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](http://ncatlab.org/nlab/show/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?



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


