Simple integrands dense in square integrable integrands

Question

we defined $\Pi_2=\Pi_2(\{X_t\})$ as the class of predictable processes $\{H_t\}$ such that $||H||_x:=(E[\int_0^{\infty}H_s^2d[X]_s]^{\frac{1}{2}}<\infty$

and $\mathcal{M}^2$ as the set of all $L^2$ bounded continuous martingales endowed with $||X||_2:=(\text{sup}_{t\geq0}E[X_t^2])^{\frac{1}{2}}$

We first showed that the stochastic integral defined for simple integrands maps a subset of $\Pi_2$ isometrically to $\mathcal{M}^2$.

Now we want to show that this subset is dense in $\Pi_2$. We write this theorem like this.

I have some questions to the proof. The proof is here and I marked the points in colours where I'm referring to in the questions.

1. Why is $M^2E[[X^i]_T]\leq M^2 \text{sup}_{t\geq 0} E[(X_t^i)^2]$? ($[X^i]_T$ is the variance process)
2. Why follows $||X||_2 < \infty$?
3. Is this a typo and instead of $f$ it means $0$?
4. Is dropping the finiteness assumption meant in the way that we want our $T$ to be infinitive?

I'm so thankful for your answers!

Answer 1

1. Why is $M^2E[[X^i]_T]\leq M^2 \text{sup}_{t\geq 0} E[(X_t^i)^2]$? ($[X^i]_T$ is the variance process)

By the definition of the quadratic variation (... you call it "variance process"...) we have $$ \mathbb{E}([X]_T^i) = \mathbb{E}((X_T^i)^2)- \mathbb{E}((X_0^i)^2) \leq \mathbb{E}((X_T^i)^2).$$ Hence, in particular, $$ \mathbb{E}([X]_T^i) \leq \sup_{t \geq 0} \mathbb{E}((X_t^i)^2).$$

2. Why follows $||X||_2 < \infty$?

It doesn't follow; it's an assumption.

3. Is this a typo and instead of $f$ it means $0$?

Yes.

4. Is dropping the finiteness assumption meant in the way that we want our $T$ to be infinitive?

No. Until this point the author was assuming that $H$ is bounded (in the sense that $|H(t,\omega)| \leq M$ for some uniform constant $M$). The final step is to extend the result to processes which need not be bounded but which "only" satisfy the assumption of square integrability (see the statement of the theorem).