(a) Well, the idea of your proof is correct, but there a several things which I would like to point out:
- There is some abuse of notation in it. So, for example, every probabilist will know what you mean by $$\{(t,\Omega); h(t) \in B\},$$ but (at least from my point of view) that's not a nice way to put it. Just write $$\{(t,\omega); \omega \in \Omega, h(t) \in B\},$$ it's not really more effort to do so and it is rigorous.
- You write "... for some $\bar{B} \in \mathcal{B}_{\mathbb{R}_+}$"; well, why not simply given an explicit expression for the set? What about $\bar{B} = h^{-1}(B)$?
- You write "Since I can generate $(\bar{B},\Omega)$ by using [...]"; how can you generate it? Please be a bit more precise... (e.g. "Since the Borel $\sigma$-algebra $\mathcal{B}_{\mathbb{R}_+}$ is generated by $\{(a,b]; 0<a<b\}$ ...").
I would phrase it like that: Since $h$ is Borel-measurable, there exists a sequence of simple functions $(s_n)_{n \in \mathbb{N}}$ of the form, $$s(t) = \sum_{j=1}^n c_j 1_{(a_j,b_j]}(t),$$ such that $s_n \to f$ as $n \to \infty$. Since each $s_n$ is obviously predictable (it is left-continuous), $h$ is predictable as a pointwise limit of predictable functions.
(b) Actually, you show that the composition of two measurable functions $S: (\Omega_1,\mathcal{A}_1) \to (\Omega_2,\mathcal{A}_2)$ and $T: (\Omega_2,\mathcal{A}_2) \to (\Omega_3,\mathcal{A}_3)$ is measurable. Since this is well-known, there is no need to prove this, but nevertheless your proof is correct.
(c) Define $$\Sigma := \{A \in \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R}); (t,\omega) \mapsto 1_A(t,X_t(\omega)) \, \text{is predictable}\}.$$ Show that $\Sigma$ is a $\sigma$-algebra. Conclude from $\mathcal{B}(\mathbb{R}_+) \times \mathcal{B}(\mathbb{R}) \subseteq \Sigma$ that $$\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{B}(\mathbb{R}) = \Sigma.$$ Finally, approximate (similar as in (a)) any Borel-measurable function $g: \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}$ by simple functions and conclude that $Z_t$ it is predictable as the pointwise limit of predictable functions.
There are at least two reasons. I am certain of the second, only 60% of the first one, so please keep this in mind.
1. Look at equation (2.7); $F_t$ is a stopped version of the integral of the progressively measurable process.
It should follow that (at least restricted to locally bounded/integrable progressively measurable processes) that the integral itself is progressively measurable.
To the best of my understanding, an arbitrary stopped version of a measurable and adapted process does not necessarily have to be either measurable or adapted (or at least can fail one of the other two).
On the other hand, any progressively measurable function stopped at a stopping time is again progressively measurable. To the best of my understanding, the definition of progressively measurable also implies that progressively measurable processes are exactly the processes which are still adapted after stopping at deterministic times (and even progressively measurable still since deterministic times are stopping times).
2. I am 100% certain of this one. Look at the set $A$. As a result of its definition and the definition of progressive measurability, measurable and adapted is in general insufficient for $A$ to be measurable with respect to the product sigma-algebra; we need progressive measurability for this to be true. If $A$ is not measurable with respect to the product sigma-algebra, then Fubini is not applicable and we cannot say anything about the cross-sections of $A$, and therefore we cannot prove that there exists a sequence of simple predictable processes converging to $X_t$ in $L^2$.
Also, as a final note, even if adapted and measurable were sufficient for this proof (although I believe they are not), we still would only want to focus on progressively measurable processes anyway.
That is because the semimartingale stochastic integral only accepts locally bounded progressively measurable processes as integrands, so even if we could approximate in $L^2$ a wider class of functions by simple processes, we still would not be able to use that approximation in defining/generalizing stochastic integrals, so there would be no point in considering it.
Again though, I am fairly certain that progressive measurability is necessary for the proof to go through for either one of the two reasons mentioned above.
Best Answer
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.