MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T$ be a commutative monoid, written additively. The set $T$ is equipped with a canonical pre-order, defined by $s \le t$ when there exists $s' \in T$ so that $s + s' = t$. Consequently, $T$ may be equipped with the specialization topology for this pre-order, where the closed sets are those which are downward-closed. Note that $T$ is typically not Hausdorff, since the closure of a singleton is its down-set: $\overline{\{t\}} = t\!\downarrow\, := \{ s : s \le t \}$.

Let $\mathcal B(T)$ denote the Borel $\sigma$-algebra with respect to this topology. In this way, every commutative monoid is canonically a measurable space.

Equipped with the $\sigma$-algebra $\mathcal B(T)$, does every commutative monoid $T$ admit a (non-trivial) family of translation-invariant measures?

share|cite|improve this question
If $T$ is a group, this is not a very interesting topology... – Daniel Litt Feb 12 '13 at 19:48
Of course not. That's why I said commutative monoid. – Tom LaGatta Feb 12 '13 at 23:35
up vote 4 down vote accepted

I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. Thus the weight of n is the weight of n plus the weights of all numbers less than n. Thus the weights of all numbers strictly less than n are zero. Since n is arbitrary all elements have weight 0.

share|cite|improve this answer
Why the downvote? It seems ok. – Benjamin Steinberg Feb 13 '13 at 16:32
That's odd that someone would downvote this. It's a great counterexample, and your argument is elegant. It's going to take me some time to reformulate my question so as to circumvent such obstructions. Thanks very much, @Benjamin Steinberg. – Tom LaGatta Feb 13 '13 at 19:38

If in the finer interval topology the monoid is locally compact, I suspect by the same construction that it admits a 'Haar' measure. Moreover, as the Borel sigma algebras generated by these topologies are identical, in this case yes you get a family of invariant measures parameterized by '$\mathbb{R}^+$'.

share|cite|improve this answer
That's a good point. – Tom LaGatta Feb 12 '13 at 23:36
Benjamin's example is locally compact, no? – Mariano Suárez-Alvarez Feb 13 '13 at 6:29
My example last night was wrong. I have a new example. Namely Z with max. The interval topology here is still locally compact since the order is the usual one. – Benjamin Steinberg Feb 13 '13 at 13:06
RE: Benjamin. Let's break this down in detail. Let $\Gamma := \mathbb{Z}$ (so I can keep track of the action versus the space). Then under the specialization or the interval topology, the Borel sigma algebra $\mathcal{B}_{\mathbb{Z}} := \mathcal{B}$ as you pointed out contains sigletons and hence $\mathcal{B} = 2^{\mathbb{Z}}$. Let $\Gamma$ act on $\mathbb{Z}$ by max e.g. for $\gamma \in \Gamma$ define $\gamma(m) = max(\gamma,m)$. So, $\gamma^{-1}(\{n\}) = \{m \in \mathbb{Z}: max(\gamma,m) = n \} = \{n\}$ if $\gamma < n$ ; $\{m \leq n\}$ if $\gamma = n$; $\varnothing$ if $\gamma > n$. – Tyler Bryson Feb 13 '13 at 18:27
That's slightly cleaner than how I wrote it but yes. – Benjamin Steinberg Feb 13 '13 at 19:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.