Let $M_n$ be a Markov chain on $\mathcal{X}$ with transition kernel $P$, i.e. \begin{equation} \mathrm{Pr}(M_{n+1}\in A|M_0=x) = \int_A P(x,dy). \end{equation} For each $x$ in $\mathcal{X}$ kernel $P(x,dy)$ is also a measure on $\mathcal{X}$ in the following sense: for a subset $A$ of $\mathcal{X}$ \begin{equation} P(x, A) := \int_{A} P(x, dy). \end{equation}
Suppose $\pi$ is a stationary measure for the chain, i.e. $\pi P=\pi$. My question is: as a measure, is $P(x,dy)$ absolutely continuous with respect to $\pi$?
If $\pi(A)=0$, then because $\pi$ is stationary this implies $[\pi P](A)=0$. From here, \begin{align*} [\pi P](A) & = \int P(x, A) \pi(dx). \end{align*} Knowing $P(x, A)$ is nonnegative for each $x$ implies $P(x, A)=0$ for all $x$, and this shows $P(x,\cdot)<< \pi(\cdot)$. Is this correct?