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

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    $\begingroup$ 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 $F(x)\to 1$. $\endgroup$ Commented Aug 17, 2015 at 15:33
  • $\begingroup$ @ 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. $\endgroup$
    – KGhatak
    Commented Aug 18, 2015 at 18:01
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    $\begingroup$ 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 $x_n=n$. $\endgroup$ Commented Aug 18, 2015 at 18:51
  • $\begingroup$ @ 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) $\endgroup$
    – KGhatak
    Commented Aug 18, 2015 at 20:23
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    $\begingroup$ Effectively yes. Usually uparrow $a$ means approaches $a$ from below, but in the case "$a=\infty$" there is no alternative to from below. $\endgroup$ Commented Aug 18, 2015 at 20:30

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