Independence of random variables measure theory I wish to show that for two random variables $X$ and $Y$, the condition $P(X\leq x, Y\leq y ) = P(X\leq x)P(Y\leq y)$ implies that X and Y are independent. 
I am approaching this problem from a measure theoretic perspective. So in particular I can write that $P(X\leq x) = \mu_X ((-\infty, x])$ and $P(Y\leq y) = \mu_Y ((-\infty, y])$. Also the independence condition here is that $\sigma(X)$ and $\sigma(Y)$ are independent, where these are the sigma algebras generated by the random variables.
 A: Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $\mathbb{R}$.


*

*Show that $\sigma(X) = \{X^{-1}(B) : B \in \mathcal{B}\}$.  ($\supset$ is clear.  For $\subset$, show the right side is a $\sigma$-algebra.)

*Fix $y \in \mathbb{R}$.  Let $\mathcal{L} = \{ B \in \mathcal{B} : P(X \in B, Y \le y) = P(X \in B) P(Y \le y)\}$.  Show that $\mathcal{L}$ is a $\lambda$-system.  Let $\mathcal{P} = \{ (-\infty, x] : x \in \mathbb{R} \}$.  Show that $\mathcal{P}$ is a $\pi$-system, $\mathcal{P} \subset \mathcal{L}$ and that $\sigma(\mathcal{P}) = \mathcal{B}$.  By the $\pi$-$\lambda$ theorem, $\mathcal{B} \subset \mathcal{L}$.  Conclude that for every $B \in \mathcal{B}$ and $y \in Y$, $P(X \in B, Y \le y) = P(X \in B) P(Y \le y)$.

*Fix $B \in \mathcal{B}$.  Let $\mathcal{L}' = \{C \in \mathcal{B} : P(X \in B, Y \in C) = P(X \in B) P(Y \in C)\}$.  Show that $\mathcal{L}'$ is a $\lambda$-system, and proceed as before to show $\mathcal{B} \subset \mathcal{L}'$.

*We have now shown that $P(X \in B, Y \in C) = P(X \in B) P(Y \in C)$ for all $B, C \in \mathcal{B}$.  By (1), this says that $\sigma(X)$ and $\sigma(Y)$ are independent.
A: Let $\mathcal C$ denote the set of those Borel measurable sets $A$ for which
$$ \Pr(X \in A, Y \le y) = \Pr(X \in A) \Pr(Y \le y) $$
Show that $\mathcal C$ is a sigma field.  Deduce it is equal to the set of Borel measurable sets.
Next for each Borel measurable $A$, let $\mathcal D$ denote the set of those Borel measurable sets $B$ for which
$$ \Pr(X \in A, Y \in B) = \Pr(X \in A) \Pr(Y \in B) $$
and carry on as before.
