So I am working/reading through a proof that general Ito integrals have continuous versions. So that
$$I_f(t) = \int_0^t f(s,\omega)dB_s(\omega)$$
Has a continuous version in $t$.
The proof I am reading does this by considering a sequence of elementary functions $\phi_n$ that converge to $f$ in $L^2 (P)$. The author defines
$$I_n (t) = \int_0^t \phi_n (s,\omega) dB_s (\omega)$$
and states that $I_n (t)$ is continuous, but does not give the justification.
Here, $\phi_n$ is assumed to take the form
$$\phi_n = \sum_j e_j^{(n)} (\omega)\times \mathbb{1}_{\left( t \in[t_j,t_{j+1})\right) }$$
My question is why must $I_n (t)$ be continuous in $t$ when we do not know this to be the case for the more general $f$ (though $f$ is also restricted to being $\mathcal{F}_t$ adapted etc).
It is assumed throughout the book that the process $B_t$ is a continuous version of Brownian motion, so I don't find it difficult to believe, but I don't understand why it automatically holds for elementary functions and not the more general $f$?
edit
Following Saz's comment. So to show it's continuous for elementary functions, can we write:
$$|I_n (t) – I_n (s) | \leq \varepsilon$$
$$\iff \left| \sum_{t_j \in [s,t]} e_j^{(n)} \Delta B_{t_j} \right| \leq \varepsilon$$
And so we have
$$\left| \sum_{t_j \in [s,t]} e_j^{(n)} \Delta B_{t_j}\right| \leq |t-s| \max_{t_j \in [s,t]} |e_j^{(n)}| \max_{t_j \in [s,t]} |\Delta B_j|$$
And so choosing
$$|t-s| \leq \varepsilon/ \max_{t_j \in [s,t]} |e_j^{(n)}| \max_{t_j \in [s,t]} |\Delta B_j| = \delta$$
It follows that if $|t-s|\leq \delta$ then $|I_n(t)-I_n(s)|\leq \varepsilon$
Is this correct?
Best Answer
I don't see what you wrote as $$ |I_n (t) - I_n (s) | \leq \varepsilon\iff \sum_{t_j \in [s,t]} e_j^{(n)} \Delta B_{t_j} < \varepsilon. $$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by $$ I_n (t) = \sum_{j=1}^n e_j(\omega) (B_{t\wedge t_{j+1}}(\omega) - B_{t\wedge t_{j}}(\omega)). $$ We can see immediately that $I_n(t)$ is continuous in $t$ since $t\mapsto B_t(\omega)$ is continuous for all $\omega\in\Omega$.