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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.

StubbornAtom
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Heatconomics
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  • Maybe if you work with the explicit definitions of limits, you can make the transition. Take $\epsilon>0$ and show that $\lim_{x\to \infty} \int_\Omega f(x,\omega)d\omega =\int_\Omega f(\omega)d\omega$ using what you've already shown. – Keen-ameteur Oct 31 '20 at 21:57
  • @keen-ameteur thanks. Actually I think it works as follows. Prove by contradiction. If the limit of the integral in $x$ diverges from the integral of the limit in $x$, there exists some $epsilon>0$, such that for any $X$, there is $x>X$ such that the difference of the limit of the integral and the integral of the limit is more than $\epsilon$. Then construct a discrete sequence with $X=1,2,\dots$ To contradict the MCT I know is true for discrete sequences. – Heatconomics Oct 31 '20 at 23:20

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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).

Masacroso
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  • Thank you so much. Funny thing is I just thought this was a good approach for it! Happy to see it materialized! Cheers! – Heatconomics Oct 31 '20 at 23:35