Independent random variables and product spaces Let's say we have a measurable space $(\Omega,\mathcal{A},P)$ and some independent random Variables $X$ and $Y$. Then we know that $P(X+Y \le s)$ is equal to $P(\{a \in \Omega : X(a)+Y(a) \le s\})$.
Now $P_{X}$ together with the Borel sets form a measurable space, analog also with $Y$, so we construct the product space with the mass $P_{X}\otimes P_{Y}$.
The question is, how can we show that $$P(\{a \in \Omega : X(a)+Y(a) \le s\}) = P(X+Y \le s) = (P_{X}\otimes P_{Y})(\{(a,b) \in \mathbb{R^2} : a + b \le s\})$$
The first equality is just definition, but how do we prove the second equality?
 A: This gives detail on my comment: We can use a uniqueness result which is something that can be viewed as more basic than the Caratheodory extension theorem (but is along the same lines):
Lemma 1.6b: If two probability measures agree on a $\pi$-system, then they agree on the sigma algebra generated by that $\pi$-system (from Probability with Martingales by David Williams).
We have two probability measures:

*

*$\mu_1:\mathcal{B}(\mathbb{R}^2)\rightarrow [0,1]$ is defined
$$ \mu_1(C) = P[(X,Y)\in C] \quad \forall C \in \mathcal{B}(\mathbb{R}^2)$$


*$\mu_2:\mathcal{B}(\mathbb{R}^2)\rightarrow [0,1]$ is defined
$$ \mu_2(C) = (P_X\otimes P_Y)(C) \quad \forall C \in \mathcal{B}(\mathbb{R}^2)$$
Define $V$ as the set of all "rectangle-type" sets $A \times B \subseteq\mathbb{R}^2$ such that $A\in \mathcal{B}(\mathbb{R})$ and $B \in \mathcal{B}(\mathbb{R})$. It can be shown that $\sigma(V)=\mathcal{B}(\mathbb{R}^2)$ and that $V$ is a $\pi$-system, since the intersection of any finite number of rectangle sets is another rectangle set. Since $\mu_1(C)=\mu_2(C)$ for all $C \in V$ (recall that $X$ and $Y$ were defined as independent random variables) we must have $\mu_1(C)=\mu_2(C)$ for all $C \in \sigma(V)$.
A: Suppose $(X, Y)$ is a random vector that maps $(\Omega, \mathcal{A}, P)$ to $(\mathbb{R}^2, \mathscr{R}^2)$.  By definition (e.g., see Equation (20.17) in Probability and Measure), the distribution $\mu$ of $(X, Y)$ satisfies:
\begin{align}
\mu(A) = P[(X, Y) \in A], \quad A \in \mathscr{R}^2.
\end{align}
Now take $\mu = P_X \otimes P_Y$, and $A = \{(x, y): x + y \leq s\}$.
