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.
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. 


Not every locally compact semigroup admits a (locally finite) leftinvariant 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 $\mu(S)=\mu(aS)=\mu(bS)$. 


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. Update: The result I was trying to recall is from http://www.ams.org/journals/proc/196601702/S00029939196601883417/S00029939196601883417.pdf. 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. 


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 onesided 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 nonstrict 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. 

