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On the real line the absolute value plays a dual role: a metric (by letting $d(x,y)=|x-y|$ and the seed of a measure (by letting the measure of an interval between $a$ and $b$ be the distance between $a$ and $b$ and then extending to the Borel $\sigma$-algebra using standard techniques). I'm wondering how close this relationship is in general.

I think a reasonable setting for this question is a linearly ordered topological space $X$ (meaning that its topology is generated by the open intervals). We can consider two additional structures on $X$:

  1. Suppose $\mu$ is a measure defined on all Borel subsets of $X$ and assume that the measure of every nonempty interval in $X$ is positive. We can then define $d(x,y)$ to be the measure $\mu((x,y))$ of the open interval between $x$ and $y$ (provided that $x\leq y$, otherwise we swap them). Unless I made a mistake, it is straightforward to check that $d$ is a metric.

Question 1: Is $d$ compatible with the existing topology on $X$?

  1. Suppose that $X$ is metrizable with metric $d$. We can define a pre-measure $\nu$ on the ring generated by the half-open intervals $[x,y)$ by letting $\nu([x,y))=d(x,y)$ (provided that $x\leq y$). By Carathéodory's extension theorem, $\nu$ extends to a measure $\mu$ on the Borel $\sigma$-algebra of $X$.

Question 2: How nice is the measure $\mu$? Is it Radon (maybe assuming something about $X$, like local compactness)?

Question 3: If Question 1 has a positive answer, are these two definitions inverses of one another (i.e. if I start with a Borel measure, derive a metric, and then derive a measure from that, will I recover the original measure, and vice versa)?

It seems plausible to me that going from a metric to a measure and back to a metric will give the original metric back. I'm less sure about starting with a measure since Carathéodory's theorem doesn't give uniqueness without extra assumptions.

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  • $\begingroup$ This question is related but a bit more general than what I'm looking for. $\endgroup$ Commented Oct 15 at 16:49
  • $\begingroup$ Do you really have a pre-measure in part 2? It looks like $\nu$ is subadditive, but not necessarily additive. $\endgroup$ Commented Oct 15 at 17:31
  • $\begingroup$ @ChrisCulter A pre-measure is supposed to be additive for unions of pairwise disjoint sets/intervals, no? So can't we simply define the pre-measure of a disjoint union of intervals to be the sum of the pre-measures of the individual intervals? $\endgroup$ Commented Oct 15 at 18:02
  • $\begingroup$ I think in the setting of 1. it should be true that $X$ is isomorphic (as a linear order, and as a measure space) with a subset of the reals. $\endgroup$ Commented Oct 15 at 20:20

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Here's a problem with part 2. Let $X=\mathbb R$, the real numbers with the usual ordering and the usual topology. Let $\bar d(x,y)=\min(|x-y|, 1)$ be the standard bounded metric on $\mathbb R$. You can check that the metric topology generated by $\bar d$ is the same as the usual topology; that's Theorem 20.1 in Munkres 2nd edition. So $(\mathbb R, \bar d)$ is a valid input that your theory ought to handle.

However, the proposed $\nu$ has $\nu([0,1))=\nu([1,2))=\nu([0,2))=1$, which means $\nu([0,1))+\nu([1,2))\neq\nu([0,2))$, so $\nu$ is not a pre-measure.

In addition to the counterexample of $\bar d$, it's also worth considering how to handle:

  • The snowflake metrics $d_\alpha(x,y)=|x-y|^\alpha$ on $\mathbb R$ for $\alpha\in(0,1)$.
  • The ultrametric on the Cantor space $2^\omega$.

(I'm not claiming that the basic idea is infeasible. I'm just pointing out that the $\nu$ doesn't work as stated. You might need stronger hypotheses, or a more elaborate construction of $\nu$, or both.)

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