I have the following (compensated Poisson process):
$$M_t = N_t - \lambda t$$
I know that $$dM^c_t = -\lambda dt$$
Is my approach to the following true? $$\begin{align} \int_0^tM_s dM^c_s &= \int_0^t N_s ds + \int_0^t \lambda sds\\ &=0 + \frac{\lambda t^2}{2} \\ &= \frac{\lambda t^2}{2}\end{align}$$ My understanding is that given $N_t$ will be non-zero at a countable no. of points, it should follow that $$\int_0^tN_sds = 0$$
By Tonelli-Fubini: $$E\bigg[\int_0^tN_sds\bigg]=\int_0^tE[N_s]ds=\lambda\frac{t^2}{2}>0,\,\forall t>0$$ Since for $t>0$ we have $\int_0^tN_sds\geq0$ and $E[\int_0^tN_sds]>0$, then $P(\int_0^tN_sds>0)>0$.
It may be useful to illustrate why this is the case. Recall $N_t$ is càdlàg, pure-jump and nondecreasing. Define $\tau_0=\inf\{t\geq0:N_t>0\}$ and $\tau_j=\inf\{t> \tau_{j-1}:N_{t}>N_{\tau_{j-1}}\},\,j\geq 1$. Then on $[\tau_j,\tau_{j+1})$, the process is flat with value $N_{\tau_j}$. So $$\int_{\tau_j}^{\tau_{j+1}}N_sds=N_{\tau_j}(\tau_{j+1}-\tau_j)\implies\int_0^tN_sds=\sum_{n \in \mathbb{N}_0}N_{\tau_n}(\tau_{n+1}\wedge t-\tau_n\wedge t)$$ where $x\wedge y:=\min(x,y)$. So $\int_0^tN_sds=0$ on $\{t\leq \tau_0\}$ but not in general.