu_0(B_{X_0}(x, r) \cap X) = \mu_0(B_{X_0}(x,r))$, so $\mu$ is doubling on $X$.
For an explicit example, take $X_0 = [0,1]$, $d$ the usual Euclidean distance, and $\mu_0$ to be Lebesgue measure.  For each $n$, let $K_n$ be a fat Cantor set which is compact, nowhere dense, and has $\mu_0(K_n) \ge 1-1/n$.  Set $X = \bigcup_n K_n$.  Then $X$ is Borel and has full measure; in particular $X$ is dense.  But $X$ is also meager, and by the Baire category theorem a dense $G_\delta$ cannot be meager in $[0,1]$.  So $X$ is not $G_{\delta}$.
In this case we can actually prove directly that $X$ is not completely metrizable, without needing to appeal to the "standard theorem" mentioned above.  We can show that $K_n$ is nowhere dense in $X$, meaning that $X$ is meager in itself, which by Baire category is impossible for a completely metrizable space.  Since $K_n$ is compact it is closed in $X$, so we need to show it has empty interior.  Let $U$ be nonempty and open in $X$, so that $U = X \cap U_0$ for some nonempty $U_0$ which is open in $X_0 = [0,1]$.  Now $K_n$ is closed and nowhere dense in $X_0$, so $U_0 \setminus K_n$ is open and nonempty.  Since $X$ is dense, then $\emptyset \ne (U_0 \setminus K_n) \cap X = (U_0 \cap X) \setminus K_n = U \setminus K_n$ so $U$ is not contained in $K_n$.  $U$ was arbitrary, so $K_n$ has empty interior and thus is nowhere dense.
One might ask if we can do this for some wide variety of metric measure spaces $(X_0, \mu_0)$.  Let's say $(X_0, d)$ is an uncountable Polish space and  $\mu_0$ is $\sigma$-finite, Radon, atomless and has full support.  Does there necessarily exist a Borel set $X \subset X_0$ which has full measure and is not $G_\delta$?  I think a simple cardinality argument will show there must be a measurable $X$ having full measure and not $G_\delta$, but we want $X$ to actually be Borel.  I am not sure if this is always possible.
Edit. The answer to this question is yes.  Let $E \subset X_0$ be countable and dense.  Since $\mu_0$ is atomless we have $\mu_0(E) = 0$.  By regularity of $\mu_0$ there is a $G_\delta$ set $G \supset E$ with $\mu_0(G) = 0$.  $G$ is still dense so it is comeager.  Set $X = X_0 \setminus G$ so that $X$ has full measure and is Borel (indeed $F_\sigma$).  Since $\mu_0$ had full support, $X$ is also dense, but $X$ is also meager so by Baire category it cannot be $G_\delta$.
Moreover, since $G$ is comeager it is uncountable.  I believe an uncountable Borel set should contain sets of arbitrarily high Borel rank.  By taking complements we can get full-measure Borel sets of arbitrarily high Borel rank.
