Haar Measure on Locally Compact Semigroups 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.
 A: 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/1966-017-02/S0002-9939-1966-0188341-7/S0002-9939-1966-0188341-7.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. 
A: 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 $\mu(S)=\mu(aS)=\mu(bS)$.
A: 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. 
A: Existence of left invariant measures on a semigroup $S$ with definitions of $m(A)=m(xA)$ or $m({y:y \ \text{belongs to}\ xA})$ does not mean that support $m$ is a right group because it could be embedded in a right group, i.e. the direct product of a semigroup embeddable in a group and right nulls semigroup. But even the later description is not equivalent to existence of a left invariant measure on the left cancellable semigroup because it requires Malcev conditions.
