Proofing Analytic continuation and stationary increments of an exponential Family

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In U.Küchler "Exponential Families of Stochastic Processes" 1997 Theorem 4.2.1 we have the following setup. Let $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\geq0})$ be a filtered measurable space. Let $P_{\varphi}$ be a probability measure on that space. Let $P_{\varphi}^{t}$ be the restrition of $P_{\varphi}$ to $\mathcal{F}_{t}$. We assume that for the exponential family $\mathcal{P}=\{P_{\varphi},\varphi\in\Gamma\}$ we have the following representation of Radon-Nikodym-Derivatives for arbitrary $\varphi_{0}\in\Gamma$ \begin{align} \frac{dP_{\varphi}^{t}}{dP_{\varphi_{0}}^{t}}=\exp[<\varphi-\varphi_{0},B_{t}>-\phi_{t}(\varphi)+\phi_{t}(\varphi_{0})]\,\varphi\in\Gamma,t\geq 0 \end{align} Now Küchler states:

  1. If $int \Gamma \not=\emptyset$, the canonical process $B_{t}$ has independent increments under $P_{\varphi},\forall \varphi \in \Gamma$. The Laplace transfrom of the increments $B_{t}-B_{s}$ under $P_{\varphi}$ is for $\lambda \in \Gamma-\varphi$ given by \begin{align} E_{\varphi_{0}}\left[\exp[<\lambda,B_{t}-B_{s}>]|\mathcal{F}_{s}\right]&=\exp[\phi_{s}(\varphi)-\phi_{s}(\lambda+\varphi)-\phi_{t}(\varphi)+\phi_{t}(\lambda+\varphi)] \end{align}
  2. In particular, $B_{t}$ has stationary increments if and only if $\phi_{t}(\varphi)$ is of the form $tf(\varphi)$.

$B_{t}$ is a stochastic process, which is right-continuous with limits from the left and $\mathcal{F}_{t}$-adapted. And $\phi_{t}$ is a function which is positive real, non-random and right-continuous with limits from the left.

To point 1: We choose $\varphi_{0}\in \Gamma$ arbitrary. With the canonical representation from above, we know from some previous stuff (not mentioned here), that $(dP_{\varphi}^{t}/dP_{\varphi_{0}}^{t})_{t\in\mathbb{R}_{+}}$ is a $P_{\varphi_{0}}$-martingale. So for $s\leq t$: \begin{align} E_{\varphi_{0}}\left[\frac{dP_{\varphi}^{t}}{dP_{\varphi_{0}}^{t}}|\mathcal{F}_{s}\right]&=\frac{dP_{\varphi}^{s}}{dP_{\varphi_{0}}^{s}}\\ E_{\varphi_{0}}\left[\exp[<\varphi-\varphi_{0},B_{t}>+\phi_{t}(\varphi_{0})-\phi_{t}(\varphi)]|\mathcal{F}_{s}\right]&=\exp[<\varphi-\varphi_{0},B_{s}>+\phi_{s}(\varphi_{0})-\phi_{s}(\varphi)]\\ \Leftrightarrow E_{\varphi_{0}}\left[\exp[<\varphi-\varphi_{0},B_{t}-B_{s}>]|\mathcal{F}_{s}\right]&=\exp[\phi_{s}(\varphi_{0})-\phi_{s}(\varphi)-\phi_{t}(\varphi_{0})+\phi_{t}(\varphi)] \end{align} and therefore $\forall \lambda \in \Gamma-\varphi_{0}$ we get the conditional Laplace-Transform \begin{align} E_{\varphi_{0}}\left[\exp[<\lambda,B_{t}-B_{s}>]|\mathcal{F}_{s}\right]&= \exp[\phi_{s}(\varphi_{0})-\phi_{s}(\lambda+\varphi_{0})-\phi_{t}(\varphi_{0})+\phi_{t}(\lambda+\varphi_{0})] \end{align} Now comes the unclear Part: Küchler states that the conditional Laplace transform of $B_{t}-B_{s}$ under $P_{\varphi_{0}}$ is equal to a non-random function in the open set $int \Gamma-\varphi_{0}$. By analytic continuation the above stated conditional Laplace transform is non-random in its entire domain. Why we need continuation here? Cause we have already stated, that the formula holds for all $\lambda\in\Gamma-\varphi_{0}$. I could need a proof here.

To point 2. I am actually stuck at the beginning