# Locally FV processes are FV processes

Does anyone see why any process which is of Locally Finite Variation has to be of Finite Variation ?

Best regards

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## 1 Answer

Hi I think I got it please tell me if I'm wrong.

Only one implication has to be proven as the other is obvious. So let's demonstrate the contrapositive proposition of $X$ in Locally FV implies $X$ is a FV process. That is let's take $X$ not an FV process and show it cannot be a locally finite process.

As $X \not\in FV$ ,then there exists an event $A$ such that :
-$P(A)>0$
-$\exists t>0,\int_0^t |dX_s|(\omega)=+\infty$ for any $\omega \in A$.

Then for any sequence of stopping time $\tau^n$ increasing to $\infty$ almost surely, then the stopped process $X^{\tau^n}$ are such that the events $A_n=\{\omega \in \Omega,\int_0^t |dX^{\tau^n}_s|(\omega)=+\infty \} \to A$ almost surely.

This shows as $P(A)>0$ that $X$ cannot be of locally finite variation.

Best regards

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