Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Let $p \in (1, \infty)$ and $p'$ its Hölder conjugate. Let $f_n,f,g:\Omega \to \mathbb R$ be measurable functions such that $\sup_n \|f_n\|_p < \infty$ and that $f_n \to f$ $\mu$-a.e. Then $f_n \to f$ in the weak topology $\sigma(L^p, L^{p'})$.
Could you confirm if my attempt is fine?
Proof Let $\varphi$ be a subsequence of $\mathbb N$ and let $g_n := f_{\varphi (n)}$. It suffices to prove that $(g_n)$ has a subsequence that converges to $f$ in $\sigma(L^p, L^{p'})$. Notice that $L^p (\Omega)$ is uniformly convex and thus reflexive. So $(g_n)$ is relatively compact in $\sigma(L^p, L^{p'})$, i.e., there is $g \in L^p (\Omega)$ such that $g_n \to g$ in $\sigma(L^p, L^{p'})$. Clearly, $g_n \to f$ $\mu$-a.e. Then $g=f$ $\mu$-a.e. by below lemma, i.e.,
Lemma Let $p \in [1, \infty)$ and $p'$ its Hölder conjugate. Let $f_n,f,g:\Omega \to \mathbb R$ be measurable functions such that $f_n,f \in L^p(\Omega)$ for all $n$. Assume $f_n \to f$ in the weak topology $\sigma(L^p, L^{p'})$ and $f_n \to g$ $\mu$-a.e. Then $f=g$ $\mu$-a.e.
This completes the proof.
Update I have found another (but longer) approach below, i.e.,
Fix $u \in L^{p'} (\Omega)$. It suffices to prove that $\int uf_n \to \int uf$. Let $\mathcal S := \{ 1_B : B \in \mathcal F, \mu(B) < \infty\}$. Because $p' < \infty$, we get $\operatorname{span} (S)$ is dense in $L^{p'} (\Omega)$. So it suffices to assume that $u = 1_B$ with $\mu(B) < \infty$ (please see here for more details). We will prove that $1_B f_n \to 1_B f$ in $L^1$. Clearly, $1_B f_n \to 1_B f$ $\mu$-a.e.
Vitali theorem 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 $L^p(\Omega)$.
We assume $R := \sup_n \|f_n\|_p < \infty$. Fix $\varepsilon >0$. By Hölder's inequality, $$ \int_A |1_B f_n| \le \bigg ( \int_A 1_B\bigg )^{1/p'} \|f_n\|_p^p \le (\mu(A))^{1/p'} R^p. $$
It follows that $\sup_n \int_A |1_B f_n|$ will be controlled directly by $\mu(A)$. The claim then follows from Vitali theorem.
