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Let $(\Omega, \mathcal F, \mu)$ be a measure space. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Assume that $\mu(\Omega) < \infty$. Let $p \in (1, \infty)$ and $p'$ its Hölder conjugate. Let $f_n,f,g:\Omega \to \mathbb R$ be measurable such that $\sup_n \|f_n\|_p < \infty$ and that $f_n \to f$ $\mu$-a.e. Then $\|f_n - f\|_q \to 0$ for every $q \in [1, p)$.

Could you confirm if my attempt is fine?

Proof Let $R := \sup_n \|f_n\|_p < \infty$. By Hölder's inequality, $$ \|1_Bf_n\|_q \le (\mu (B))^{\frac{1}{q} - \frac{1}{p}} \|1_Bf_n\|_p \le (\mu (B))^{\frac{1}{q} - \frac{1}{p}} R^p \quad \forall B \in \mathcal F, \forall n \in \mathbb N. $$

So $$ \sup_n \int_B |f_n|^q = \sup_n \|1_Bf_n\|_q^q \le (\mu (B))^{1 - \frac{q}{p}} R^{pq} \quad \forall B \in \mathcal F. $$

Notice that $1- \frac{q}{p} >0$ because $q<p$. So we get a direct control of $\sup_n \int_B |f_n|^q$ by $\mu(B)$. The claim then follows from below Vitali theorem (where $\mu(\Omega) < \infty$ is needed), i.e.,

Vitali theorem Assume that $\mu(\Omega) < \infty$. Let $f,f_n$ be real-valued measurable 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 $L^p(\Omega)$.

Akira
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    For finite measure spaces, $f_{n}\xrightarrow{L^{p}} f$ if and only if $|f_{n}|\to f$ in measure and ${|f_{n}|^{p}}$ is uniformly integrable. And a sufficient condition for uniform integrability is $|f_{n}|$ be uniformly bounded in $L^{p+\delta}$ for some $\delta>0$. In your case , $|f_{n}|$ is uniformly bounded in $L^{q+\delta}$ such that $q+\delta=p$. – Mr.Gandalf Sauron Apr 05 '23 at 16:20

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