I have some trouble understanding changes of measure for jump processes. I guess I'm missing some important bit of the theory.
Consider a simple example. Let $N_t$ be a standard Poisson process with constant intensity $\lambda_t=1$ adapted to filtration $(\mathcal F_t)_{t\geq 0}$. We wish to change the intensity to $\mu_t=X_{t-}$ by performing a change of measure with Radon-Nikodym derivative $$ \left.\frac{d\tilde P}{dP}\right|_{\mathcal F_t} = L_t = \prod_{n\geq 1} \mu_{T_n}1_{T_n \leq t} \exp\left(- \int_0^t (1-\mu_s)\lambda_s ds \right) $$
This yields $$ L_t= \prod_{n\geq1}^{N_t} X_{T_n-} \exp\left( \int_0^t X_{s-}ds - t\right). $$ This is an equivalent change of measure if $\mathbb E L_1 = 1$. My question is: is this an equivalent change of measure? And if yes, what is wrong with the following argumentation:
By changing measure we alter the intensity of the process $N_t$. We pass from constant $\lambda_t=1$ to non-constant $\mu_t=X_{t-}$, and clearly $P( \lambda_. >1)=0$ but $\tilde P( \mu_.>1)>0$ if we assume $X_0>0$.