In page 199 of the paper A Martingale approach to Infinite Systems of Interacting Processes one reads:
Bearing in mind that there might be a typo in $|\theta|^2_c$ as mentioned in How to take this derivative? A question concerning a paper from Holley Stroock 1976 and assuming that we still can find such a sequence of approximations by simple functions, then
It makes sense to continue reading
Question How do we prove that (1.4) and (1.5) are still $P$ Martingales
Atempt: let $s = 0$ for simplicity's sake. To prove that (1.4) is a martingale, we consider a sequence of simple $\theta^n$ such that $$\Bbb{E} [\int_0^T|\theta(u) - \theta^{(n)}(u)|^2_{c(u)} \, du] \to 0 \tag 1$$ and write
$M^n_t = \int_0^t \theta^n(u)\, d\tilde{\alpha}(u)$
(1.5) for simple function $\theta^n$ implies
$\left<M^n\right >_t = \Bbb{E}[(M^n_t)^2] = 4 \int_0^t |\theta^n(u)|^2_{c(u)} \, du$
Therefore (1) implies that $M^n$ converges uniformly to $M^\infty$ in $L^2(P)$. Now, we would like to see that $M^\infty$ is a martingale.
Let $t_2 \geq t_1$ be fixed. Let $\phi$ be any bounded continuous function $t_1$ measurable. If we show that
$$\Bbb{E}[\phi M^\infty_{t_2}] = \Bbb{E}[\phi M^\infty_{t_1}]$$
we will have shown that $M^\infty$ is a martingale.
To show this we compute $$\Bbb{E}[\phi M^\infty_{t_2}] - \Bbb{E}[\phi M^\infty_{t_1}]=\\ =\Bbb{E}[\phi (M^\infty_{t_2} - M^n_{t_2}] +\Bbb{E}[\phi M^n_{t_2}] - \Bbb{E}[\phi M^n_{t_1}] + \Bbb{E}[\phi (M^n_{t_1}-M^\infty_{t_1})] = \\ = \Bbb{E}[\phi (M^\infty_{t_2} - M^n_{t_2}] + \Bbb{E}[\phi (M^n_{t_1}-M^\infty_{t_1})] \\ \leq \|\phi \| \Bbb{E}[(|M^\infty_{t_2} - M^n_{t_2}|^2] + \|\phi \| \Bbb{E}[ (|M^\infty_{t_1} - M^n_{t_1}|^2] \to 0 $$
To see the last step, we compute
$$ \Bbb{E}[(|M^\infty_{t} - M^n_{t}|^2] = \Bbb{E}[\liminf_m(|M^m_{t} - M^n_{t}|^2] \leq \liminf_m \Bbb{E}[(|M^m_{t} - M^n_{t}|^2] \leq \liminf_m \Bbb{E}[\int_0^t|\theta^m(u) - \theta^n(u)|^2_{c(u)}]\, du \xrightarrow[n \to \infty]{} 0 $$
This proves (1.4) is a martingale
The question that remais is to prove that $(1.5)$ is also a martingale
idea: using the same notation as above, we define $$N^n = (M^n)^2 - \left<M^n\right>_t$$ since $M^n \to M^\infty$ and (1) convergence of $N^n$ follows and we define $$N^\infty = (M^\infty)^2 - \left[M^\infty\right]_t $$
where
$$\left[M^\infty\right]_t = \int_0^t |\theta(u)|_{c(u)}\, du$$
To see that $N^\infty$ is a martingale, it suffices to prove that $N^n$ is uniformly bounded in $L^2(P)$ since this implies uniform integrability of $N^n$ and therefore it implies that $N^\infty$ is a martigale.
To prove uniform boundedness in $L^2$ we write
$$\Bbb{E} [(N^n)^2] \leq 2\Bbb{E} [(M^n)^4] + 2\Bbb{E} [(\left<M^n\right>_t)^2] $$ since $(a + b)^2 \leq 2 a^2 + 2b^2$
Use Burkholder-Davis-Gundy inequality: $$\Bbb [(M^n)^4] \leq C_2 \Bbb [(\left<M^n\right>_t)^2] $$
this yields $$\Bbb{E} [(N^n)^2] \leq (2C_2 + 2)\Bbb{E} [(\left<M^n\right>_t)^2] $$
the quetsion then is: can we bound uniformly $\Bbb{E} [(\left<M^n\right>_t)^2] = \Bbb{E} [(4\int_0^t |\theta^n(u)|_{c(u)}\, du)^2]$?
Any comments?
Yes compute $$\Bbb{E} \bigg[(4\int_0^t |\theta(u)|_{c(u)}\, du)^2\bigg] \leq 4 t\Bbb{E} \bigg[\int_0^t |\theta(u)|_{c(u)}^2\, du\bigg] = K <\infty $$
taking $\theta^{(n)}\uparrow \theta$ we obtain an uniform bound $$ \Bbb{E} \bigg[(4\int_0^t |\theta^{(n)}(u)|_{c(u)}\, du)^2\bigg] <K$$