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.



Existence of doubling non-Polish metric measure spaces
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.


