# Stochastic integral density of simple functions no1

I am trying to understand proposition 2.6 page p.134 from Karatza's book Brownian motion and stochastic calculus.

If $M$ is continuous square integrable martingale on $(\Omega, \mathcal{F}, P)$ and the mapping $t \rightarrow <M,M>_t(\omega)$ is absolutely continuous.

Consider the space $( [0, \infty) \times \Omega, \mathcal{B}[0, \infty) \times \mathcal{F})$. For a measurable, and $F_t$ - adapted process $X = (X_t, F_t)$ we define: $$[X]_T^2 = \mathbb{E} \int_0^T X(t, \omega)^2 d<M,M>_t$$ and the pseudometric, $$[X-Y] = \sum_{n=0}^{\infty}\frac{\min(1, [X-Y]_n)}{2^n}$$

then the space of the simple processes $L_0$ is dense (under this metric)in $L$ the space of measurable and adapted processes $X$ that $[X]_T < \infty$.

First of all he proves that for that $X \in L$ there is a sequence $X^m$ of $L_0$, satisfying, $$\sup_{T>0}\lim_{m \rightarrow \infty} \mathbb{E}\int_{0}^{T}|X_t^m - X_t|^2 dt = 0$$ Now, he states that we can find a subsequence of $X^m$, say $X^{k(m)}$ such that the following set, $$\left \{ (t, \omega) \in [0, \infty) \times \Omega)| \lim_{m \rightarrow \infty}X^{k(m)} = X \right \}^c$$ has product measure $0$. So the first question of mine is whether anyone can see that. I know how to extract a subsequence which converges a.s $P \times \lambda_{[0,T]}$ on $[0, T] \times \Omega)$to $X$ but not on $[0, \infty) \times \Omega)$. Any ideas?

Question2

He continues saying that the absolute continuity of $t \rightarrow <M,M>_t(\omega)$ and the bounded convergence theorem imply $[X^{k(m)} - X] \rightarrow 0$, as $m \rightarrow \infty$. Any ideas why?

Thanks