What you write is true in some ways, though I don't like your choice of notation which can hide a multitude of traps for the unwary. The product sample space $\Omega_{X,Y}$ (note the comma between $X$ and $Y$) is indeed the Cartesian product of $\Omega_{X}$ and $\Omega_{Y}$, that is, for example, if $\Omega_{X} = \{a, b\}$ and $\Omega_{Y} = \{c, d\}$, then $\Omega_{X,Y} = \Omega_X\times \Omega_Y$ is a set of 4 ordered pairs of real numbers, specifically $$\Omega_{X,Y} = \Omega_X\times \Omega_Y = \big\{(a, c),\, (a,d),\, (b,c),\, (b,d)\big\}.$$ Now, the non-trivial sigma algebras $\mathcal F_{X}$ and $\mathcal F_{Y}$ defined on $\Omega_{X}$ and $\Omega_{Y}$ are their respective power sets \begin{align} \mathcal F_{X} &= 2^{\Omega_X} = \big\{\emptyset,\, \{a\},\, \{b\},\, \Omega_X\big\}\\ \mathcal F_{X} &= 2^{\Omega_Y} =\big\{\emptyset,\, \{c\},\, \{d\},\, \Omega_Y\big\} \end{align} but the product sigma algebra $\mathcal F_{X,Y}$ is not $\mathcal F_{X}\times \mathcal F_Y$, that is, $$\mathcal F_{X,Y} = 2^{\Omega_{X,Y}} \neq 2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y.$$ Note, for example, that $2^{\Omega_{X,Y}}$ contains the event $\big\{(a,c),\, (b,d)\big\}$ since this set is a subset of the product sample space $\Omega_{X,Y}$ whereas $2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y$ does not.

Turning to probabilities, it is an "obvious" choice to set the probability of the singleton event $\{(x,y)\}$ (here $x \in \Omega_X, y \in \Omega_Y$) to $P(\{x\})P(\{y\})$ which effectively assumes physical independence (implying statistical independence) of the two experiments, but there is no need for such an assumption since the only requirement is that the joint probability measure $P_{X,Y}$ have marginal measures $P_X$ and $P_Y$ respectively. If, say, $P_X(\{a\}) = P_Y(\{c\}) = p$,then the assignments $$\begin{array}{ll} P_{X,Y}(\{(a,c)\}) = p^2 & P_{X,Y}(\{(a,d)\}) = p(1-p)\\ P_{X,Y}(\{(b,c)\}) = p(1-p) & P_{X,Y}(\{(b,d)\}) = (1-p)^2\\ \\ P_{X,Y}^\prime(\{(a,c)\}) = p & P_{X,Y}^\prime(\{(a,d)\}) = 0 \\ P_{X,Y}^\prime(\{(b,c)\}) = 0 & P_{X,Y}^\prime(\{(b,d)\}) = 1-p \end{array}$$ both correspond to the same marginal distribution $$P_X(\{a\}) = P_Y(\{c\}) = p, ~P_X(\{b\}) = P_Y(\{d\}) = 1-p.$$

In short, it is not enough to glibly write

they correspond to a sample space $\Omega_{XY}$ in the probability space $(\Omega_{XY}, \mathcal{F}_{XY}, P_{XY})$

without thinking a little bit as to what exactly is meant by $\mathcal{F}_{XY}$ and $P_{XY}$, or, in more careful notation, $\mathcal F_{X,Y}$ and $P_{X,Y}$

What you write is true in some ways, though I don't like your choice of notation which can hide a multitude of traps for the unwary. The product sample space $\Omega_{X,Y}$ (note the comma between $X$ and $Y$) is indeed the Cartesian product of $\Omega_{X}$ and $\Omega_{Y}$, that is, for example, if $\Omega_{X} = \{a, b\}$ and $\Omega_{Y} = \{c, d\}$, then $\Omega_{X,Y} = \Omega_X\times \Omega_Y$ is a set of 4 ordered pairs of real numbers, specifically $$\Omega_{X,Y} = \Omega_X\times \Omega_Y = \big\{(a, c),\, (a,d),\, (b,c),\, (b,d)\big\}.$$ Now, the non-trivial sigma algebras $\mathcal F_{X}$ and $\mathcal F_{Y}$ defined on $\Omega_{X}$ and $\Omega_{Y}$ are their respective power sets \begin{align} \mathcal F_{X} &= 2^{\Omega_X} = \big\{\emptyset,\, \{a\},\, \{b\},\, \Omega_X\big\}\\ \mathcal F_{Y} &= 2^{\Omega_Y} =\big\{\emptyset,\, \{c\},\, \{d\},\, \Omega_Y\big\} \end{align} but the product sigma algebra $\mathcal F_{X,Y}$ is not $\mathcal F_{X}\times \mathcal F_Y$, that is, $$\mathcal F_{X,Y} = 2^{\Omega_{X,Y}} \neq 2^{\Omega_X}\times 2^{\Omega_Y} = \mathcal F_X\times \mathcal F_Y.$$ Note, for example, that $2^{\Omega_{X,Y}}$ contains the event $\big\{(a,c),\, (b,d)\big\}$ since this set is a subset of the product sample space $\Omega_{X,Y}$ whereas $2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y$ does not.

