Currently I am self-studying Measure Theory before studying Measure-Theoretic Probability Theory. The following is a problem I solved recently from Folland's Real Analysis, in particular Problem(2.1.3). I have read a few solutions online, including on this site, but they are different from mine, some quite complicated, and I would appreciate it if someone could confirm that my simple argument is correct.
In the proof, I use the implicit assumption made by Folland that the $\sigma$-algebra on the range space for a real-valued function is always understood to be the Borel $\sigma$-algbera $\mathcal{B}_{\mathbb{R}}$ unless specified otherwise. I make use of two propositions from Folland's book:
Proposition(2.6): If $f,g : X \rightarrow \mathbb{C}$ are $\mathcal{M}$-measurable, then so are $f + g$ and $fg$.
Proposition(2.7): If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued $\mathcal{M}$-measurable functions on $(X, \mathcal{M})$, then the functions$$g_1(x) = \sup_j f_j(x),\quadg_2(x) = \inf_j f_j(x)\\g_3(x) = \lim \sup f_j(x),\quadg_4(x) = \lim \inf f_j(x)$$are all $\mathcal{M}$-measurable. If $f(x) = \lim_{j \rightarrow \infty} f_j(x)$ exists for every $x \in X$, then $f$ is $\mathcal{M}$-measurable.
Problem: Let $(X, \mathcal{M}, \mu)$ be a measure space. If ${f_n}$ is a sequence of $\overline{\mathbb{R}}$-valued $\mathcal{M}$-measurable functions on $X$, then $\{x\ |\ \lim f_n(x) \text{ exists}\}$ is a measurable set.
Proof: Let $f_i = \lim \inf f_n(x)$, $f_s = \lim \sup f_n(x)$, and let $F = \{x\ :\ |f_i(x)| < \infty \text{ or } |f_s(x)| < \infty\}$. Then define $$g(x) = \chi_F(x)(f_i(x) - f_s(x)) + (1 - \chi_F(x))\beta$$ for some choice of $\beta \in \mathbb{R}_{>0}$. We write $\chi_F$ to denote the indicator function on the set $F$, and as usual, we assume the convention that $0 \cdot \pm \infty = 0$. By Proposition(2.7), both $f_i$ and $f_s$ are $\mathcal{M}$-measurable functions. $\mathbb{R}$ is a Borel set, meaning that both $f_i^{-1}(\mathbb{R}), f_s^{-1}(\mathbb{R}) \in \mathcal{M}$, and hence $F = f_i^{-1}(\mathbb{R}) \cup f_s^{-1}(\mathbb{R})$ is measurable. Therefore $\chi_F$ is a simple $\mathcal{M}$-measurable function, and now by repeated application of Proposition(2.6), we conclude that $g$ is $\mathcal{M}$-measurable too.Moreover, notice that our definition of $g$ ensures that it is well defined in the case where both $f_i$ and $f_s$ take infinite values. Finally, observe that$$ \{x\ |\ \lim f_n(x) \text{ exists}\}\ =\ g^{-1}(\{0\})$$since $\lim f_n(x)$ exists if and only if $\lim \inf f_n(x) = \lim \sup f_n(x)$. The set $\{0\}$ is a Borel set, and therefore $g^{-1}(\{0\}) \in \mathcal{M}$, since $g$ is $\mathcal{M}$-measurable. That is the set $\{x\ |\ \lim f_n(x) \text{ exists} \}$ is a measurable set.