I am interested in understanding how to prove the following theorem, where all integrals are understood in the sense of Darboux (Riemann).
Theorem: Suppose that $f: \Omega \to \mathbb{R}$ is such that for all $x \in \Omega \subset \mathbb{R}^n$, $f(x) \geq 0$. Also, let $\{ \Omega_n \}_{n \in \mathbb{N}}$ and $\{ \Omega'_n \}_{n \in \mathbb{N}}$ be exhaustion of $\Omega$ (to be defined in a moment), and for each $\Omega_n$ and $\Omega'_n$, $f$ is integrable on them. Then, $ \lim \limits_{n \to \infty} \int_{\Omega_n} f$ exists if and only if $ \lim \limits_{n \to \infty} \int_{\Omega'_n} f$ exists. Moreover,
$$ \lim \limits_{n \to \infty} \int_{\Omega_n} f = \lim \limits_{n \to \infty} \int_{\Omega'
_n} f.$$
Let me now define what is an exhaustion of an arbitrary (not necessarily bounded) set $\Omega \subset \mathbb{R}^n$.
Definition: An exhaustion of $\Omega \subset \mathbb{R}^n$ is a sequence of regions $\{ \Omega_n \}_{n \in \mathbb{N}}$ such that each $\Omega_n$ is closed and bounded, and for each $k$, $\Omega_k \subset \Omega_{k+1}$, and $\bigcup \limits_{n=1}^{\infty} \Omega_n = \Omega$.
I have tried to do the following so please let me know if you think all steps are reasonable or if you have a different strategy in mind. I also have some particular questions about some of the steps so I appreciate help with those as well.
As $f(x) \geq 0$ for all $x \in \Omega$ and because $\Omega_n \subset \Omega$, for all $n$, $f(x) \geq 0$ for all $x \in \Omega_n$. So, for each $n$, as $f$ is integrable on $\Omega_n$, we have the following inequality.
$$ \int \limits_{\Omega_n} f \leq \int \limits_{\Omega_{n+1}} f $$
So, $\{ \int \limits_{\Omega_n} f \}_{n \in \mathbb{N}}$ is a monotonically increasing sequence of real numbers. Let us assume that it is bounded, so supremum $L$ for values in the sequence exists and, therefore, sequence converges (to supremum).
Now, if $\{ \Omega'_n \}_{n \in \mathbb{N}}$ is another exhaustion of $\Omega$, consider a particular set $\Omega_k'$ for some $k$. For each $n$, $f$ is integrable on both $\Omega_k'$ and $\Omega_n$ (assumption of theorem). My idea is to consider sequence of sets $\{ \Omega_k' \cap \Omega_n \}_{n \in \mathbb{N}}$.
Question 1: If $f$ is integrable on $\Omega_k'$ and $\Omega_n$, can I conclude that $f$ is integrable on $\Omega_k' \cap \Omega_n$? If $f$ was continuous, then I believe because $\Omega_k'$ and $\Omega_n$ are both closed and bounded then $\Omega_k' \cap \Omega_n$ is also closed and bounded and, therefore, compact. Then, $f$ would be also uniformly continuous on this intersection and so $f$ would be integrable. However, here we are not assuming that $f$ is continuous, so I am not sure how to proceed.
Let us for now assume that indeed for all $k$ and $n$, $f$ is integrable on $\Omega_k' \cap \Omega_n$. In that case, as $\Omega_k' \cap \Omega_n \subset \Omega_n$ and $f(x) \geq 0$ for all $x \in \Omega_n$ we get that for all $k,n$:
$$ \int \limits_{\Omega_k' \cap \Omega_n} f \leq \int \limits_{\Omega_n} f. $$
Moreover, I believe that $\{ \Omega_k' \cap \Omega_n \}_{n \in \mathbb{N}}$ is an exhaustion of $\Omega_k'$ so that because $f$ is integrable on $\Omega_k' \subset \Omega$ (assumption of theorem and definition of exhaustion), we have:
$$ \lim \limits_{n \to \infty} \int \limits_{\Omega_k' \cap \Omega_n} f = \int \limits_{\Omega_k'} f.$$
Question 2: Is $\{ \Omega_k' \cap \Omega_n \}_{n \in \mathbb{N}}$ an exhaustion of $\Omega_k'$?
