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I'm trying to understand the proof Novikov Condition and other works in this field (Kazamaki etc) and for this, i have to understand the Girsanov Theorem, which is also a part of the proof. In Karatzas,Shreve -Brownian Motion And Stochastic Calculus we have:

5.1 Theorem (Girsanov (1960), Cameron and Martin (1944))
Let $W$ be a Brownian Motion and $X$ be a continuous local martingale on a filtered probability space $(\Omega, F, (F_t)_{t\geq 0} ), P)$, such that $Z_t(X) := \exp (X - \frac{1}{2} [X])$ is a martingale. Under this condition the defined process $\tilde{W}_t := W_t - \int_0^tX_s ds$ is for every fixed $T \geq 0$ a Brownian Motion on $(\Omega, F_T, F_{0 \leq t\leq T}, P_T)$, where $P_T(A):=E(1_A Z_T(X))$ for all $A \in F_T$.

5.2 Corollary
Let $W$ be the coordinate mapping process on $\Omega := C[0,\infty)$, so that $F_{\infty}^W:= \mathcal{B}(C[0,\infty))$. Let $P$ be Wiener measure on $(\Omega,F_{\infty}^W)$. Let $X$ be a continuous locale martingale. If $Z(X)$ is a martingale, then there is a unique probability measure $\tilde{P}$ satisfying $\tilde{P}_{|F_T} = P_T$, and $\tilde{W}$ as above is a Brownian Motion on $(\Omega, F_{\infty}^W, (F_t^W)_{t\geq 0},\tilde{P})$.

As stated in the same book, we can't just set $T= \infty$ in 5.1, to get 5.2 for a arbitrary probability space. But a lot of works in this field don't mention that and work with 5.1 on arbitray spaces. Is there maybe a argument i don't know? Thanks!

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  • $\begingroup$ Thanks. My problem is: As we can see in 5.2 Corollary $Z$ does not have to be UI martingale, but just an martingale, to let $Q$ as defined be a measure. But this is only true for a class of measure spaces, for example the Wiener Space. In generally cases $Z$ has to be a UI martingale. But a lot of works in this field just define $Q$ without asuming UI or a special case of measure space. $\endgroup$
    – jekodo
    Commented Nov 12, 2019 at 22:04
  • $\begingroup$ No, in $5.2$ the processes $Z$ and $\tilde{W}$ are not stopped. They are processes on timeline $[0,\infty)$. $\endgroup$
    – jekodo
    Commented Nov 13, 2019 at 16:12

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