# Continuity of the Ito integral

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 $0=t_0<t_1<t_2<\cdots<t_n$. 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 $t$.
So, did I understand you correct? For each $\omega\in \Omega$, the function $t\mapsto I_n(t,\omega)$ is continuous only on intervals $(t_j,t_{j+1}]$ - but not on the whole set $[0,T]$. – Ilya Feb 22 '12 at 13:47
No, it is continuous for all $t$. Check my formula carefully... – Byron Schmuland Feb 22 '12 at 13:49
I see, you used the running $t$ in the last term, while in my formula I've assumed that there should be $B_{t_j}$ instead of $B_t$. On the other hand, I am not sure which formula is meant in the book - in the second string of formula $(3.2.2)$ in the proof of theorem it is written that $$\int\limits_t^s \phi_ndB = \sum\limits_{t\leq t_j\leq t_{j+1}\leq s}e_j\Delta B_j$$ and $\Delta B_j$ was usually standing for $B_{t_{j+1}} - B_{t_j}$... – Ilya Feb 22 '12 at 13:59
... but anyway, I guess the confusion only comes from the notation because $I_n$ is defined as an Ito integral for an elementary functions which was before given only for a fixed upper argument $T$. Thanks for your answer - that should be what was meant in the proof, I believe. – Ilya Feb 22 '12 at 14:02