= Solution
\b[Yes.] The preceding calculation established $\mathbb E L_\tau=-2ab/(b-a)<\infty$. For every $n$, the stopped <martingale> satisfies
$$
|M_{n\wedge\tau}|
\leq |S_{n\wedge\tau}|+L_{n\wedge\tau}
\leq C+L_\tau=:H,
\qquad C=\max\{-a,b\}.
$$
Here $H$ is a single nonnegative <integrable random variable> that bounds the whole family. Thus, for any $K>0$,
$$
\sup_n\mathbb E\!\left[|M_{n\wedge\tau}|\mathbf1_{\{|M_{n\wedge\tau}|>K\}}\right]
\leq\mathbb E[H\mathbf1_{\{H>K\}}]\longrightarrow0.
$$
The last limit follows from the <dominated convergence theorem>. This is precisely <uniform integrability>. It also gives $M_{n\wedge\tau}\to M_\tau=|S_\tau|-L_\tau$ with <convergence in L1>, and therefore $\mathbb EM_\tau=0$, consistently with the preceding answer. A common bound on the individual <expected values> alone would not have sufficed; the <integrable> dominating <random variable> is what supplies control of every tail.
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