Proof - Limits of CDF

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
$P\left(\bigcup_{n=1}^\infty A_n\right) = \lim_{_m\to\infty} P\left(\bigcup_{n=1}^m A_n\right)=\lim_{_m\to\infty}P(A_m)$

• On the book proof, it might have been useful to add that for any real $x$, there is an integer $n_x$ such that $n_x\le x\lt n_x+1$, and that $F(n_x)\le F(x)\le F(n_x+1)$. Let $x\to\infty$. Then $F(n_x)$ and $F(n_x+1)\to 1$, so by squeezing $F(x)\to 1$. Commented Aug 17, 2015 at 15:33
• @ André Nicolas - Great stuff. Can't explain how much I appreciate it as I'm self studying these topics. Hope you could shed some light on the 2nd part of my post about the validity of my proof. Commented Aug 18, 2015 at 18:01
• I do not see the point of the double-limit, which complicates matters, Take a sequence $(x_n)$ going to $\infty$, let $A_n$ be the event $X\le x_n$. Then the union of the $A_n$ has probability $1$. Work exactly as in the book proof, but without the restriction $x_n=n$. Commented Aug 18, 2015 at 18:51
• @ André Nicolas - Thanks a lot. Btw, is $y\uparrow\infty$ is same as $y\to\infty$ for any $y\in\mathbb{R}$? (I believe yes) Commented Aug 18, 2015 at 20:23
• Effectively yes. Usually uparrow $a$ means approaches $a$ from below, but in the case "$a=\infty$" there is no alternative to from below. Commented Aug 18, 2015 at 20:30