Why predictable processes?

Solution 1:

Being predictable (as in being measurable with respect to the predictable $\sigma$-algebra) always implies being progressive (as in being measurable with respect to the progressive $\sigma$-algebra), but the other implication only holds in special cases, such as when e.g. the underlying filtration is the usual completion of the filtration generated by a Brownian motion.

There are good reasons for restricting yourself to predictable integrands. In general, stochastic integral processes with e.g. progressive integrands and local martingale integrators will not be local martingales (generally, such integrals will be difficult to define at all except for simple progressive integrands). For an example of this, let $N$ be a standard Poisson process and let $M_t = N_t - t$. Then $M$ is a local martingale (in fact, a true martingale).

We then have $$ \int_0^t N_s d M_s =\int_0^t N_{s-} dM_s +\int_0^t \Delta N_s d M_s \\ =\int_0^t N_{s-} dM_s + \sum_{0<s\le }(\Delta N_s)^2 =\int_0^t N_{s-} dM_s + N_t. $$ Here, the first term is a local martingale (because it's the integral of a left-continuous and adapted, hence predictable, process, with respect to a local martingale), while the second term is not a local martingale. Therefore, the integral $\int_0^t N_s d M_s$ is not a local martingale. In essence, the reason for this is that $N$ is progressive (because it is cadlag and adapted) but not predictable.

One of the main properties which makes it possible to obtain the local martingale property for stochastic integrals of predictable processes with respect to local martingales is the following result: If $X$ is cadlag and predictable, then the jumps of $X$ can be covered with a countable sequence of predictable stopping times. And predictable stopping times plays well with martingales.

For more on all this, see e.g. the books "Difffusions, Markov processes and martingales" by Rogers & Williams, or "Semimartingale theory and Stochastic Calculus" by He, Wang and Yan.