Metropolis-Hastings kernel in measure theory

Question

I'm facing difficulties in formulating the Metropolis-Hastings kernel for a specific problem where I need to sample from a probability distribution involving both discrete and continuous degrees of freedom (e.g., a state $s=(x,c)$ representing the position $x$ of a particle and its colour $c$).

1. Is it safe to assume that this probability has a density $p$ with respect to some generic measure $\mu$ on the space $(S,\Sigma)$?
2. Can I define the proposal kernel $Q$ to be $Q(s,A)$ for $A\subset S$? E.g. say I draw my proposal $s'=(x',c')$ by sampling $x'$ from a gaussian and $c'$ from a Bernoulli.
3. Can I assume this proposal kernel has density $q$ with respect to the same measure $\mu$? I'd say this measure is neither the Lebesgue nor the counting measure as the measure space is neither discrete nor continuous.
4. Can I then write the Metropolis-hastings ratio as $$ r(x,x')=\frac{p(x')q(x',x)}{p(x)q(x,x')}, $$ that is, using the densities with respect to $\mu$ and not the measures?

Answer 1

You asked another question that seems somewhat likely to be closed for not being research-level mathematics. I will include my posted answer to that here because that one may get closed.

One does not call a "measurable space" $(S,\Sigma)$ "discrete" or "continuous", although one may call a particular measure on that space "discrete" or "continuous".

You seem to suggest that the $x$ component of $s=(x,c)$ may take values in $\mathbb R^n$ and $c$ may be a Bernoulli random variable, i.e. $c\in\{0,1\}.$ In that case the proposal kernel would typically have a density with respect to the product measure whose two factors are Lebesgue measure on $\mathbb R^n$ and counting measure on $\{0,1\}.$ So your point #2 seems unproblematic.

Appendix: How I answered that other question

A discrete probability distribution is one made up entirely of point masses; i.e. there is some set $S_1\subseteq S$ for which, for every $s\in S_1,$ the probability $P\{s\}$ assigned to the set $\{s\}$ is positive, and $\sum_{s\in S_1} P\{s\} = 1.$

A probability measure $P$ is a mixture – a weighted average – of a discrete probability distribution and a distribution with no discrete part precisely if $\sum_{s\in S_1} P\{s\}<1$ where $S_1 = \{ s\in S : P\{s\}>0\} \ne\varnothing. $

A probability distribution has no discrete part if for all $s\in S,$ $P\{s\}=0.$

In all of the above I avoided the word "continuous." More on that below.

There is nothing in your description that implies that a probability measure on your space has any discrete part.

If $X$ is a random variable taking values in the space you describe, certainly the $c$ component of $X$ – call it $c(X)$ – is a random variable whose probability distribution is discrete.

But that doesn't mean $X$ itself has a positive probability of being at any one point.

"Continuous" can mean having a density, which in many contexts is equivalent to being absolutely continuous with respect to some underlying measure. Having a density is always a sufficient condition for absolute continuity.

But sometimes "continuous" means having a continuous cumulative distribution function. That is a sufficient condition for having no discrete part, but does not entail that there is a density function.