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Take an adapted process to a given non tricky filtration and dependent on two "variables", omega (event, not really a variable) and $t$ (time).

By definition adaptation means only measurability with respect to the filtration.

I am unable to prove with discrete sums and limits that the quadratic variation of the integral over $[0, t]$ of the adapted process, at a given omega, is zero.

I can't find any proof of it in books whereas plenty of them just write that the adapted process times $dt^2 = 0$ as if the formal proof (not this recipe) had been given ahead.

The proof is easy when the adapted process is continuous or bounded over $[0, t]$ but when only measurable, I am stuck.

One finds often the CLAIM of a zero quadratic variation of the drift imposed to the original brownian when proving Girsanov. But not the PROOF with discrete sums, partition steps and limits, if ever.

Any one can help ?

Adam Chalumeau
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Gilles D
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  • If $X$ is some integrable measurable process, then the integral $t \mapsto Y(t) := \int_0^t X_s(\omega)$ is continuous and bounded on compact sets for each fixed $\omega$. These two properties are the key to prove that $Y$ has quadratic variation zero. See also this closely related question – saz May 22 '19 at 05:43
  • Problem is that adapted is weaker than integrable – Gilles D May 23 '19 at 00:12
  • A measurable function can lead to an infinite integral, measurability is with respect to R bar, by which an infinite integral is a defined integral ! Girsanov is always presented with a process solely measurable, not integrable – Gilles D May 23 '19 at 00:23

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