In this case, by previous inequalities, we have the following.
$$ \int \limits_{\Omega_k'} f = \lim \limits_{n \to \infty} \int \limits_{\Omega_k' \cap \Omega_n} f \leq \lim \limits_{n \to \infty} \int_{\Omega_n} f$$
This bound is independent of $k$ and so sequence $\{ \int \limits_{\Omega_k'
} f \}_{k \in \mathbb{N}}$ which is also monotonically increasing is bounded by $\lim \limits_{n \to \infty} \int_{\Omega_n} f$ and, thus, converges to the supremum of values $\{ \int \limits_{\Omega_k'} f \}_{k \in \mathbb{N}}$. In particular, it means the following.
$$ \lim \limits_{n \to \infty} \int \limits_{\Omega_n'} f \leq \lim \limits_{n \to \infty} \int \limits_{\Omega_n} f $$
Repeating argument by replacing $\Omega_k'$ with $\Omega_k$ and $\Omega_n$ with $\Omega_n'$, we also conclude the following, and, therefore conclude the theorem.
$$ \lim \limits_{n \to \infty} \int \limits_{\Omega_n} f \leq \lim \limits_{n \to \infty} \int \limits_{\Omega_n'} f $$
Question 3: Is the remainder of the proof correct?
Best Answer
Your proof is essentially right; the only thing which you have to justify is why for each $k\in\Bbb{N}$, you have the equality \begin{align} \lim \limits_{n \to \infty} \int \limits_{\Omega_k' \cap \Omega_n} f = \int \limits_{\Omega_k'} f. \end{align}
In other words, you have to justify why the integral as previously defined for a bounded set over which $f$ is Riemann-integrable coincides with this new definition as a limit of integrals over exhaustions.
By writing the integrals of $f$ over bounded sets as integrals, over a large rectangle, of $f$ multiplied by appropriate characteristic functions, one immediately sees that this equality amounts to proving that volumes of rectifiable sets satisfy what is known as continuity from above. More generally, the result also follows from either the dominated convergence theorem or the monotone convergence theorem; both these are theorems are valid in the setting of Riemann integrals as well, and an elementary proof is provided in the article Taking Limits Under The Integral Sign by F.Cunningham.
Note that when I speak of "elementary proof", I simply mean that not much prerequisites are required to follow along, but I would still classify this as a hard theorem to prove, simply because one doesn't have the proper foundation to prove it systematically using measure theory (which honestly sounds like abstract nonsense, but makes life a billion times easier).
Some remarks about the article: Lemma 1 (Arzela's lemma) is a very technical (but intuitively plausible) lemma, and is where the difficult lies. It would follow easily if one knew some measure theory. Particularly, by letting $\lambda$ be the Lebesgue measure, \begin{align} \lambda\left(\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}F_m\right)=\lim_{n\to\infty}\lambda\left(\bigcup_{m=n}^{\infty}F_m\right)\geq \liminf_{n\to\infty}\lambda(F_n)\geq \epsilon>0; \end{align} thus the set $\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}F_m$ is not empty, which says exactly that there exists a point $x$ which lies in infinitely many of the $F_n$'s (all the hardwork goes into constructing the Lebesgue measure, but once this is done, a lot of things become simpler). Lemma 2 is I think a straight-forward quantification (using nothing more than upper/lower sums) of the following statement "if a function is non-negative and has positive integral then it must be relatively large on a substantial portion of the domain". Finally, the proof of the dominated convergence theorem itself follows quite trivially from these two lemmas (as explained in the single paragraph proof on page 181).