Turning to probabilities, it is an "obvious" choice to set the probability of the singleton event $\{(x,y)\}$ (here $x \in \Omega_X, y \in \Omega_Y$) to $P(\{x\})P(\{y\})$ which effectively assumes physical independence (implying statistical independence) of the two experiments, but there is no need for such an assumption since the only requirement is that the joint probability measure $P_{X,Y}$ have marginal measures $P_X$ and $P_Y$ respectively. If, say, $P_X(\{a\}) = P_Y(\{c\}) = p$,then the assignments $$\begin{array}{ll} P_{X,Y}(\{(a,c)\}) = p^2 & P_{X,Y}(\{(a,d)\}) = p(1-p)\\ P_{X,Y}(\{(b,c)\}) = p(1-p) & P_{X,Y}(\{(b,d)\}) = (1-p)^2\\ \\ P_{X,Y}^\prime(\{(a,c)\}) = p & P_{X,Y}^\prime(\{(a,d)\}) = 0 \\ P_{X,Y}^\prime(\{(b,c)\}) = 0 & P_{X,Y}^\prime(\{(b,d)\}) = 1-p \end{array}$$ both correspond to the same marginal distribution $$P_X(\{a\}) = P_Y(\{c\}) = p, ~P_X(\{b\}) = P_Y(\{d\}) = 1-p.$$

In short, it is not enough to glibly write

they correspond to a sample space $\Omega_{XY}$ in the probability space $(\Omega_{XY}, \mathcal{F}_{XY}, P_{XY})$

without thinking a little bit as to what exactly is meant by $\mathcal{F}_{XY}$ and $P_{XY}$, or, in more careful notation, $\mathcal F_{X,Y}$ and $P_{X,Y}$

What you write is true in some ways, though I don't like your choice of notation which can hide a multitude of traps for the unwary. The product sample space $\Omega_{X,Y}$ (note the comma between $X$ and $Y$) is indeed the Cartesian product of $\Omega_{X}$ and $\Omega_{Y}$, that is, for example, if $\Omega_{X} = \{a, b\}$ and $\Omega_{Y} = \{c, d\}$, then $\Omega_{X,Y} = \Omega_X\times \Omega_Y$ is a set of 4 ordered pairs of real numbers, specifically $$\Omega_{X,Y} = \Omega_X\times \Omega_Y = \big\{(a, c),\, (a,d),\, (b,c),\, (b,d)\big\}.$$ Now, the non-trivial sigma algebras $\mathcal F_{X}$ and $\mathcal F_{Y}$ defined on $\Omega_{X}$ and $\Omega_{Y}$ are their respective power sets \begin{align} \mathcal F_{X} &= 2^{\Omega_X} = \big\{\emptyset,\, \{a\},\, \{b\},\, \Omega_X\big\}\\ \mathcal F_{X} &= 2^{\Omega_Y} =\big\{\emptyset,\, \{c\},\, \{d\},\, \Omega_Y\big\} \end{align} but the product sigma algebra $\mathcal F_{X,Y}$ is not $\mathcal F_{X}\times \mathcal F_Y$, that is, $$\mathcal F_{X,Y} = 2^{\Omega_{X,Y}} \neq 2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y.$$ Note, for example, that $2^{\Omega_{X,Y}}$ contains the event $\big\{(a,c),\, (b,d)\big\}$ since this set is a subset of the product sample space $\Omega_{X,Y}$ whereas $2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y$ does not.

