In Ullman's Introduction to Automata, Languages and Computation (1979):
We formally denote a finite automaton by a 5-tuple $(Q, \Sigma, \delta, q_0, F)$, where $Q$ is a finite set of states, $\Sigma$ is a finite input alphabet, $Q_0$ in $Q$ is the initial state, $F \subseteq Q$ is the set of final states, and $\delta$ is the transition function mapping $Q \times \Sigma$ to $Q$. That is, $\delta(q, a)$ is a state for each state $q$ and input symbol $a$.
A Moore machine is a six-tuple $(Q, \Sigma, \Delta, \delta, \lambda, q_o)$, where $Q$, $\Sigma$, $\delta$ and $q_0$ are as in the DFA. $\Delta$ is the output alphabet and $\lambda$ is a mapping from $Q$ to $\Delta$ giving the output associated with each state.
A Mealy machine is also a six-tuple $M = (Q, \Sigma, \Delta, \delta, \lambda, q_o)$, where all is as in the Moore machine, except that $\lambda$ maps $Q \times \Sigma$ to $\Delta$. That is, $\lambda(q, a)$ gives the output associated with the transition from state $q$ on input $a$.
Why do both Moore machines and Mealy machines not have a set of final states, like $F$ in a DFA?
Do DFA's with outputs in general not have a set of final states?
Without a set of final states, how do they decide whether to accept an input string?