The pure jump part of Levy process and Martingale Assume $X_{t}$ be a Levy process with generating triplet $(\sigma, \gamma, \nu)$. Here $\nu$ is the measure on $R$ satisfying 
 $$ \int_{R}\min( 1,y^{2})\nu(dy)<\infty $$ 
According to the standard decomposition of Levy process, $X_{t}$ can be written as 
$$
   X_{t}= \gamma t + \sigma B_{t} + N_{t} 
$$
Here $B_{t}$ is the standard Brownian and $N_{t}$ is the pure jump Levy process defined as 
$$ N_{t}=\lim_{\epsilon\rightarrow 0}( \sum_{s\leq t, |\Delta X_{s}|\geq \epsilon } \Delta X_{s} - t\int_{1 \geq |y|\geq \epsilon} y\nu(dy)) $$
And $\Delta X_{s}=X_{s}-X_{s-}$. 
Now my question is how to make $N_{t}$ a martingale. I know the concept of compensator. But it seems to put some condition upon $\nu$. Is the following process an integral  martingale ?
$$ N_{t}=\lim_{\epsilon\rightarrow 0}( \sum_{s\leq t, |\Delta X_{s}|\geq \epsilon } \Delta X_{s} - t\int_{ |y|\geq \epsilon} y\nu(dy)) $$
 A: For $U \subseteq \mathbb{R}$, $0 \notin \bar{U}$, we set
$$X^U_t := \sum_{s \leq t} \Delta X_s \cdot 1_U(\Delta X_s).$$
Then $(X_t^U)_{t \geq 0}$ is a Lévy process and, if $\int_U y \, d\nu(y)<\infty$, then $$\mathbb{E}X_t^U = t \cdot \int_U y \, d\nu(y).$$
In particular, for $U(\varepsilon) := (\varepsilon,1)$ this implies that
$$X_t^{U(\varepsilon)} - t \int_{U(\varepsilon)} y \, d\nu(y) \tag{1}$$
is a martingale. If we further assume that $(X_t)_{t \geq 0}$ has finite first moments, then $\int_{|y| \geq 1} y \, d\nu(y)<\infty$ and we see that
$$X_t^{\mathbb{R} \backslash [-1,1]} - t \cdot \int_{\mathbb{R} \backslash [-1,1]} y \, d\nu(y) \tag{2}$$
is a martingale. Hence,
$$\begin{align*} N_t - \int_{\mathbb{R} \backslash [-1,1]} y \, \nu(dy) &= \lim_{\varepsilon \to 0} \left( \sum_{s \leq t, |\Delta X_s| \geq \varepsilon} \Delta X_s - t \int_{\varepsilon \leq |y|} y \, \nu(dy) \right) \\ &= \lim_{\varepsilon \to 0} \bigg[ \bigg(X_t^{U(\varepsilon)} - t \int_{U(\varepsilon)} y \, d\nu(y)  \bigg) +  \bigg(X_t^{\mathbb{R} \backslash [-1,1]} - t \cdot \int_{\mathbb{R} \backslash [-1,1]} y \, d\nu(y)  \bigg) \bigg] \end{align*}$$
is indeed a martingale.

Alternative solution If $(X_t)_{t \geq 0}$ is a Lévy process with finite first moments, then the decomposition $$X_t = \gamma t + \sigma B_t +N_t$$ shows that $(N_t)_{t \geq 0}$ is a Lévy process with finite first moments. Using the independence and stationarity of the increments, it is not diffult to see that $(N_t-\mathbb{E}N_t)_{t \geq 0}$ is a martingale and $$\mathbb{E}N_t = t \cdot \mathbb{E}N_1 = t \int_{|y| \geq 1} y \, d\nu(y).$$
