For your first question, if $X$ and $Y$ are two boolean locale then $X \times Y$ is boolean only if $X$ or $Y$ is discrete. So unless $\neg \neg A$ or $\neg \neg B$ are discrete, $\neg \neg A \times \neg \neg B$ and $\neg \neg (A \times B)$ cannot be isomorphic because one is boolean and the other is not.

For you second question, the answer is yes in a lot a situations, but for completely stupid reasons that will more probably show you that you are not asking the good question: basically all the set you mention in you question are empty !

Indeed, If $A$ is the locale corresponding to a separable metric space with no isolated point, (for example $A =\mathbb{R}$ or $A=\mathbb{R}^n$), then there exist no probability measure on $\neg \neg A$.

In particular probablity measure on $\neg \neg \mathbb{R}$ does not correspond to meansure absolutely continuous with respect to the Lebesgue measure. There is indeed a boolean locale that I'm going to all $\mathbb{R}^{Lebesgue}$ endowed with a morphism to $\mathbb{R}$ such that probability measure on $\mathbb{R}^{Lebesgue}$ corresponds to probablity measure on $\mathbb{R}$ absolutely continuous with respect to the Lebesgue measure but this locale have not much to do with $\neg \neg \mathbb{R}$.

I might be wrong on that, but I don't think that $\mathbb{R}^{Lebesgue}$ is a sublocale of $\mathbb{R}$ (the map to $\mathbb{R}$ is a monomorphism, but I don't think it is an inclusion)

More precisely, $\mathbb{R}^{Lebesque}$ is exactly the exemple of pointless topos introduce in SGA4: its opensublocales are the measurable subsets of $\mathbb{R}$ modulo equality almost every where.

For your first question, if $X$ and $Y$ are two boolean locale then $X \times Y$ is boolean only if $X$ or $Y$ is discrete. So unless $\neg \neg A$ or $\neg \neg B$ are discrete, $\neg \neg A \times \neg \neg B$ and $\neg \neg (A \times B)$ cannot be isomorphic because one is boolean and the other is not.

For you second question, the answer is yes in a lot a situations, but for completely stupid reasons that will more probably show you that you are not asking the good question: basically all the set you mention in you question are empty !

Indeed, If $A$ is the locale corresponding to a separable metric space with no isolated point, (for example $A =\mathbb{R}$ or $A=\mathbb{R}^n$), then there exist no probability measure on $\neg \neg A$.

In particular probablity measure on $\neg \neg \mathbb{R}$ does not correspond to meansure absolutely continuous with respect to the Lebesgue measure. There is indeed a boolean locale that I'm going to all $\mathbb{R}^{Lebesgue}$ endowed with a morphism to $\mathbb{R}$ such that probability measure on $\mathbb{R}^{Lebesgue}$ corresponds to probablity measure on $\mathbb{R}$ absolutely continuous with respect to the Lebesgue measure but this locale have not much to do with $\neg \neg \mathbb{R}$.

I might be wrong on that, but I don't think that $\mathbb{R}^{Lebesgue}$ is a sublocale of $\mathbb{R}$ (the map to $\mathbb{R}$ is a monomorphism, but I don't think it is an inclusion)

More precisely, $\mathbb{R}^{Lebesgue}$ is exactly the example of pointless topos introduced in SGA4: its opensublocales are the measurable subsets of $\mathbb{R}$ modulo equality almost every where.

For your first question, if $X$ and $Y$ are two boolean locale then $X \times Y$ is boolean only if $X$ or $Y$ is discrete. So unless $\neg \neg A$ or $\neg \neg B$ are discrete, $\neg \neg A \times \neg \neg B$ and $\neg \neg (A \times B)$ cannot be isomorphic because one is boolean and the other is not.

For you second question, the answer is yes for a lot a situations, but for completely stupid and uninteresting reasons that will more probably show you that you are not asking the good question:

If $A$ is the locale corresponding to a separable metric space with no isolated point, (for example $A =\mathbb{R}$ or $A=\mathbb{R}^n$), then there exist no probability measure on $\neg \neg A$.

In particular probablity measure on $\neg \neg \mathbb{R}$ does not correspond to meansure absolutely continuous with respect to the Lebesgue measure. There is indeed a boolean locale that I'm going to all $\mathbb{R}^{Lebesgue}$ endowed with a morphism to $\mathbb{R}$ such that probability measure on $\mathbb{R}^{Lebesgue}$ corresponds to measure on $\mathbb{R}$ absolutely continuous with respect de the Lebesgue measure but this locale have not much to do with $\neg \neg \mathbb{R}$.

I might be wrong on that, but I don't think that $\mathbb{R}^{Lebesgue}$ is a sublocale of $\mathbb{R}$ (the map to $\mathbb{R}$ is a monomorphism, but I don't think it is an inclusion)

More precisely, $\mathbb{R}^{Lebesque}$ is exactly the exemple of pointless topos introduce in SGA4: its opensublocales are the measurable subsets of $\mathbb{R}$ modulo equality almost every where.

For your first question, if $X$ and $Y$ are two boolean locale then $X \times Y$ is boolean only if $X$ or $Y$ is discrete. So unless $\neg \neg A$ or $\neg \neg B$ are discrete, $\neg \neg A \times \neg \neg B$ and $\neg \neg (A \times B)$ cannot be isomorphic because one is boolean and the other is not.

For you second question, the answer is yes in a lot a situations, but for completely stupid reasons that will more probably show you that you are not asking the good question: basically all the set you mention in you question are empty !

Indeed, If $A$ is the locale corresponding to a separable metric space with no isolated point, (for example $A =\mathbb{R}$ or $A=\mathbb{R}^n$), then there exist no probability measure on $\neg \neg A$.

In particular probablity measure on $\neg \neg \mathbb{R}$ does not correspond to meansure absolutely continuous with respect to the Lebesgue measure. There is indeed a boolean locale that I'm going to all $\mathbb{R}^{Lebesgue}$ endowed with a morphism to $\mathbb{R}$ such that probability measure on $\mathbb{R}^{Lebesgue}$ corresponds to probablity measure on $\mathbb{R}$ absolutely continuous with respect to the Lebesgue measure but this locale have not much to do with $\neg \neg \mathbb{R}$.

I might be wrong on that, but I don't think that $\mathbb{R}^{Lebesgue}$ is a sublocale of $\mathbb{R}$ (the map to $\mathbb{R}$ is a monomorphism, but I don't think it is an inclusion)

More precisely, $\mathbb{R}^{Lebesque}$ is exactly the exemple of pointless topos introduce in SGA4: its opensublocales are the measurable subsets of $\mathbb{R}$ modulo equality almost every where.