How do we need to define the acceptance ratio of the Metropolis-Hastings algorithm for a vanishing proposal density?

Question

I've seen that different authors define the acceptance probability $\alpha$ of the Metropolis-Hastings algorithm with target distribution density $p$ and proposal kernel density $q$ differently; some define $$\alpha(x,y):=\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\\color{red}1&\text{, otherwise}\end{cases}\tag1$$ and others define $$\alpha(x,y):=\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\\color{red}0&\text{, otherwise}.\end{cases}\tag2$$

So, is one of them wrong or does it simply not matter which one we use?

Answer 1

While the Markov chain $(X_t)$ is producing values such that $p(x_t)=0$ it has clearly not "reached" stationarity, so it should keep moving over the state space. Any way to move around until the support of $p$ is reached is valid, including (1) of course.

However, solution (2) is clearly wrong since, if $p(x)=0$, the chain will never move away from its current value.

Note also that$$p(x)q(x,y)=0$$implies $$p(x)=0\qquad \text{a.s.}$$when $y$ is a realisation from the density $q(x,\cdot)$.