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I'm reading on Haar measure and we know that every locally compact group admits a Haar measure, is the same true for semigroups? if not, is there a class of semigroups that admits a Haar measure?

Thank you for the help.

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up vote 12 down vote accepted

Not every locally compact semigroup admits a (locally finite) left-invariant measure. In fact, this has nothing to do with any sort of analytic technicalities and already fails for finite semigroups. For example, consider $S=\{a,b\}$ with $ab=a^2=a$ and $ba=b^2=b$. Then $aS=\{a\}$ and $bS=\{b\}$, so no (finite) measure on $S$ can have $\mu(S)=\mu(aS)=\mu(bS)$.

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If memory serves a locally compact semigroup admits a left invariant Borel measure iff it has a minimal ideal which is a direct product of a group and a right zero semigroup.

The measure is supported on the minimal ideal.


The result I was trying to recall is from

A locally finite Borel measure $\mu$ on a locally compact inverse semigroup $S$ is said to be left invariant if $\mu(s^{-1}B)=\mu(B)$ for all Borel sets $B$ where $s^{-1}B=\{x\in S\mid sx\in B\}$.

It is proved in the above paper that if $S$ is locally compact and left translations are proper (i.e. $s^{-1}K$ is compact when $K$ is compact), then $S$ has a left invariant measure iff it has a unique minimal left ideal. This is the same as saying it has a minimal ideal which is a direct product of a group and a right zero semigroup. You can take as the measure the product of Haar measure on the semigroup with your favorite locally finite Borel measure on the right zero semigroup.

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What about the positive reals with usual addition and usual topology? – Yemon Choi Sep 3 '14 at 23:50
Maybe this is the result for compact semigroups. Also invariant measure for a semigroup means the measure of the inverse image of a left translation is the same. – Benjamin Steinberg Sep 4 '14 at 0:38
@YemonChoi, your example is not a counterexample because preimages of Borel sets can change measure. – Benjamin Steinberg Sep 4 '14 at 1:21
My confusion was that I thought left invariant meant that images under translation had the same measure, rather than inverse images – Yemon Choi Sep 4 '14 at 2:52
@YemonChoi, people have also studied your variant of an invariant measure. I seem to recall that under minor hypotheses such a measure of global support exists when precisely if you embed in a group, but I am not sure. – Benjamin Steinberg Sep 4 '14 at 10:56

The natural generalization of Haar measure onto semigroups doesn't feel harmonious (as even the material already provided in this thread seems to indicate). Thus let me introduce two other notions: strict and liberal. When a semigroup is actually a group then its liberal measures (henceforth the strict too) are invariant, as seen below. By invariance I mean a one-sided invariance (I don't know which side though since I am ambidextrous :-)

NOTATION Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ c\in G.\ $ Then let $\ m_c : G\rightarrow G\ $ be defined by: $\ m_c(x) = x\cdot c\ $ for every $\ x\in G$.
Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ \mu\ $ be a $\sigma$-measure in $G$ (with its respective $\sigma$-algebra). Measure $\mu$ is called liberal $\ \Leftarrow:\Rightarrow\ $ for every measurable set $\ A\subseteq G\ $ such that $\ m_c^{-1}(m_c(A))\ = A\ $, and for every $\ c\in G,\ $ there exists a measurable $\ B\subseteq G\ $ such that $\ m_c(A)\subseteq B\ $ and $\ \mu(A)\ge\mu(B)$.

Also, measure $\ \mu\ $ is called strict $\ \Leftarrow:\Rightarrow\ $ for every measurable $\ A\subseteq G\ $ there exists a measurable $\ B\subseteq G\ $ such that $m_c(A)\subseteq B\ $ and $\ \mu(A)\ge \mu(B)$.

Thus every strict measure is liberal is strict (while there are simple examples of non-strict liberal measures). Nevertheless, even in the liberal case, they are all invariant in the case of any group. Next, in the case of finite semigroups, cardinality is a strict measure. It'd be awesome (:-) to connect the general liberal and strict measures with the structure of finite semigroups. And in the case of (infinite) topological semigroups we would get a lot to think about it seems.

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measure $μ$ is called strict, if for every measurable $A⊆G$ there exists a measurable $B⊆G$ such that $mc (A)⊆B$ and $ μ(A)≥μ(B)$? for any $c$? Or for some $c$? – David Chan May 18 '15 at 12:19

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