# Does every commutative monoid admit a translation-invariant measure?

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?

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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

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.

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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 '$\mathbb{R}^+$'.

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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