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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 $t$.

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Assume $0=t_0. The stochastic integral, for $t_j, 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 $t$.

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    @Ilya ah! I think I know why now, since this some sequence approximating it on $[S,t]$ which is in some sense a new integral and the integral is independent of approximation sequence!2017-09-18