Independence of Random Variables and Distribution Functions Let $X_1, X_2,\ldots$ be random variables on $(\Omega, \mathcal{A}, \mathbb{P})$.
If $\mathbb{P}(X_1 \leq x, X_2 \leq y)=\mathbb{P}(X_1 \leq x)\mathbb{P}(X_2 \leq y)$ for all $x,y \in \mathbb{R}$.
Show that $X_1$ and $X_2$ are independent.
Really struggling as how to do this inclusion.
This may help:

 A: Standard argument uses the following two ingredients:

Proposition. The collection $\mathcal{A} = \{ (-\infty, x] : x \in \Bbb{R} \}$ is a $\pi$-system that generates the Borel $\sigma$-algebra on $\Bbb{R}$

and 

Dynkin's $\pi$-$\lambda$ Theorem. Let $\mathcal{D}$ be any Dynkin system that contains a $\pi$-system $\mathcal{A}$. Then $\mathcal{D}$ also contains $\sigma(\mathcal{A})$.

Here, a Dynkin system (or $\lambda$-system) $\mathcal{D}$ is a collection of subsets of $\Omega$ that is closed in both complement and disjoint limit:


*

*$\Omega \in \mathcal{D}$,

*If $A \in \mathcal{D}$, then so is $A^{c} \in \mathcal{D}$.

*If $A_{n} \in \mathcal{D}$ are disjoint, then $\cup_{n=1}^{\infty}A_{n} \in \mathcal{D}$.


(This definition is not the same as the classical definition as collections which are closed under subtraction and monotone limit, but they are indeed equivalent. See here.) Then we can prove the following theorem (from Section 2.1, Durrett):

Proposition. If $\mathcal{A}_{n}$ are independent $\pi$-systems, then $\sigma(\mathcal{A}_{n})$ are also independent.

Proof. Let $\mathcal{D}_{1}$ be defined as
$$\mathcal{D}_{1} = \{ A \in \sigma(\mathcal{A}_{1}) : A, A_{2}, \cdots, A_{n} \text{ are independent for all } A_{2} \in \mathcal{A}_{2}, \cdots, A_{n} \in \mathcal{A}_{n} \}.$$
By using the countable additivity of $\Bbb{P}$, we easily check that $\mathcal{D}_{1}$ is a Dynkin system that contains $\mathcal{A}_{1}$, hence $\mathcal{D}_{1} = \sigma(\mathcal{A}_{1})$ by the Dynkin's $\pi$-$\lambda$ Theorem. This shows that $\mathcal{A}_{2}, \cdots, \mathcal{A}_{n}, \sigma(\mathcal{A}_{1})$ are independent. Now apply this result to prove the independence of $\mathcal{A}_{3}, \cdots, \mathcal{A}_{n}, \sigma(\mathcal{A}_{1}), \sigma(\mathcal{A}_{2})$ and so on until we reach the desired conclusion. ////
Now you can combine this theorem with the proposition above to conclude that

Corollary. If $\Bbb{P}(X_{1} \leq x_{1}, \cdots, X_{n} \leq x_{n}) = \Bbb{P}(X_{1} \leq x_{1}) \cdots \Bbb{P}(X_{n} \leq x_{n})$ for any $x_{1}, \cdots, x_{n} \in \Bbb{R}$, then $X_{n}$ are independent.

A: You want to prove that for all intervals, right? Here I will use the semi-closed intervals of the type (a,b] (it does not matter witch kind of interval you will use because all of then generates the Borel $\sigma$-field )
$$P(X \in (a,b],Y \in (c,d])=P(X \in (a,b])\, P(Y \in (c,d]).$$
But then, you just need to remember, 
\begin{align*}
P(X \le x,Y \in (c,d]) & = P(X \le x,Y \le d) - P(X \le x,Y \le c)\\ 
&= P(X \le x)[P(Y \le d) - P(Y \le c)] \\
&= P(X \le x)\cdot P(Y \in (c,d]);& \forall x,c,d \in \mathbb{R}, c<d 
\end{align*}
Now, you just have to do the same with the $X$ part.
