I am reading "Diffusion, Markov Processes and Martingales" by Roger/Williams, which state the following version of the monotone class theorem applied to bounded elementary processes $b\mathcal{E}$.
Later (section 27) they define the norm
and define the stochastic integral through the ito isometry. In the last step the Ito isometry should be extended from the elementary processes to as much as possible in $L^2(M)$. They just say that from Lemma 6.5 it follows that $\overline{b\mathcal{E}} = L^2(M)$.
In my opinion at least $(i)$ and $(ii)$ do not hold for $d[M]_s = ds$ (Brownian Motion case), since $\int_0^\infty c^2 ds = \infty$ for $c \not= 0$.
Also how come Lemma 6.5 only gives us that the bounded predictable processes are in $\mathcal{H}$ and now we just take the whole $L^2(M)$ (no boundedness assumption anymore).
Best Answer
First, it is important to notice that (not unusually) Roger and Williams have, at the point you are looking at, restricted their integrator $M$ to be a continuous $L^2$-bounded martingale. This means that at this stage, Brownian motion is not an allowed integrator, since it is not $L^2$-bounded. The punchline will be that by localisation, once we have the theory for $L^2$-bounded martingales, we will be able to get it for a wider class that includes Brownian motion.
This is important since it means that the class of martingales they consider satisfies $\mathbb{E}([M]_\infty) < \infty$. This will imply that constant functions are in $L^2(M)$, for example.
Now for the main body of your question. Let $U$ be the set of elementary processes in $L^2(M)$ (coinciding with the notation of Rogers/Williams). We want to show that $\overline{U} = L^2(M)$.
First we check that the space of bounded previsible process $b\mathcal{E}$ satisfies $b \mathcal{E} \subseteq \overline{U}$.
For this, we use their lemma 6.5 (the monotone class theorem). First, constant functions are in $\overline{U}$ since $c 1_{[0,n]} \to c$ in $L^2(M)$ as $n \to \infty$.
For the second condition, for arbitrary $\varepsilon$ we can pick an $N$ large enough that $|H_n(s,\omega) - H(s,\omega)| \leq \varepsilon$ for all $n \geq N$, $s \in (0,\infty)$ and $\omega \in \Omega$. Hence $$\|H_n - H\|_{L^2(M)}^2 \leq \mathbb{E} \bigg [ \int_0^\infty \varepsilon d [M]_s \bigg] = \varepsilon \mathbb{E}([M]_\infty).$$ That is, the uniform convergence implies convergence in $L^2(M)$ so $H \in \overline{U}$ also since $\overline{U}$ is closed.
Finally, it is an easy exercise to use the DCT to see that $\overline{U}$ satisfies condition $3$ in Lemma 6.5 also.
Therefore, $b \mathcal{E} \subseteq \overline{U}$. You are right that general elements of $L^2(M)$ need not be bounded so we aren't quite finished. The point is that a general element of $L^2(M)$ can be approximated by functions in $b \mathcal{E}$.
The crudest way one might try to do this is to take $\phi \in L^2(M)$ and just brutally cut it off at height $N$. That is, let $$\phi_N = \begin{cases} N \qquad |\phi| \geq N \\ \phi \qquad \text{otherwise} \end{cases}$$ Then $\phi_N$ is certainly bounded and it follows by the DCT that $\phi_N \to \phi$ in $L^2(M)$. Hence $L^2(M) \subseteq \overline{b \mathcal{E}} \subseteq \overline{U}$ as desired.