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.

  • $\begingroup$ To avoid trivialities, you probably want to assume that $\mu$ has full support. $\endgroup$ – Nate Eldredge Apr 3 '15 at 20:10
  • $\begingroup$ Can't you do something silly like the following? Take, say, $[0,1]$ with Euclidean distance and Lebesgue measure, and let $X \subset [0,1]$ be a Borel subset of full measure which is not $G_\delta$, hence not completely metrizable. Then let $\mu$ be the restriction of Lebesgue measure to $X$. $\endgroup$ – Nate Eldredge Apr 3 '15 at 20:17
  • $\begingroup$ For instance, let $K_n \subset [0,1]$ be a fat Cantor set which is closed, nowhere dense, and has Lebesgue measure at least $1-1/n$. Then let $X = \bigcup_n K_n$. $X$ is Borel and has full measure, hence $X$ is dense in $[0,1]$, but $X$ is also meager in $[0,1]$ so it cannot be $G_\delta$. $\endgroup$ – Nate Eldredge Apr 3 '15 at 20:29
  • $\begingroup$ Given that you've answered my question I wouldn't call that silly at all! I'd thought of such a construction (taking a non-$G_\delta$ Borel set of full measure) but wasn't sure how to construct one, so clearly my basic measure theory needs some work. Feel free to submit your example as a solution if you want some internet points. $\endgroup$ – user14166 Apr 4 '15 at 18:55

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy