This questions is very short. Under the Girsanov Theorem assumptions we have two equivalent probability measures $\mathbb P$ and $\mathbb Q$ and a measurable space $(\Omega,\mathcal F)$, right? We have that $$\mathcal E(Z)=\frac{d\mathbb Q}{d\mathbb P}$$ for some contiuous local martingale $Z$ in $(\Omega,\mathcal F,\mathbb P)$ and $\mathcal E (\cdot)$ is the exponential local martingale operator.
I want to proof that for a semi-martingale $X$ we have $H\bullet X$ is the same in both the spaces $(\Omega,\mathcal F,\mathbb P)$ and $(\Omega,\mathcal F,\mathbb Q)$. I have read a couple of books (e.g. Revuz-Yor) but they are all very fast in this. It seems that they somehow just conclude it. I found a detailed proof in Jean-François Le Gall's book Brownian Motion etc. but even there it is too fast in my opinion. In page 136 the author says more or less the following. Let us denote $(H\bullet X)_\mathbb P$ to emphasize the space. We can write $X=M+V$ with $M$ a continuous local martingale in $(\Omega,\mathcal F,\mathbb P)$ and $V$ a process of bounded variation. We know that by Girsanov we can write $X=M'+V'$ with $M'=M-\langle M,Z\rangle $ a $\mathbb Q$ local martingale.
So $M'$ is a semi-martingale and I know how to integrate that in $(\Omega,\mathcal F ,\mathbb P)$ $$(H\bullet M')_\mathbb P=(H\bullet M)_\mathbb P-\langle (H\bullet M)_\mathbb P,Z\rangle $$ Girsanov says that $(H\bullet M')_\mathbb P$ is a $\mathbb Q$-local-martingale. But then Jean-François Le Gall concludes that every $\mathbb Q$-local-martingale $N'$ has bracket process with $(H\bullet M')_\mathbb P$ equal to $$\langle (H\bullet M')_\mathbb P,N'\rangle =H\bullet \langle M,N'\rangle =H\bullet \langle M',N'\rangle \tag{A} $$ I know that for $N$ a $\mathbb P$-local-martingale we have $$H\bullet \langle M,N\rangle =\langle (H\bullet M)_\mathbb P,N\rangle $$ It seems that such formula is used but I don't know why it would hold for $N'$.
Question. Why does (A) hold?
After thinking and extensive discussing I have an answer. By applying Girsanov the other way we conclude that every $\mathbb Q$-local-martingale $N'$ can be written as $N'=N-\langle N,Z\rangle $ for some $N$ a $\mathbb P$-local-martingale. So we have \begin{align} \langle (H\bullet M')_\mathbb P,N'\rangle=\langle (H\bullet M)_\mathbb P-\langle (H\bullet M)_\mathbb P,Z\rangle ,N'\rangle =\langle (H\bullet M)_\mathbb P,N'\rangle \end{align} Now we use the decomposition that we have \begin{align} \langle (H\bullet M')_\mathbb P,N'\rangle=\langle (H\bullet M)_\mathbb P,N-\langle N,Z\rangle \rangle = \langle (H\bullet M)_\mathbb P,N\rangle =H\bullet\langle M,N\rangle \end{align} Now we can use the decompositin again for $N'$ and $M'$ to get the conclusion.