Why does $\mu(\{f_n \to f\})=\mu(\bigcap_{\epsilon \in \mathbb Q^+}\bigcup_{n\in\mathbb N}\bigcap_{m\ge n}\{\mid f_m-f\mid<\epsilon\})$ holds?

Question

If $f_n \to f$ a.e, i.e. $\forall \epsilon >0 \exists n_0\in\mathbb N \,\forall x\in N^c \forall n\ge n_0:\mid f_n(x)-f(x)\mid <\epsilon$.

Why does $\mu(\{f_n \to f\})=\mu(\bigcap_{\epsilon \in \mathbb Q^+}\bigcup_{n\in\mathbb N}\bigcap_{m\ge n}\{\mid f_m-f\mid<\epsilon\})$ holds?

Answer 1

$x\in\bigcap_{\epsilon\in\mathbb Q^+}\bigcup_{n\in\mathbb N}\bigcap_{m\geq n}\{|f_m-f|<\epsilon\}$ is exactly the statement that for every positive $\epsilon\in\mathbb Q$ an integer $n$ exists such that for every $m\geq n$ we have $|f_m(x)-f(x)|<\epsilon$.

This is evidently equivalent with the same statement with the only difference that $\mathbb Q$ is replaced by $\mathbb R$, and that statement on its own is exactly the statement that $\lim_{n\to\infty}f_n(x)=f(x)$ or equivalently $x\in\{f_n\to f\}$.

So we have:$$\{f_n\to f\}=\bigcap_{\epsilon\in\mathbb Q^+}\bigcup_{n\in\mathbb N}\bigcap_{m\geq n}\{|f_m-f|<\epsilon\}$$and consequently:$$\mu(\{f_n\to f\})=\mu(\bigcap_{\epsilon\in\mathbb Q^+}\bigcup_{n\in\mathbb N}\bigcap_{m\geq n}\{|f_m-f|<\epsilon\})$$