Are there improvements on Dana Angluin's algorithm for learning regular sets
In his answer  on cstheory.SE, Lev Reyzin directed me to Robert Schapire's thesis that improves the bound to $O(n^2 + n\log m)$ membership queries in section 5.4.5. The number of counterexample queries remains unchanged. The algorithm Schapire uses differs in what it does after a counterexample query.

Sketch of the improvement
At the highest level, Schapire forces $(S,E,T)$ from Angluin's algorithm to have the extra condition that for a closed $(S,E,T)$ and each $s_1, s_2 \in S$ if $s_1 \neq s_2$ then $row(s_1) \neq row(s_2)$. This guarantees that $|S| \leq n$ and also makes the consistency property of Angluin's algorithm trivial to satisfy. To ensure this, he has to handle the results of a counterexample differently.
Given a counterexample $z$, Angluin simply added $z$ and all its prefixes to $S$. Schapire does something more subtle by instead adding a single element $e$ to $E$. This new $e$ will make $(S,E,T)$ to be not closed in Angluin's sense and the update to get closure with introduce at least one new string to $S$ while keeping all rows distinct. The condition on $e$ is:
$$\exists s, s' \in S, a \in \Sigma \quad \text{s.t} \quad row(s) = row(s'a) \; \text{and} \; o(\delta(q_0,se)) \neq o(\delta(q_0,s'ae))$$
Where $o$ is the output function, $q_0$ is the initial state, and $\delta$ the update rule of the true 'unknown' DFA. In otherwords, $e$ must serve as a witness to distinguish the future of $s$ from $s'a$.
To figure out this $e$ from $z$ we do a binary search to figure out a substring $r_i$ such that $z = p_ir_i$ and $0 \leq |p_i| = i &lt; |z|$ such that the behavior of our conjectured-machine differs based on one input character. In more detail, we let $s_i$ be the string corresponding to the state reached in our conjectured machine by following $p_i$. We use binary search (this is where the $\log m$ comes from) to find an $k$ such that $o(\delta(q_0,s_kr_k)) \neq o(\delta(q_0,s_{k+1}r_{k+1})$. In other words, $r_{k+1}$ distinguishes two states that our conjectured machines finds equivalent and thus satisfies the condition on $e$, so we add it to $E$.