Let $f_n:D\to \mathbb{R}$ be measurable for all $n$ and suppose that $f_n(x)\leq f_{n+1}(x)$ for all $n$ and for all $x\in D$. If $\lim_{n\to\infty}f_n(x)=f(x)\forall x\in D$, show that $f$ is measurable.
Thoughts/Attempt:
I know that monotone functions are Lebesgue measurable.
Since $f_n$ is measurable for all $n$, $\{x\mid f_n(x)>\alpha\}=f_n^{-1}((\alpha,\infty))$ is measurable as well, for all $\alpha\in\mathbb{R}$
I know that the union, complement, and intersection of measurable sets is measurable, there might be some way to use these by rewriting the preimage into terms of unions, intersections, etc... but I'm not seeing how to rewrite it like that (if this is even possible?)
Maybe something along the lines of the following:
Each $f_n$ is measurable, so I could have a union of the functions from $f_1,f_2,\ldots, f_n$, and this union would be measurable.
$$\bigcup_k\bigcup_{n}f_n^{-1}\left(\left(-\infty,\alpha-\frac{1}{k}\right)\right)$$
But I'm not even interested in all $n$, because only the $n\geq N$ matter since $f_n\to f$, so I can change this to:
$$\bigcup_k\bigcup_{n\geq N}f_n^{-1}\left(\left(-\infty,\alpha-\frac{1}{k}\right)\right)$$
I have the union of $k$'s since I'm trying to use the information in the question regarding the monotone functions. Since I have a union, for each $k$, my $f_n\leq f_{n+1}$.
I'm having trouble getting any further, and I'm not even sure if this line of thinking is the right way to go. Any suggestions would be appreciated. Thanks.
You already stated that since each $f_n$ is measurable, $\{x\mid f_n(x)>\alpha\}$ is measurable for all $n \in \mathbb N$ and all $\alpha\in\mathbb{R}$.
For each $x \in D$, $f_n(x)$ is an increasing sequence converging to $f(x)$, therefore
$$ f(x)>\alpha \Longleftrightarrow f_n(x)>\alpha \text{ for some } n \in \mathbb N \, . $$
It follows that for each $\alpha\in\mathbb{R}$, $$ \{x\mid f(x)>\alpha\} = \bigcup_{n \in \mathbb N} \{x\mid f_n(x)>\alpha\} $$ is measurable as a countable union of measurable sets. Therefore $f$ is measurable.