The answer here makes sense to me, in the sense that the stochastic integral does not preserve the local martingale property if the integrand is not predictable, but I am confused as to why. I have been taught the following definition of the stochastic integral:
We first define the quadratic variation for a RCLL martingale, $M$, to be $$ Q(M,t) \equiv \lim_{n \rightarrow \infty} \sum_{i=1}^\infty(M_{\sigma_{i+1} \wedge t}-M_{\sigma_{i} \wedge t})^2$$ where $\sigma_0 \equiv 0$ and $$ \sigma_{i+1} \equiv \inf \left(t > \sigma_i : |M_t - M_{\sigma_i}| \ge 2^{-n} \text{ or } |M_{t-} - M_{\sigma_i}| \ge 2^{-n} \right) $$ This limit is shown to converge uniformly on compact sets almost surely from first principles in this excellent pedagogical paper. The limit is shown to be non-decreasing, RCLL, etc. Now for any bounded RCLL martingale $M$, and any (predictable, but where does this argument rely on predictability?) process $H$ such that $$\mathbb{E} \left(\int_0^\infty H_s^2 Q(M,ds) \right) < \infty $$ we define the following mapping from the Hilbert space of bounded RCLL martingales to $\mathbb{R}$:
$$J(N) \equiv \mathbb{E} \left(\int_0^\infty H_s Q(M,N,ds) \right),$$
where $Q(M,N,t)$ is the quadratic covariation $$Q(M,N,t) \equiv \frac{1}{2} \Big(Q(M+N,t) - Q(M,t) - Q(N,t) \Big).$$
Since $J$ is a linear map, Riesz representation gives us a unique element $L$ which is in the space of bounded RCLL martingales which is such that $$\textbf{(1)} \qquad \qquad J(N) = \mathbb{E}(L_{\infty} N_{\infty})$$ and we define this $L$ to be the stochastic integral of $H$ with respect to $M$.
I know that it is typical (as in Protter's text) to define the stochastic integral first and then obtain its quadratic variation afterward, hence my confusion. Since $L$ is by construction a RCLL martingale, this seems to contradict the notion that lack of predictability of $H$ is a problem with respect to preserving the (local) martingale property of the stochastic integral, so my questions are as follows: where in this argument is the predictability (or frankly, because my knowledge is so utterly and depressingly poor, adaptedness) of $H$ used? What am I missing/where does everything fall apart?
Any and all help in understanding this would be massively appreciated, and I thank you for even just reading through this ''mini -essay''.
As mentioned in Revuz-Yor pg.120, the main reason that we require $H_{s}$ to be progressively measurable, is because we want the integral against an adapted increasing process $A_{s}$
$$H\cdot A:=\int H_{s}dA_{s}$$
to remain adapted and in turn preserve martingale too. So as mentioned in (2.2) Theorem part (a), if we don't have that $H\cdot A$ is a martingale, then we can't do the uniqueness argument i.e. we want
$$\langle L-L',L-L'\rangle=0$$
to give us $L=L'$. Then they also use Riesz-representation theorem to get a representative.
So in the above construction, one needs to add some measurability assumption for $H$ in order to ensure that we get a martingale back.
This is also shown here in lemma 2 https://almostsuremath.com/2010/01/03/the-stochastic-integral/ using the monotone class theorem.
See also Le Gall Brownian Motion, Martingales, and Stochastic Calculus Proposition 5.3 and Theorem 5.4, in particular on pg. 100
So since work over this quotient space, we have to make sure that the representative we obtain is well-defined even if we switch to some other equivalent element.
In fact, the Bichteler-Dellacherie theorem and here say that for semimartingales this the most general construction possible.
Theorem 1 (Bichteler-Dellacherie) For a cadlag adapted process $X$, the following are equivalent.
$X$ is a semimartingale.
For each ${t\geq 0}$, the set given by $$\displaystyle \left\{\int_0^t\xi\,dX\colon\xi{\rm\ is~ simple~ predictable },\ \vert\xi\vert\le1\right\} $$ is bounded in probability.
$X$ is the sum of a local martingale and an FV process.