= Solution
Let $C_t$ be the cumulative claim amount and $U_t=u+ct-C_t$ the surplus in the <classical risk model>. The supplied <adjustment coefficient> satisfies $\lambda(M_X(R)-1)=cR$. The <Compound Poisson process> has independent increments, and the <random-sum transform identity> gives
$$
\mathbb E[e^{R(C_t-C_s)}]=\exp((t-s)\lambda(M_X(R)-1)).
$$
It follows that $Z_t=\exp(R(C_t-ct))$ is a nonnegative <martingale> with $Z_0=1$: conditional on the past, the multiplicative increment has <expected value> one. This is the <exponential surplus martingale>.
Let $\tau=\inf\{t\ge0:U_t<0\}$ be the ruin <stopping time>. Surplus decreases only at claim arrivals, so on $\{\tau\le t\}$ the deficit at ruin gives $C_\tau-c\tau>u$ and $Z_\tau>e^{Ru}$. The <optional stopping theorem> at the bounded <stopping time> $\tau\wedge t$ yields
$$
1=\mathbb E Z_{\tau\wedge t}\ge\mathbb E[Z_\tau\mathbf1_{\{\tau\le t\}}]
\ge e^{Ru}\mathbb P(\tau\le t).
$$
As $t$ increases without bound, the events $\{\tau\le t\}$ increase to eventual ruin. Hence the <ultimate ruin probability> satisfies the <Lundberg inequality>
$$
\boxed{\psi(u)=\mathbb P(\tau<\infty)\le e^{-Ru},\qquad u\ge0.}
$$
The proof uses only bounded-time stopping; it does not presume that the <martingale> can be stopped with equality of <expected values> at an unbounded ruin time.
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