Definition of Measurable Space: An ordered pair $(\Omega, \mathcal{F})$ is a measurable space if $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$.

Definition of Measure: Let $(\Omega, F)$ be a measurable space, $μ$ is an non-negative function defined on $\mathcal{F}$ (that is $\mu: \mathcal{F} \to [0, +\infty]$). If $\mu(\emptyset) = 0$ and $\mu$ is countably additive (that is $A_n \in \mathcal{F}$, $n \geqslant 1$, $A_n \cap A_m = \emptyset$, $n \neq m \Rightarrow \mu(\cup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu(A_n)$) then $\mu$ is a measure on $(\Omega, \mathcal{F})$.

Definition of Measure Space: Let μ is a measure on $(\Omega, \mathcal{F})$ then $(\Omega, \mathcal{F}, \mu)$ is a measure space.

Definition of Metric Space: A metric space is an ordered pair $(M,d)$ where $M$ is a set and $d$ is a metric on $M$, i.e., a function $$d \colon M \times M \rightarrow \mathbb{R}$$ such that for any $x, y, z \in M$, the following holds:

  1. $d(x,y) \ge 0$ (non-negative),

  2. $d(x,y) = 0\, \iff x = y\ $, (identity of indiscernibles),

  3. $d(x,y) = d(y,x)\ $, (symmetry),

  4. $d(x,z) \le d(x,y) + d(y,z)$ (triangle inequality).

Is a measure space $(\Omega, \mathcal{F}, \mu)$ necessarily a metric space? What's the relationship between them?

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    $\begingroup$ There is no relationship between these two notions. $\endgroup$ – Crostul Aug 19 '15 at 15:46
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    $\begingroup$ @Crostul: sad story o(╯□╰)o $\endgroup$ – Bear and bunny Aug 19 '15 at 15:50
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    $\begingroup$ You can always define a Borel measure in a metric space using the sigma-algebra generated by open sets. $\endgroup$ – Gary. Aug 19 '15 at 15:52
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    $\begingroup$ @Bear and bunny : Yes, in a metric space $(X,d)$, the topology is generated by open balls $B(x,r)$ , meaning that a set $S$ is open in $X$ if for every $x$ in $S$ there is a ball $B(x,r); r>0$ so that $B(x,r) \subset S$. $\endgroup$ – Gary. Aug 19 '15 at 15:58
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    $\begingroup$ @Bear and bunny: If you mean that there is no intrinsic measure assigned to a space, then this is correct. And it is an interesting question whether fa measure space can be seen as the sigma-algebra of a metric space defined on the underlying set. Is that what you meant? $\endgroup$ – Gary. Aug 19 '15 at 16:07

The best I can think of, are: Given a metric space $(X,d)$ , we can assign sigma-algebras.

1) Borel Measure: This is the sigma algebra generated by the open sets generated by the open balls in the metric.

Or, 2)Looking at some of the linked questions are that of a Hausdorff measure associated with a metric space:


But yours is an interesting question: given a measure triple (X, A, $\mu$), where $A$ is a sigma algebra and $X$ is the underlying space, can this be the Borel algebra resulting from a metric space? I don't have a full answer but some obvious requirements are that $X$ must be metric , or at least metrizable. Still, while outside of the scope of your question, one can define measures on non-metric, non-metrizable spaces using the Borel sigma algebra.

  • $\begingroup$ I saw some light. Must upvote it. $\endgroup$ – Bear and bunny Aug 19 '15 at 16:20
  • $\begingroup$ Thanks, let me know of any additional question/comment. $\endgroup$ – Gary. Aug 19 '15 at 16:25

There is a theory of "metric measure spaces" which are metric spaces with a Borel measures, ie., a measure compatible with the topology of the metric space. It has a big literature that is well represented online.

There is also the trivial answer, that if you don't demand compatibility of the measure and the metric, there is no particular relation between them. A wider neighborhood has bigger measure, and easy things like that, but not necessarily anything nontrivial.

  • $\begingroup$ Compatibility....emmm, a good word, make sense. $\endgroup$ – Bear and bunny Aug 19 '15 at 16:40

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