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 as 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.

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.

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

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 as 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.

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 as 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.