In the classic Crámer-Lundberg Model $$U_t=u + ct - S_t$$ Defining $$g(r)=\lambda (M_U (r) - 1) - cr$$ where $M_U$ is the moment generating function of $U$ and the adjustment coefficient R is going to be constructed as $g(R)=0$
We kow that $E[e^{-r(U_t - U_s)}]=e^{(t-s)g(r)}$ for $s<t$
And to show that $E(e^{-rU_t - g(r)t})=e^{-rU_s - g(r)s}$ (is a martingale) for $s<t$.
I found in some probability theory lectures online: \begin{align} E(e^{-rU_t - g(r)t}|F_s)&= E(e^{-r(U_t - U_s)})e^{-rU_s - (\lambda(M_U (r) - 1)-cr)t}\\ &= E(e^{-r(U_t - U_s)})e^{-rU_s - \lambda(M_U (r) - 1)-crs}\\ &=e^{-rU_s - g(r)s} \end{align}
Yet I can't see how we do these steps, and the first one doesn't make sense to me at all.
Doing: \begin{align} E(e^{-rU_t - g(r)t}|F_s)&=E(e^{-r(U_t - U_s) - rU_s - g(r)t}|F_s)\\ &= e^{(t-s)g(r)} e^{rU_s} E(e^{-g(r)t}|F_s) \end{align}
I think I get closer to the answer, but I can't show the first step (that they are independent so I can do the product of expectations) and if $E(e^{-g(r)t}|F_s)$ is martingale (which I also can't prove), I would get $E(e^{-g(r)t}|F_s)=e^{-g(r)s}$ and a wrong answer. What am I supposed to do?
first of all, when you say that $M_U$ is the moment generating function of $U$, you are not being precise! $U$ is a stochastic process. What you probably mean is that $M_U$ is the moment generation function of $U_1$. Then we also have that \begin{align}\tag{1} \label{1} \mathbb{E}\bigl[ \exp(-r\cdot U_t)\bigr] = \exp(g(r) \cdot t), \quad \text{for all} \ t \geq 0. \end{align}
Second, the function $t \mapsto e^{-g(r)t}$ does not involve any random variables, hence it is a deterministic function. In particular we have that $$ \mathbb{E}\bigl[ e^{-g(r)t} \mid \mathcal{F}_s \bigr] = \mathbb{E}\bigl[ e^{-g(r)t} \bigr] = e^{-g(r)t}. $$ Also, $t \mapsto e^{-g(r)t}$ is not a martingale because it is a determinisitic function in $t$ which is not constant.
Edit: new part
Properties of the stochastic process $U$
From the definition of $U$ we see that the only random element in it is the process $S$ which is a compound Poisson process. This means that there exists a Poisson process $N$ and an i.i.d. sequence $X$ such that $$ S_t= \sum_{i=1}^{N_t}X_i. $$ Per definition, $N$ has independent and stationary increments. It is then an easy exercise to show that $S$ inherits these properties from $N$, meaning that $S$ also has independent and stationary increments. It then follows that $U$ also has independent and stationary increments.
Martingale part
From here on we use the stationary and independent increments of the Compound Poisson process: these properties are handed on to the process $U$, hence we have that \begin{align} \mathbb{E}\bigl[ \exp(-r\cdot U_t)\mid \mathcal{F}_s \bigr] = \mathbb{E}\Bigl[ \exp\bigl(-r\cdot (U_t-U_s)\bigr) \cdot \exp(-r\cdot U_s)\mid \mathcal{F}_s \Bigr]. \end{align} Now $\exp(-r\cdot U_s)$ is $\mathcal F_s$ measurable hence we can take it out of the conditional expectation. Furthermore, $U_t-U_s$ is independent from $\mathcal{F}_s$, therefore the conditional expectation is equal to the "normal" expectation: we arrive at \begin{align} \mathbb{E}\Bigl[ \exp\bigl(-r\cdot (U_t-U_s)\bigr) \cdot \exp(-r\cdot U_s)\mid \mathcal{F}_s \Bigr] = \mathbb{E}\Bigl[ \exp\bigl(-r\cdot (U_t-U_s)\bigr) \Bigr] \cdot\exp(-r\cdot U_s). \end{align} Next we will use that $U_t-U_s$ has the same distribution as $U_{t-s}$. If two random variables have the same distribution they also have the same expectation, hence we get: \begin{align} \mathbb{E}&\Bigl[ \exp\bigl(-r (U_t-U_s)\bigr) \Bigr] \exp(-r U_s)= \mathbb{E}\Bigl[ \exp\bigl(-r\cdot U_{t-s}\bigr) \Bigr] \exp(-r U_s) \\[6pt] &\stackrel{(\ref{1})}{=}\exp\bigl(g(r) \cdot (t-s)\bigr) \exp(-r U_s). \end{align} To sum up, we have shown that \begin{align} \tag{2} \label{2} \mathbb{E}\bigl[ \exp(-r U_t)\mid \mathcal{F}_s \bigr]=\exp\bigl(g(r) \cdot (t-s)\bigr) \exp(-r U_s). \end{align}
From here it is easy to show the martingale property: \begin{align} \mathbb{E}&\bigl[ \exp\bigl(-r U_t -tg(r)\bigr)\mid \mathcal{F}_s \bigr] = \mathbb{E}\bigl[ \exp(-r U_t)\mid \mathcal{F}_s \bigr]\exp\bigl(- tg(r)\bigr)\\[6pt] & \stackrel{(\ref{2})}{=} \exp\bigl(g(r) \cdot (t-s)\bigr) \exp(-r U_s)\exp\bigl(- tg(r)\bigr) \\[6pt]& = \exp\bigl(-r U_s -sg(r)\bigr). \end{align}