Assumptions:
- $(\mathsf{X}, \mathcal{X})$ is a measurable space.
- $M_n$ and $L_{n-1}$ are Markov probability kernels for $n=2, \ldots, P$.
- $\mu_n$ be probability measures on $(\mathsf{X}, \mathcal{X})$ for each $n=2, \ldots, P$.
- $\nu$ be a probability measure on $(\mathsf{X}, \mathcal{X})$ with $\mu_1 \ll \nu$.
Does $$ \mu_n(dx_n)\prod_{k=1}^{n-1} L_{k}(x_{k+1}, dx_{k}) \ll \nu(dx_1)\prod_{k=1}^{n-1}M_{k+1}(x_k, dx_{k+1}) \qquad \forall\, n=2, \ldots, P $$ imply this? $$ \mu_n(dx_n) L_{n-1}(x_n, dx_{n-1}) \ll \mu_{n-1}(dx_{n-1}) M_n(x_{n-1}, dx_n) $$ I have a feeling it has to do with the chain rule of Radon-Nikodym derivatives and transitivity.
Attempt
Since the equation below holds $$ \mu_n(dx_n)\prod_{k=1}^{n-1} L_{k}(x_{k+1}, dx_{k}) \ll \nu(dx_1)\prod_{k=1}^{n-1}M_{k+1}(x_k, dx_{k+1}) \qquad \forall\, n=2, \ldots, P, $$ then there must exist a Radon-Nikodym derivative $$ R_n(x_{1:n}) = \frac{d(\mu_n \otimes \prod_{k=1}^{n-1} L_{k})}{d(\nu \otimes \prod_{k=1}^{n-1}M_{k+1})}(x_{1:n}) \qquad \forall\, n=2, \ldots, P. $$