in the book "ruin probabilities" by Asmussen and Albrecher, in Chapter II Thm. 1.2, the following statement is made:
Let $\{X_t\}$ be a Lévy process and $\alpha\in\mathbb{R}$. If $\mathbb{E}[e^{\alpha X_t}]$ is finite for all $t>0$, then it holds $\mathbb{E}[e^{\alpha X_t}]=e^{t\kappa(\alpha)}$ for some $\kappa(\alpha)\in\mathbb{R}$ and the process $e^{\alpha X_t-t\kappa(\alpha)}$ is a martingale.
In the proof they mention you have to choose $\kappa(\alpha)=\log\mathbb{E}[e^{\alpha(X_1-X_0)}]$.
Now I get, why the mentioned process is a martingale, but I somehow do not see, why $\mathbb{E}[e^{\alpha X_t}]=e^{t\kappa(\alpha)}$ should hold for that specific choice of $\kappa$. I mean for $t\in\mathbb{N}$ it's easy, but what about a general $t\in\mathbb{R}$?
Any hints are much appreciated!
Fix $\alpha>0$. Define for all $y\geq0$: $$ f(y)=E[e^{\alpha (X_y-X_0)}]$$ Then from Lévy properties we can prove:
Property 1: $$ f(yn)=f(y)^n \quad \forall y\geq0,n\in\{1,2,3,...\}$$
From Property 1 alone we can prove:
Property 2: $$f(y/m)=f(y)^{1/m} \quad \forall y\geq0,m\in \{1,2,3,...\}$$
Now from Properties 1 and 2 we get for any rational number $n/m>0$ (for $n,m$ positive integers): $$ f(n/m)=f(1)^{n/m}$$ However properties of Lévy imply $f$ is continuous and so for any $y\geq0$ we can approximate $y$ with rational numbers to conclude $$f(y)=f(1)^y$$