For a cdf, defined as $F(x)=P(X\le x)$, in order to prove $\lim\limits_{x\,\uparrow\,\infty}F(x)=1$, I've two concerns: (1) Some concern about a proof from a book, and (2)Validity of a proof that I've chalked out. Would really appreciate your help on this.
(1) Concern for a proof from a book
The proof goes like this
$\Omega=\mathbb{R}=(-\infty,\infty)=\bigcup_{n=1}^\infty(-\infty,n]$
$P(\Omega)=1$, and
$P\left(\bigcup_{n=1}^\infty (-\infty, n]\right) = \lim_{_m\to\infty} P((-\infty, m])$ $\qquad$_...by ref (A). see below_ $\qquad\qquad\qquad\qquad=\lim_{_m\to\infty} P(X\le m) = \lim_{_m\to\infty} F(m)$
Equating left and right side, we get $\lim_{_m\to\infty} F(m) = 1$
My concern here is that both $n$ and $m$ $\in\mathbb{N^+}$ where as the domain of a cdf must be real number i.e. $x \in\mathbb{R}$ by definition. Thus above proof is valid for $x$ $\in\mathbb{N^+}$ but not for $x\in\mathbb{R}$. Am I missing something!!
(2) Validity of my proof
If we define intervals $(-\infty,x_1]$, $(-\infty,x_2]$.. such that $x_1, x_2,..\in\mathbb{R}$, and $x_1\lt x_2\lt ..\lt x_n..\lt x$ and then we've non-decreasing events (intervals) where $x_n\uparrow x$. Further, if we make $x\to\infty$, then by the above setting we see that as $n\to\infty$ we've $x\to\infty$.
Now we can proceed as,
$\Omega=\mathbb{R}=(-\infty,\infty)=\bigcup_{n=1}^\infty(-\infty,x_n]$
$P(\Omega)=1$, and
$P\left(\bigcup_{n=1}^\infty (-\infty, x_n]\right) = \lim_{n\to\infty} P((-\infty, x_n])$ $\qquad\qquad\qquad$_...step (I)_ $\qquad\qquad\qquad\qquad=\lim_{x_n\uparrow x,\; x\to\infty} P((-\infty, x])$$\qquad\qquad\qquad\qquad$_...step (II)_
$\qquad\qquad\qquad\qquad= \lim_{x\uparrow\infty} P(X\le x) = \lim_{x\uparrow\infty} F(x) $ $\qquad\qquad$_...step (III)_
My concerns are:
i. Is this proof valid?
ii. Is the deduction of step (II) from (I) valid/math-rigor (especially the limit part)?
iii Is $y\uparrow\infty$ is same as $y\to\infty$ for any $y\in\mathbb{R}$?
References
(A)If events $A_n$ are non-decreasing in the sense $A_n\subset A_{n+1}$, then
$P\left(\bigcup_{n=1}^\infty A_n\right) = \lim_{_m\to\infty} P\left(\bigcup_{n=1}^m A_n\right)=\lim_{_m\to\infty}P(A_m)$