I would like to prove that $Z_t$ (a counting process) is finite almost surely.
I give all the context for completeness but the main point of doubt is highlighted in bold.
We consider $0<T_1<…<T_n<..$ the sequence of jump event and $\lambda(t)$ the intensity function at time $t$. We define the poisson random measure (PRM) associated to the sequence of jump event by
$$ M(\omega)(A)= \sum_{i\geq 0}\delta_{T_i(\omega)}(A),\quad A\in\mathcal{B}(\mathbb{R}^{2}) $$
I will skip the $\omega$ in the sequel for the PRM. Now we define the set
$$ A_t = \{ (x_1,x_2) : 0\leq x_1\leq t, 0\leq x_2\leq\lambda(x_1)\} $$
By precising that $\lambda(t) = \mu + \sum_{n : T_n <t}h(t-T_n)$ with $\mu>0$ and $h$ positive and integrable, we define the counting process by
$$ Z_t = M(A_t) $$
That is
$$ Z_t = \int_{[0,t]}\int_{0}^{\infty}1_{z\leq\lambda(s)} M(ds,dz) $$
I take for granted the following equality that I am not able to prove for now
$$ \int_{0}^{t\wedge T_n}\left(\int_{(0,s)}h(s-u)dZ_u\right)ds= \int_{0}^{t\wedge T_n} h(s) Z_{t\wedge T_n - s}ds $$
Proof : First we use a well know result of integration with respect to a Poisson random measure with intensity measure given be the Lebesgue measure that is :
$$ \forall f\in\mathcal{M}(\mathbb{R}^{2}), \forall A\in\mathcal{B}(\mathbb{R}^{2}) : \mathbb{E}\left(\int_{A} f(x,y)M(dx,dy)\right) = \int_{A_1}\int_{A_2} f(x,y)dxdy $$
where $\mathcal{M}(\mathbb{R}^{2})$ is the set of Borel functions defined on $\mathbb{R}^{2}$.
We apply this result with the function $1_{z\leq\lambda(s)}$ to obtain that
$$ \mathbb{E}[Z_t] = \int_{0}^{t}\lambda(s)ds $$
It can be proven that $Y_t = Z_t - \int_{0}^{t}\lambda(s)ds$ is a martingale and thus the stopped process $Y_{t}^{T_n}$ also. From this and the fact that $\lambda(t) = \mu + \sum_{n : T_n <t}h(t-T_n)$ , we infer that
$$ \mathbb{E}[Z_{t}^{T_n}] = \mathbb{E}\left(\mu(t\wedge T_n) + \int_{0}^{t\wedge T_n}\left(\int_{[0,s]}h(s-u)dZ_{u}\right)ds\right) = \mathbb{E}\left(\mu(t\wedge T_n) + \int_{0}^{t\wedge T_n}h(s)Z_{t\wedge T_n - s}ds\right) $$
Using the fact that $t\geq t\wedge T_n$, $(t-s)\wedge T_n \geq t\wedge T_n - s$ and the fact that $h(.)Z_{.}$ is positive we get
$$ \mathbb{E}[Z_{t}^{T_n}]\leq \mu t + \int_{0}^{t}h(t-s)\mathbb{E}[Z_{s}^{T_n}]ds $$
Now by integrability of $h$ and with the introduction of the set $\{ h(t-s) \geq A_n\}$ with $A_n$ a sequence that goes to $\infty$ we can find a $n_0$ such that $\int_{0}^{\infty}h(t-s)1_{h(.)\geq A_{n_0}}ds\leq\frac{1}{2}$. This gives, using the fact that $\mathbb{E}[Z_{t}^{T_n}]$ is increasing in $t$
$$ \mathbb{E}[Z_{t}^{T_n}]\leq \mu t + \frac{1}{2}\mathbb{E}[Z_{t}^{T_n}] + A_{n_{0}}\int_{0}^{t} \mathbb{E}[Z_{s}^{T_n}]ds $$
This gives
$$ \mathbb{E}[Z_{t}^{T_n}]\leq 2\mu t + A_{n_{0}}\int_{0}^{t} \mathbb{E}[Z_{s}^{T_n}]ds $$
And the Gronwall lemma gives
$$ \mathbb{E}[Z_{t}^{T_n}]\leq 2\mu te^{A_{n_0}t} $$
Clearly the right hand side is finite.
The doubt I have is in this argument
The sequence $T_n$ is increasing and positive so it converges almost surely. By notice that $\{Z_t \geq n \} = \{ T_n \leq t\}$ it can be shown that $T_n$ converges almost surely to $\infty$ : otherwise there exists a subset of measure strictly positive on which $Z_t =\infty$ contradicting the inequality above.
Thus, $T_n\wedge t$ converges increasingly to $t$ as $n$ goes to $\infty$ so $Z_{t}^{T_n}\to Z_{t}$ by left continuity of the trajectories and an application of monotone convergence theorem shows that
$$ \mathbb{E}[Z_t] \leq 2\mu te^{A_{n_0}t} $$
And $Z_t$ is almost surely finite.
I would like to know if the last step of the proof consisting of the use of the monotone convergence theorem is okay.
My argument is that the left continuity is enough to ensure, a.s, that $Z_{t}^{T_n}\to Z_{t}$ (in an increasing way since it is a counting process) and thus pass to the limit.
Thank you a lot !