Lévy Process existence of the expectation of the supremum of the past process. 
Given a Lévy Process $X_{t}$ in $\mathbb{R}^{d}$, with $X_{t}^{*}:=\sup_{s\in[0,t]}|X_{s}|$.
I want to show, that for $t>0$ with $E[|X_{t}|]<\infty$ for  $t>0$, then $E[X_{t}^{*}]<\infty$.

Attempt:
I know with Etemadi's inequality, it holds that for $a,b>0$ it holds that
$$
P[X_{t}^{*}>a+b]\leq \frac{P[|X_{t}|>a]}{P[X_{t}^{*}\leq b/2]}
$$
We can choose due to the càdlàg property of $X$ the $b>0$ such that $P[X_{t}^{*}\leq b/2]>0$.
That looks like we can use it.
Hope you can help me out.
 A: Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable $X$, will also be useful:
$$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X > r) \, dr.$$
Because of the monotonicity of $r \mapsto \mathbb{P}(X > r)$ this implies 
$$b \sum_{k=0}^{\infty} \mathbb{P}(X > (k+1)b) \leq \mathbb{E}(X) \leq b \sum_{k=0}^{\infty} \mathbb{P}(X > kb) \tag{1}$$
for any $b>0$.

Now back to your problem: Choose $b>0$ such that $\mathbb{P}(X_t^* \leq b/2)>0$ and set $c:= 1/\mathbb{P}(X_t^* \leq b/2)$. Then, by Etemadi's inequality (for $a=kb$),
$$\mathbb{P}(X_t^* > k b) \leq c \mathbb{P}(|X_t| > (k-1)b)$$
for any $k \in \mathbb{N}$ (see this question for a proof). Summing over $k$ yields
$$\sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb) \leq c \sum_{k=0}^{\infty} \mathbb{P}(|X_t|>kb). \tag{2}$$
Since, by assumption $X_t \in L^1$, it follows from $(1)$ that
$$\sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb) \stackrel{(2)}{\leq} c+ c\sum_{k=1}^{\infty} \mathbb{P}(|X_t|>kb) \stackrel{(1)}{\leq} c+\frac{c}{b} \mathbb{E}(|X_t|)<\infty.$$
Using again $(1)$, we conclude
$$\mathbb{E}(X_t^*) \leq b \sum_{k=0}^{\infty} \mathbb{P}(X_t^* >kb) \leq b + b \sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb)<\infty.$$
