Nonregular language satisfying the pumping lemma
= Nonregular language satisfying the pumping lemma
Let $K\subseteq\mathbb N$ be undecidable and
$$
L=\{wab^n:w\in\{a,b\}^*,\ n\in K\}\cup b^*.
$$
Pumping its first symbol always stays in $L$, so it satisfies the pumping lemma with pumping length one. Yet $L\cap ab^*=\{ab^n:n\in K\}$ is not regular, because a finite automaton for it would decide $K$.