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Given an Ito-process $X(t)$, $t\in[0,T]$

$$X(t)=X_{0}+\int_{0}^{t}F(s)ds + \int_{0}^{t}G(s)dW(s)$$

with $F\in \mathbb{L}^{1}(0,T)$ and $G\in\mathbb{L}^{2}(0,T)$. It is now often claimed that this representation is unique (up to indistinguishability of the processes $F$ and $G$). By linearity it is enough to assume the case $X=0$ and to show that $X_{0}=0$, $F=0$ and $G=0$. Taking $t=0$, it follows immediately that $X_{0}=0$. I now thought one could just use simply the fact that

$$\int_{0}^{t}G(s)dW(s)$$

is a martingale as well as to use Ito-isometry to prove this. If we have shown that $F=0$. Then $$ 0 = \mathbb{E}\left[\left(\int_{0}^{t}G(s)dW(s) \right)^{2}\right]=\mathbb{E}\left[\int_{0}^{t}G^{2}(s)ds \right]$$ by Ito-isometry. Thus $$\int_{0}^{t}G^{2}(s)ds=0$$ almost surely. Can I now already conclude that $G=0$ up to indistinguishability?

Additionally, how do I show that $F=0$?

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  • $\begingroup$ You can invoke Doob-Meyer decomposition to prove it right away or adapt the proof of that to your case. $\endgroup$ – Calculon Mar 25 '18 at 6:26
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You want to show that

$$\int_0^tF(s)\,ds = \int_0^tG(s)\,dW_s$$

implies that both $F$ and $G$ are indistinguishable from $0$. The process on the left has bounded variation. The process on the right is a continuous local martingale that is equal to $0$ at time $0$. Continuous local martingales of bounded variation are constant processes. So both integrals are equal to $0$ at all times. Hence both $F$ and $G$ are equal to $0$ at all times.

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