Turning to probabilities, it is an "obvious" choice to set the probability of the singleton event $\{(x,y)\}$ (here $x \in \Omega_X, y \in \Omega_Y$) to $P(\{x\})P(\{y\})$ which effectively assumes physical independence (implying statistical independence) of the two experiments, but there is no need for such an assumption since the only requirement is that the joint probability measure $P_{X,Y}$ have marginal measures $P_X$ and $P_Y$ respectively. If, say, $P_X(\{a\}) = P_Y(\{c\}) = p$,then the assignments $$\begin{array}{llll} P_{X,Y}(\{(a,c)\} = p^2 & P_{X,Y}(\{(a,d)\} = p(1-p) & P_{X,Y}(\{(b,c)\} = p(1-p) & P_{X,Y}(\{(b,d)\} = (1-p)^2\\ P_{X,Y}^\prime(\{(a,c)\} = p & P_{X,Y}^\prime(\{(a,d)\} = 0 & P_{X,Y}^\prime(\{(b,c)\} = 0 & P_{X,Y}^\prime(\{(b,d)\} = (1-p) \end{array}$$ both correspond to the same marginal distribution $$P_X(\{a\}) = P_Y(\{c\}) = p, ~P_X(\{b\}) = P_Y(\{b\}) = 1-p.$$

In short, it is not enough to glibly write

they correspond to a sample space $\Omega_{XY}$ in the probability space $(\Omega_{XY}, \mathcal{F}_{XY}, P_{XY})$

without thinking a little bit as to what exactly is meant by $\mathcal{F}_{XY}$ and $P_{XY}$, or, in more careful notation, $\mathcal F_{X,Y}$ and $P_{X,Y}$

What you write is true in some ways, though I don't like your choice of notation which can hide a multitude of traps for the unwary. The product sample space $\Omega_{X,Y}$ (note the comma between $X$ and $Y$) is indeed the Cartesian product of $\Omega_{X}$ and $\Omega_{Y}$, that is, for example, if $\Omega_{X} = \{a, b\}$ and $\Omega_{Y} = \{c, d\}$, then $\Omega_{X,Y} = \Omega_X\times \Omega_Y$ is a set of 4 ordered pairs of real numbers, specifically $$\Omega_{X,Y} = \Omega_X\times \Omega_Y = \big\{(a, c),\, (a,d),\, (b,c),\, (b,d)\big\}.$$ Now, the non-trivial sigma algebras $\mathcal F_{X}$ and $\mathcal F_{Y}$ defined on $\Omega_{X}$ and $\Omega_{Y}$ are their respective power sets \begin{align} \mathcal F_{X} &= 2^{\Omega_X} = \big\{\emptyset,\, \{a\},\, \{b\},\, \Omega_X\big\}\\ \mathcal F_{X} &= 2^{\Omega_Y} =\big\{\emptyset,\, \{c\},\, \{d\},\, \Omega_Y\big\} \end{align} but the product sigma algebra $\mathcal F_{X,Y}$ is not $\mathcal F_{X}\times \mathcal F_Y$, that is, $$\mathcal F_{X,Y} = 2^{\Omega_{X,Y}} \neq 2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y.$$ Note, for example, that $2^{\Omega_{X,Y}}$ contains the event $\big\{(a,c),\, (b,d)\big\}$ since this set is a subset of the product sample space $\Omega_{X,Y}$ whereas $2^{\Omega_X}\times 2^{\Omega_X} = \mathcal F_X\times \mathcal F_Y$ does not.

Turning to probabilities, it is an "obvious" choice to set the probability of the singleton event $\{(x,y)\}$ (here $x \in \Omega_X, y \in \Omega_Y$) to $P(\{x\})P(\{y\})$ which effectively assumes physical independence (implying statistical independence) of the two experiments, but there is no need for such an assumption since the only requirement is that the joint probability measure $P_{X,Y}$ have marginal measures $P_X$ and $P_Y$ respectively. If, say, $P_X(\{a\}) = P_Y(\{c\}) = p$,then the assignments $$\begin{array}{ll} P_{X,Y}(\{(a,c)\}) = p^2 & P_{X,Y}(\{(a,d)\}) = p(1-p)\\ P_{X,Y}(\{(b,c)\}) = p(1-p) & P_{X,Y}(\{(b,d)\}) = (1-p)^2\\ \\ P_{X,Y}^\prime(\{(a,c)\}) = p & P_{X,Y}^\prime(\{(a,d)\}) = 0 \\ P_{X,Y}^\prime(\{(b,c)\}) = 0 & P_{X,Y}^\prime(\{(b,d)\}) = 1-p \end{array}$$ both correspond to the same marginal distribution $$P_X(\{a\}) = P_Y(\{c\}) = p, ~P_X(\{b\}) = P_Y(\{d\}) = 1-p.$$

In short, it is not enough to glibly write

they correspond to a sample space $\Omega_{XY}$ in the probability space $(\Omega_{XY}, \mathcal{F}_{XY}, P_{XY})$

without thinking a little bit as to what exactly is meant by $\mathcal{F}_{XY}$ and $P_{XY}$, or, in more careful notation, $\mathcal F_{X,Y}$ and $P_{X,Y}$