I'm running the Metropolis-Hastings algorithm on a product space $\tilde E:=I\times E'$, where $I$ is a finite nonempty set and $E'=\bigcup_{i\in I}E'_i$ with $E'_i:=[0,1)^{d_i}$ for $i\in I$.
Given the current state $\tilde x=(i,x')$, I'm perturbing $\tilde x$ to $\tilde z=(k,z')\sim\tilde U(\tilde x,\;\cdot\;)$, where $\tilde U$ is a transition kernel on $\tilde E$. Moreover, I'm sampling $j\sim\tilde T(\tilde z,\;\cdot\;)$, where $\tilde T$ is a transition kernel from $\tilde E$ to $I$.
Now, my final proposal is given by $(j,y')$, where $y':=\varphi_j^{-1}(y)$ and $y:=\varphi_k(z')$. The mappings $\varphi_l$, $l\in I$, are measurable surjections from $E'_l$ to a probability space $(E,\mathcal E,\mu)$ and are invertible outside a $\mu$-null set $N$.
Everything is constructed in a way ensuring that the $y$ in the proposal scheme described above lies outside of $N$ with probability $1$. However, in practice, it still happens that I encounter $y\in N$. In that case, the acceptance probability of the Metropolis-Hastings algorithm will be $1$.
What am I supposed to do now? I need to accept an ill-formed state $(j,y')$, since $y'$ cannot be computed. I'd like to choose $y'=x'$ in this case, but is this possible in the Metropolis-Hastings framework? This would correspond to some kind of "Dirac transition" (which has no density) ...
