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 with $\mu(\Omega) < \infty$. Let $f,f_n$ be real-valued measurable functions such that $(f_n) \subset L^p(\Omega)$ where $p \in [1, \infty)$ and that $f_n \to f$ $\mu$-a.e. Assume that for each $\varepsilon >0$ there is $\delta>0$ such that $\sup_n \int_A |f_n|^p < \varepsilon$ for every $A \in \mathcal F$ with $\mu(A) < \delta$. Prove that $f_n \to f$ in $L^p(\Omega)$.
Could you confirm if my below proof is fine?
My attempt It suffices to prove that $(f_n)$ is a Cauchy sequence in $L^p(\Omega)$. We fix $\varepsilon >0$. There is $\delta>0$ such that $\sup_n \int_A |f_n|^p < \varepsilon 2^{-p}$ for every $A \in \mathcal F$ with $\mu(A) < \delta$. By Egorov theorem (where $\mu(\Omega) < \infty$ is needed), there is $A \in \mathcal F$ such that $\mu(A) < \delta$ and $f_n \to f$ uniformly on $A^c := \Omega \setminus A$. There is $N >0$ such that for all $m,n \ge N$, $$ \begin{align} & \sup_{\omega \in A^c} |f_n(\omega) - f_m (\omega)| \\ \le{} & \sup_{\omega \in A^c} |f_n(\omega) - f (\omega)| + \sup_{\omega \in A^c} |f_m (\omega) - f (\omega)| \\ <{} &\varepsilon. \end{align} $$
We have $$ \|f_n-f_m\|_p^p = \int_A |f_n-f_m|^p + \int_{A^c} |f_n-f_m|^p. $$
For all $a,b \in \mathbb R$, $$ |a+b|^p \le (2 \max\{a,b\})^p = 2^p (\max\{a,b\})^p \le 2^p(|a|^p+|b|^p). $$
Then for all $m,n \ge N$, $$ \begin{align} \|f_n-f_m\|_p^p &\le 2^p \int_A (|f_n|^p + |f_m|^p) + \int_{A^c} |f_n-f_m|^p \\ &\le 2^p (\varepsilon 2^{-p+1}) + \varepsilon \\ &= 3 \varepsilon. \end{align} $$
This completes the proof.
