= Solution
{c}
The first entrance time $\tau_K$ is a <stopping time>, because $\{\tau_K\leq n\}=\bigcup_{j=0}^n\{M_j\leq-K\}\in\mathcal F_n$. Before this time $M_n>-K$; at entrance, the <bounded increments> give $M_{\tau_K}>-K-L$. Thus the <stopped martingale>
$$
N_n=M_{n\wedge\tau_K}+K+L
$$
is a <nonnegative martingale>. Here the <stopped process> is a <martingale> since its increment is $\mathbf1_{\{\tau_K>n\}}(M_{n+1}-M_n)$, whose <conditional expectation> given $\mathcal F_n$ is zero.
The <martingale convergence theorem> states that a <nonnegative martingale> converges <almost surely> to a finite <integrable random variable>; indeed its <expectations> are constant, giving the required bound in the <L1 norm>. Apply this theorem to $(N_n)$. On $\{\tau_K=\infty\}$ the <stopped process> equals the original one at every time. Hence
$$
\boxed{M_n\text{ converges to a finite limit almost surely on }\{\tau_K=\infty\}.}
$$
The overshoot bound is why the assumption of <bounded increments> matters.
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