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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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    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 $[0,T]$.2012-02-22
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    No, it is continuous for all $t$. Check my formula carefully...2012-02-22
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    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 $B_{t_{j+1}} - B_{t_j}$...2012-02-22
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    ... 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.2012-02-22
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    @Ilya so in general, if we approximated any integrand and wanna integrate up to $t$ in the intervall $[S,T]$, we just end at some subintervall and then integrate to $B_{t}$ within that intervall and skip the rest?2017-09-18
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    @user21312: not sure I understand what do you mean2017-09-18
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    @Ilya still? Ill rewrite again if its to bad :/ O guess part od the question is if one should consider a completely new approximation and integral or do as I mentioned above.2017-09-18
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    @user21312 I think here we are only talking about simple functions, so no approximations: everything is precise. Next, all the integrands are constant in $t$ besides the last one, where $t$ is the upper limit (this part was not clear to me in the book, I thought the upper limit is fixed there), and indeed integrates to $B_t - B_{t_{j-1}}$. What that your question?2017-09-18
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    @Ilya well I was thinking about what he calls "elementary" functions but I guess thats the same thing as simple. And such functions are used to approximate any integrand. So suppose I have some approximation of an arbitrary integrand from $[S,T]$. Then I wanna integrate this function to just $t. Can I use the same sequence of elementary functions and just cut the sum of at $B_{t}$?2017-09-18
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    @user21312: Yes.2017-09-18
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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