Let $f_n: D \rightarrow \mathbb{R}$ be measurable for each n, and suppose that $f_n(x) \leq f_{n+1}(x)$ $\forall n$ and each $x \in D$. If $f(x)=lim_{n \rightarrow \infty}f_n(x)$ exists (in $\mathbb{R}$) for each $x \in D$, prove that f is measurable.
So the statement says that we have a point-wise convergence of a series of measurable functions and we want to show that the limit function is also measurable. To this extent, it will suffice to show that given any open $U \subset \mathbb{R}$, $f^{-1}(U)$ is measurable in $D$.
I am quite new to measure theory and this problem has me scratching my head. I'm trying to think about how to get the fact that for any $\epsilon$>$0$ there exists an $N$ s.t. $\forall$$n$>$N$ $|f_n(x)-f(x)|<\epsilon$ to help me prove something about the measurability of $f(x)$...
Any insight is appreciated!
Hint:
Since each $f_n$ is measurable, the sets $\{f>r\}=\{x:f(x)>r\}$ with $r\in\mathbb{R}$ is measurable. I claim that $$\{\lim_{n\to \infty}f_n=f>r\}=\bigcup_{n}\{f_n>r\}\;.$$ Since $f_n$ is increasing, $f_n(x)\leq \lim_{n\to \infty}f_n(x)=f(x)$ for all $x$. Hence, for each $n$ $$\{f_n>r\}\subset \{\lim_{n\to \infty}f_n=f>r\}\;.$$ Can you show the other containment and finish the proof from there?