Let $(\Omega, \mathcal F_\infty, (\mathcal F_t)_{t\geq 0}, \mathbb P)$ be a probability space and $X = (X_t)_{t\geq 0}$ be a process such that there exists a sequence of finite stopping time $ (T_n)_{n\geq 0} $ satisfying $T_n \uparrow\infty$ point-wise and
$$ E\left[\int^{T_n}_0 X_t^2dt \right] < \infty, \forall n \geq 0. $$
Let $B = (B_t)_{t\geq 0}$ be a $\mathcal F$- Browninan motion. I would like to define the stochastic integral $ \left( \int^t_{0}X_sdB_s \right)_{t\geq0} $ such that it has a continuous paths. This can be done by seeing $G^n := X 1_{[0,T_n]}$ satisfies
$$ E\left[\int^{\infty}_0 (G^n_t)^2 dt \right] < \infty $$ so the stochastic integral $\left( \int^t_{0}G^n_sdB_s \right)_{t\geq0} $ is well defined and can be chosen to have continuous sample paths. Now for a fixed $t>0$, we define
$$ \int^t_{0}X_sdB_s (\omega) := \int^t_{0}G^n_sdB_s (\omega), \quad \forall \omega \in \{T_n \geq t \}. $$
I would like to see how this definition is consistent. To do so, I check by a well-known lemma, for $n \leq m$
$$ \int^t_{0}G^n_sdB_s = \int^{t}_0X_s 1_{[0,T_n]}(s)dB_s = \int^{t\wedge T_n}_0X_s 1_{[0,T_m]}(s)dB_s = \left(\int^c_{0}G^m_sdB_s \right)|_{c=t\wedge T_n}. (*) $$
where the last two equalities are in $L^2$ (hence a.s). To prove it is consistent, I have to verify
$$ \int^t_{0}G^n_sdB_s (\omega) = \int^t_{0}G^m_sdB_s (\omega), \quad \forall \omega \in \{T_n \geq t \}. $$
I dont see why this is true because every stochastic process is chosen to be the limit in $L^2$ (for each $t$) then pick a continuous version defined for all $\omega \in \Omega$ but the the last two equations in ($*$) above are true only almost surely...
I believe for each $n$, $(\int^t_{0}G^n_sdB_s)_{t\geq 0}$ is already defined for each $\omega$ beforehand. How do I make rigorous sense of this? Thank you very much for your help.
Let $$ I_n(t,\omega) := \int^t_{0} G^n_sdB_s(\omega), \quad \forall n \geq0, \ t \geq 0, \ \omega \in \Omega, $$
Then we have for each $n$, $I_n$ has continuous sample paths and $\mathcal F$-adapter. From the lemma you mentioned, we see
$$ I_n(t,\omega) = I_m(t\wedge T_n(\omega), \omega) \quad a.e, \ \forall t \geq 0, \ \forall n \geq0, \ \forall m>n $$
The two processes $I_n(t)$ and $I_m(t\wedge T_n)$ are both continuous so we can deduce that
$$ \mathbb P(\Omega_{n,m})=1,\quad \Omega_{n,m}:= \{ \omega: I_n(t,\omega) = I_m(t\wedge T_n(\omega), \omega), \quad \forall t \geq 0 \} $$
It's easy to see that $\Omega_{n,m} $ is measurable, let $\Omega_n := \bigcap_{m \geq n+1} \Omega_{n,m}$ and $\Omega_0= \bigcap_{n\geq 0} \Omega_n$ then $\mathbb P(\Omega_0) = 1$. Fix $t\geq 0$, we define for each $\omega \in \Omega_0$,
$$ I(t,\omega):= I_n(t,\omega), \quad \forall \omega \in \{ T_n \geq t \} \cap \Omega_0, $$
and set $ I(t,\omega):= 0 $ for $\omega \in \Omega_0^c$. We need to check if this definition is consistent on $\Omega_0$ or not. Clearly, for $\omega \in \Omega_0$, we have
$$ \forall n \geq 0, \forall m \geq n, \quad I_n(t,\omega) = I_m(t\wedge T_n(\omega), \omega), \quad \forall t \geq 0, $$ so if $\omega \in \{T_n \geq t\} \cap \Omega_0$ for some fixed $t$ and a given $n$ then $ I_n(t,\omega) = I_m(t, \omega), \ \forall m \geq n+1$, thus it is consistent and well defined.
Next, we check if $I(t)$ is continuous. This is true on $\Omega^c_0$. Fix $\omega \in \Omega_0$ and let $T>0$ be an arbitrary time. Then there is $K$ such that $\omega \in \Omega_0 \cap \{T_K \geq T \}$ since $T_n \uparrow \infty$. Therefore,
$$I(T,\omega) = I_K(T, \omega). $$ Moreover $ \{T_K \geq T \} \subset \{T_K \geq t \}, \ \forall t \leq T$, so $\omega \in \Omega_0 \cap \{T_K \geq t \}, \ \forall t \leq T$ and hence
$$ I(t,\omega) = I_K(t, \omega), \quad \forall t \leq T $$
so $ [0,T] \ni t \mapsto I(t,\omega) $ is continuous because $I_K$ is.
Finally, we check if $I(t)$ is $\mathcal F_t$-measurable. From the definition of $I(t)$ above we see that, for any fixed $t$,
$$ I(t,\omega) = \lim_{n\rightarrow \infty} I_n(t,\omega), \quad \forall \omega \in \Omega_0, $$
and therefore, it is an almost sure (sequential) limit of $\mathcal F_t$-measurable random variables, hence if $\mathcal F = (\mathcal F_t)_{t\geq 0}$ is complete, then $I(t)$ is $\mathcal F_t$-measurable. We finish the construction here.