Monotone Convergence Theorem applied to the limit of a real-valued function 
I have a given function $f(x,\omega)\geq 0$, where
$x\in\mathbb{R}$ is a parameter and $\omega\in\Omega$, some sampling
space. $f(x,\omega)$ is increasing in $x$ for any $\omega$. I am interested in showing that
$$\lim_{x\to\infty}\int_\Omega
f(x,\omega)d\omega=\int_\Omega \lim_{x\to\infty} f(x,\omega)d\omega.$$

What I have done thus far:
Pick any sequence $\{x_n\}_{n\in\mathbb{N}}$, such that $x_n\uparrow\infty$ when $n\to\infty$. Define $f_n(\omega)\equiv f(x_n,\omega)$ for every $n\in\mathbb{N}$, and $f(\omega)\equiv \lim_{n\to\infty}f_n(\omega)=\lim_{x\to\infty}f(x,\omega)$. Then, as $f_n(\omega)\geq 0$ and $f_n(\omega)\uparrow f(\omega)$, the Monotone Convergence Theorem implies that
$$\lim_{n\to\infty}\int_\Omega f_n(\omega)d\omega=\int_\Omega f(\omega)d\omega.$$
What I don't know is how to go back from having the MCT working for every sequence $\{x_n\}_{n\in\mathbb{N}}$, such that $x_n\uparrow\infty$ when $n\to\infty$, to working for $x\to\infty$, as the latter technically is an uncountable index. I found this post but I find the answers highly unprecise with the details. Any help is greatly appreciated.
 A: Let $I_x:=\int_{\Omega }f(x,\omega )\mathop{}\!d \omega $, then if for all sequences $\{x_n\}$ that diverges to infinity you have that $\lim_{n\to \infty }I_{x_n}=I$, for some $I\in \mathbb{R}$, then $\lim_{x\to \infty }I_x=I$.
Proof: if it would not be true then, by the logical negation of a convergent functional limit, there exists some $\epsilon >0$ such that for any $\delta >0$ exists some $x_\delta >\delta $ such that $|I_{x_\delta }-I|>\epsilon $. Now taking $\delta =n$ for each $n\in \mathbb N $ we have a sequence $\{x_n\}$ that diverges to infinity but $|I_{x_n}-I|>\epsilon $ for all $n$, what contradicts the assumption that $\lim_{n\to \infty }I_{x_n}=I$.$\Box$
By last its easy to see that if $x<y\implies I_x\leqslant I_y$ for all $x,y\in \mathbb{R}$ then $\lim_{n\to \infty }I_{x_n}=\sup_{x\in \mathbb{R}}I_x$ for all sequences $\{x_n\}$ diverging to infinity (you can prove it again by contradiction, assuming that $\lim_{n\to \infty }I_{x_n}<\sup_{x\in \mathbb{R}}I_x$ and using the definition of supremum).
