In my opinon the Language recognised is this:$(a + b)^* (ab)^$but the solution provided is: $(a + b)^* (a + b)^$Did I missunderstand the acceptance condition or is the solution wrong?My reasoning is:As far as I understood a Muller Automaton does accept a run if all the set of all infinitely often visited states can be found as acceptance condition.E.g. if a run visits inf. often the states p & r and finitely often q then the set { p , r } has to be one of the set's of the acceptance condition.I think in this case, that a word has to bounce inf. often between 2 of the 3 states (but is not allowed to visit the 3rd state inf. often). That means that in the end the word has to alternate between a & b inf. often after doing finitely often whatever it want's to do=> $(a + b)^*$ ( = do what ever you want to do finitly often and in the end) $(ab)^$ (= iterate between a and b)The given solution differs from my solution by accepting e.g. the following word:$aba^$ 
Which language is accepted by this Muller Automaton?
automata;regular expressions;buchi automata
The solution given is incorrect.The first clue that this might be the case is that $(a+b)^*(a+b)^\omega$ is a strange way to write anything: any finite number of $a$s and/or $b$s, followed by infinitely many $a$s and/or $b$s is just infinitely many $a$s and/or $b$s, so the given language is just a strange way of writing $(a+b)^\omega$.Muller automata accept if the set of states visited infinitely often is one of the specified sets.To see that the automaton doesn't accept $(a+b)^\omega$, consider the word $a^\omega$. This is a member of $(a+b)^\omega$ but it's not accepted by the automaton, since the run on that word visits every state infinitely often, and $\{p,q,r\}$ is not an accepting set.If exactly two of the automaton's states are visited infinitely often, then there must be a point at which the third state is never visited again. From that point onwards, the automaton can only alternate between the remaining two states, so any accepted word must end $(ab)^\omega$. While the automaton is still visiting all three states, it can read in any word so, yes, the accepted language is $(a+b)(ab)^\omega$.