Measurable set limit If $\forall n \in  ℕ$ ,  $  f_n: (X,M) \rightarrow (\overline{\mathbb{R}},B) $ are measurable. (where X is any space, M is a sigma algebra, B is Borel sigma algebra)
Prove that the set $A = \{x\in X: \exists lim f_n(x)$ as $n \rightarrow \infty \} \in M$
 A: Let $A = A_0 \cup A_1$, where $A_0 = \{x\in X : \lim f_n(x) < \infty\}$ and $A_1 = \{x\in X : \lim f_n(x) = \infty\}$. Then $x \in A_0$ if and only if for every $j\in \Bbb N$, there exists an $N \in \Bbb N$ such that $|f_m(x) - f_n(x)| < \frac{1}{j}$ for all $m, n \ge N$. Thus $$A_0 = \bigcap_{j = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{m = N}^\infty \bigcap_{n = N}^\infty A_0(m,n,j)$$ where $A_0(m,n,j) := \{x\in X : |f_m(x) - f_n(x)| < 1/j\}$. Since the $f_n$ are measurable, each of the sets $A(m,n,j) := \{x\in X : |f_m(x) - f_n(x)| < 1/j\}$ belong to $M$. Since $M$ is closed under countable unions and intersections, $A_0 \in M$. 
Now $x \in A_1$ if and only if to each $j\in \Bbb N$, there corresponds an $N \in \Bbb N$ such that $|f_n(x)| > j$ for all $n \ge N$. Thus $$A_1 = \bigcap_{j = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{n = N}^\infty A_1(n,j)$$ where $A_1(n,j) := \{x\in X : |f_n(x)| > j\}$. Since each of the functions $f_n$ are measurable, the sets $A_1(n,j)$ belong to $M$. As $M$ is closed under countable unions and intersections, $A_1 \in M$. So $A$, being the union of two members of $M$, belongs to $M$.
