Prove that $\lim_{x \rightarrow \infty, y \rightarrow \infty} F_{XY}(x, y) = 1. $

Question

Prove that $\lim_{x \rightarrow \infty, y \rightarrow \infty} F_{XY}(x, y) = 1. $

Attempt:

$\lim_{x \rightarrow \infty, y \rightarrow \infty} F_{XY}(x, y) = \lim_{x \rightarrow \infty, y \rightarrow \infty} P(X \leq x, Y \leq y) = P(X \leq \infty, Y \leq \infty) = P(\{R_X\}\cap\{R_Y\}) = P(R_{XY}) = 1. $

Where $R_X$ denotes the range of $X$, $R_Y$ denotes the range of $Y$, and $R_{XY}$ denotes the range of the bivariate r.v. $(X, Y)$.

Official proof:

Since $\{X \leq \infty, Y \leq \infty\} = S$, we have

$$ P(\{X \leq \infty, Y \leq \infty\}) = F_{XY}(\infty, \infty) = P(S) = 1. $$

I want to know if my proof is valid. I think so because, well, suppose it's a univariate problem, then $P(R_X) = P(S) = 1.$ But in the bivariate case, I'm unsure if $R_X \cap R_Y = R_{XY}$. Looks like a Lemma I'd have to prove separately. But I like my proof better than just saying that $\{X \leq \infty, Y \leq \infty\} = S$, because this last equation isn't very clear in my head.

Edit:

Okay, I have already found a counter example to $R_X \cap R_Y = R_{XY}$.

So suppose my attempt is, then

$\lim_{x \rightarrow \infty, y \rightarrow \infty} F_{XY}(x, y) = \lim_{x \rightarrow \infty, y \rightarrow \infty} P(X \leq x, Y \leq y) = P(X \leq \infty, Y \leq \infty) = P(R_{XY}) = 1. $

Answer 1

Take any sequence $(x_n,y_n)\nearrow(\infty,\infty)$. Then $$ \lim_{n\to\infty}F_{X,Y}(x_n,y_n)=\lim_{n\to\infty}\mathsf{P}(X\le x_n,Y\le y_n)=\mathsf{P}\!\left(\bigcup_{n\ge 1}\{X\le x_n,Y\le y_n\}\right)=1. $$

Answer 2

Something like that $$ \mathbb{P}(\mathbb{R}^2) = 1 = \int_{\mathbb{R}^2} dF_{X,Y}(x,y) $$ where $$ \lim_{(x,y) \to (\infty, \infty) } F_{X,Y}(x,y) = \int_{\mathbb{R}^2} dF_{X,Y}(x,y) = \int \int f_{X,Y}(x,y) dx dy = 1 $$