TITLE: Continuity of the  Ito integral
QUESTION [5 upvotes]: Let $B_t$ be a 1-dimensional Brownian motion. I am following "Stochastic Differential Equations" by Bernt Øksendal. On the page 32 (it is displayed in the link I've put) there is a proof of existence of continuous version of the Ito integral. There is stated that for a function
$$
\phi_n(t,\omega) = \sum\limits_j e_j(\omega)\cdot 1_{[t_j,t_{j+1})}(t)
$$
its integral
$$
I_n(t,\omega) = \int\limits_0^t\phi_n(s,\omega)dB_s(\omega)
$$
is continuous in $t$.
From what I seen, I think that
$$
I_n(t,\omega) = \sum\limits_{t_j\leq t}e_j(\omega)(B_{t_{j+1}} - B_{t_j})
$$
which does have jumps. So I wonder, how can it be continuous in $t$.

REPLY [7 votes]: Assume $0=t_0<t_1<t_2<\cdots<t_n$. The stochastic integral, for $t_j<t\leq t_{j+1}$,
is given by 
$$I_n(t,\omega)=\sum_{i=0}^{j-1} e_{i}(\omega)\,[B_{t_{i+1}}(\omega)-B_{t_i}(\omega)]+e_j(\omega)\,[B_t(\omega)-B_{t_j}(\omega)],$$
which is continuous in $t$.