Let $(\Omega, \mathcal{F}, (\mathcal{F}_t), \Bbb{P})$ be a filtered probability space. Net $N$ be a poisson process with parameter $\lambda>0$. Let $h$ be a bounded measurable function and define $L_t:=\exp\left(\int_0^t h(s) ~dN_s-\lambda \int_0^t e^{h(s)}-1~ds\right)$ for $t\geq 0$. I want to check that $(L_t)_{t\geq 0}$ is a martingale.
I am a bit stuck since I don't know where to start. First of all I know that we define $\int_0^t h(s) ~dN_s:=\sum_{s\leq t} h(s) \Delta N_s$ where $\Delta N_s:=N_s-N_{s-}$ for $N_{s-}:=\lim_{h\rightarrow 0} N_{s-h}$ the left limit at point $s$. I also know that if I consider the compensated poisson process $M_t:=N_t-\lambda t$ then $\int_0^t h(s) ~dM_s:=\int_0^t X_s~dN_s-\lambda \int_0^t X_sds$ is a martingale if the process $(X_s)$ is left continuous, adapted and $\Bbb{E}(\int_0^t |X_s|ds)<\infty$. Therefore I thought about adding and substracting $\lambda \int_0^t X_sds$ in $L_t$ to use this fact. But the problem is that a priori $h(s)$ is neither adapted nor left continuous. Thus I'm a bit lost what to do. Could someone help me?
With the abbreviations \begin{align}\tag1 Y_t&:=\exp\left(\textstyle\int_0^th(s)\,dN_s\right)\,,\\[2mm] Z_t&:=\exp\left(\textstyle-\lambda\int_0^te^{h(s)}-1\,ds\right)\,,\tag2 \end{align} we have $L_t=Y_t\,Z_t\,.$ By the integration by parts formula for semimartingales we have $$ dL_t=Y_{t-}\,dZ_t+Z_{t-}\,dY_t+d[Y,Z]_t\,. $$ Observe that $Z$ is continuous and of finite variation and that $Y_t$ changes only by jumps. To indicate the latter I shall use the notation $dY_t=\Delta Y_t\,.$ By the properties of $Y$ and $Z$ their covariation is zero and the integration by parts formula becomes $$\tag3 dL_t=Y_{t-}\,dZ_t+Z_t\,\Delta Y_t\,. $$ (The integral form of the second term is $\sum\limits_{s\le t}Z_s\Delta Y_s\,.$)
Observe that \begin{align}\tag4 \Delta Y_t&=Y_{t-}\,(e^{h(t)}-1)\Delta N_t\,,\\[2mm]\tag5 dZ_t&=-Z_t\,\lambda\,(e^{h(t)}-1)\,dt\,. \end{align} Therefore, \begin{align}\tag6 dL_t&=-Y_{t-}\,Z_t\,\lambda \,(e^{h(t)}-1)\,dt+Z_t\,Y_{t-}(e^{h(t)}-1)\,\Delta N_t\\[2mm]\tag7 &=L_{t-}(e^{h(t)}-1)(\Delta N_t-\lambda \,dt)\,. \end{align} Using again $dN_t=\Delta N_t$ this can be written as \begin{align}\tag8 L_t=1+\textstyle \int_0^tL_{s-}\,dM_s \end{align} where $M_t$ is the martingale $$\tag9 M_t=\int_0^t(e^{h(s)}-1)\,(dN_s-\lambda\,ds)\,. $$