Consider the stochastic integral of a process $H$ with respect to the local martingale $M$: $$ (H\bullet M)_t = \int_{[0,t]} H_s\,\mathrm d M_s. $$
We know that when $H$ is predictable and sufficiently integrable, then $H\bullet M$ is a local martingale. Moreover, it is also well-known that when $H$ is not predictable, then $H\bullet M$ need not be a local martingale. This answer gives a nice example demonstrating this fact. On the other hand, when $M$ also happens to be continuous, then we are able to also define $H\bullet M$ for progressive processes $H$ (cf. Karatzas and Shreve).
This naturally makes it important to identify where exactly in the construction of the stochastic integral predictability of the integrand is important. Unfortunately, I can't see where predictability plays a role here. Can anyone help clarify this?
Context and Background
A typical construction of the stochastic integral is to first define the integral for simple predictable processes. It is straightforward to show that when $H$ is simple predictable, then $H\bullet M$ is a local martingale. Standard arguments also imply that any predictable process is the limit of simple predictable ones.
Then, for a general predictable process $H$ (again, assuming sufficient integrability), we fix a sequence of simple predictable processes $\{ H^n\}$ with $H^n \to H$, and define the integral $H\bullet M=\lim H^n \bullet M$. (One can show that $H \bullet M$ does not depend on our choice of approximating sequence and is thus well-defined.) $H\bullet M$ inherits the (local) martingale property from its approximating sequence.
It seems to me that this procedure works just as well even though $H$ were not necessarily predictable, but simply a càdlàg adapted process, even for general (i.e. not necessarily continuous) local martingales.
What am I missing?
I know I am glossing over quite a few details here, since I don't want to make this post much longer than necessary. I can fill in the details as needed. For reference, the construction I have in mind is the one in Cohen and Elliott (2015).
Best Answer
It might be helpful to take a look at the discrete martingale transform.
If $(C_k)_{k \in \mathbb{N}_0}$ is predictable, i.e. $C_k$ is $\mathcal{F}_{k-1}$-measurable for each $k$, then $C \bullet M$ is a martingale (...assuming that everything is nicely integrable). This corresponds, essentially, to the fact that the stochastic integral of a predictable simple process w.r.t to a (time-continuous) martingale is a martingale (..again, provided that everything is nicely integrable). If the process is not predictable, then the martingale property fails, in general, to hold. Since martingales have constant expectation, the condition
$$0 = \sum_{j=1}^n \mathbb{E}(C_j (M_j-M_{j-1}))$$
is necessary for $C \bullet M$ being a martingale. Since this needs to hold for all $n$, we actually need
$$0 = \mathbb{E}(C_n (M_n-M_{n-1})), \qquad n \in \mathbb{N}.$$
If $C$ is not predictable there is no reason why this should be true. The difference $M_n-M_{n-1}$ has expectation zero but since we are multiplying it with something which can be correlated with $M$, the product will, in general, fail to have zero expectation. For instance, we could choose $C_n := \frac{1}{2} (M_{n-1}+M_n)$ and see that
$$\mathbb{E}(C_n (M_n-M_{n-1}) = \frac{1}{2}( \mathbb{E}(M_n^2)-\mathbb{E}(M_{n-1}^2)) = \frac{1}{2} (\mathbb{E}\langle M \rangle_n-\mathbb{E}\langle M \rangle_{n-1})$$
where $\langle \cdot \rangle$ denotes the quadratic variation. The expression on the right-hand side is, in general, strictly positive. For instance, if $M$ is a "discretized" Brownian motion, then it equals $1/2$. This is exactly the phenomenon which we observe while studying the the Stratonovich integral (see the comment by @TheBridge).