In the paper A Martingale approach to Infinite Systems of Interacting Processes one reads:
I wasn't able to prove using Theorems 1.1 and 1.2 that (1.3) is a $P$ martingale.
context: Here is theorem (1.1)
and here is theorem (1.2) (along with some relevant definitions)
Attempt We see that, since $\theta(t)$ is a simple function, $\theta(t-)$ remains constant for the time intervals $I_j = (j/n, (j+1)/n]$. So it is natural to apply item (iv) of theorem (1.1) since the term $\int_s^{t\vee s}2c_k(u)\alpha_k(u)\theta_k(u)\, du$ cancel out. We end up with $$X_\theta^s(t) = \prod_j \exp\bigg[\sum_k\theta_k(j/n) \alpha_k((t \wedge (j+1)/n)\vee s) -\alpha_k(s \vee (j)/n)\\ - \int_{I_j \cap [s,t] } c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, d\bigg] $$
But how do we show from here that $X_\theta^s(t)$ is a martingale? When does theorem (1.2) comes into play?
To simplify notation, assume without loss of generality that $s \in I_m$ and $t_2 \in I_{M_1}$ and $t_1 \in I_{M_2}$ with $t_1 \leq t_2$
then for $i = 1,2$ we have
$$X_\theta^s(t_i) = \exp\bigg[\sum_{k}\theta_k(m/n)\bigg( \alpha_k((m+1)/n) -\alpha_k( s)\bigg)\\ - \int_{s}^{(m+1)/n} c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, d\bigg] \\ \prod_{j = m + 1}^{M_i-1} \exp\bigg[\sum_{k}\theta_k(j/n)\bigg( \alpha_k((j+1)/n) -\alpha_k( j/n)\bigg)\\ - \int_{j/n}^{(j+1)/n} c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, du\bigg]\\ \exp\bigg[\sum_{k}\theta_k(M_i/n)\bigg( \alpha_k(t) -\alpha_k( M_i/n)\bigg)\\ - \int_{M_i/n}^{t} c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, du\bigg] $$
Now we use $Q^{\omega,j}$ the r.c.p.d. $P \mid \mathcal{F}_{j/n}$ for $j = M_2, M-1,\ldots, M_1$ and theorem 1.2 to obtain
$$\Bbb{E}[X_\theta^s(t_2)\mid \mathcal{F}_{t_1}] = X_\theta^s(t_1). $$
Indeed, note that from Theorem 1.2
$$\Bbb{E}\Bigg[\exp\bigg[\sum_{k}\theta_k(M_2/n)\bigg( \alpha_k(t) -\alpha_k( M_2/n)\bigg)\\ - \int_{M_i/n}^{t} c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, du\bigg]\mid \mathcal{F}_{M_2/n}\Bigg] = 1 $$
This implies that
$$\Bbb{E}[X_\theta^s(t_2)\mid \mathcal{F}_{t_1}] = \Bbb{E}[\Bbb{E}[X_\theta^s(t_2)\mid \mathcal{F}_{M_2/n}]\mid \mathcal{F}_{t_1}] = \Bbb{E}\Bigg[X_\theta^s(M_2/n)\Bbb{E}\Bigg[\exp\bigg[\sum_{k}\theta_k(M_2/n)\bigg( \alpha_k(t) -\alpha_k( M_2/n)\bigg)\\ - \int_{M_i/n}^{t} c_k(u) (e^{-2\theta_k(j/n) \alpha_k(u)}-1) \, du\bigg]\mid \mathcal{F}_{M_2/n}\Bigg]\mid \mathcal{F}_{t_1}\Bigg]\\ = \Bbb{E}\Bigg[X_\theta^s(M_2/n)\mid \mathcal{F}_{t_1}\Bigg] $$
Apply the same idea successively to obtain that
$$\Bbb{E}\Bigg[X_\theta^s(M_2/n)\mid \mathcal{F}_{t_1}\Bigg] = \Bbb{E}\Bigg[X_\theta^s((M_2-1)/n)\mid \mathcal{F}_{t_1}\Bigg] = \ldots = \Bbb{E}\Bigg[X_\theta^s((M_1 + 1)/n)\mid \mathcal{F}_{t_1}\Bigg] = X_\theta^s(t_1)$$