I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 4.14 Let $(\Omega, \mathcal F, \mu)$ be a measure space such that $\mu(\Omega)<\infty$. Let $f,f_n$ be real-valued measurable functions such that $f_n \to f$ $\mu$-a.e.
- For $\alpha>0$ and $n \in \mathbb N^*$, let $$ S_n (\alpha) := \bigcup_{k \ge n} \{ |f_k-f| > \alpha\}. $$ Prove that $\mu(S_n (\alpha)) \xrightarrow{n \to \infty} 0$ for all $\alpha>0$.
- Prove that for every $\delta>0$ there is $A \in \mathcal F$ such that $\mu(A) < \delta$ and $f_n \to f$ uniformly on $A^c :=\Omega \setminus A$.
Could you confirm if my below attempt is correct?
Proof 1. Let $$ C := \{\omega \in \Omega : f_n (\omega) \xrightarrow{n \to \infty} f (\omega)\}. $$
By $\varepsilon$-$\delta$ argument (please see here for more details), we have $$ C=\bigcap_{m \in \mathbb N^*} \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} \{|f_n - f| \le 1/m\} \in \mathcal F. $$
We have $\mu(C^c)=0$, so $$ \mu \bigg ( \bigcup_{m \in \mathbb N^*} \bigcap_{N \in \mathbb N} S_N (1/m) \bigg ) = 0. $$
Then $$ \mu \bigg (\bigcap_{N \in \mathbb N} S_N (1/m) \bigg ) = 0 \quad \forall m \in \mathbb N^*. $$
Notice that $S_{N+1} (1/m) \subset S_N (1/m)$. By the continuity of finite measure from above, $$ \mu (S_{N} (1/m)) \xrightarrow{N \to \infty} 0 \quad \forall m \in \mathbb N^*. $$
- Fix $\delta > 0$. For each $n \in \mathbb N$, there is $\varphi (n) \in \mathbb N$ such that $\mu(S_{\varphi (n)} (1/n)) < \delta 2^{-n}$. Let $A := \bigcup_n S_{\varphi (n)} (1/n)$. Then $\mu(A) \le \sum_n \delta 2^{-n} \le \delta$ and $$ A^c = \bigcap_n \bigcap_{k \ge \varphi (n)} \{ |f_k-f| \le 1/n\}. $$
Fix $\varepsilon >0$. There is $N \in \mathbb N$ such that $1/N < \varepsilon$. If $k \ge \varphi(N)$ then $|f_k-f| \le 1/N < \varepsilon$ on $A^c$. This completes the proof.
