> [!theorem] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], $p \in (1, \infty)$, and $\seq{f_n} \subset L^p(X, \cm, \mu)$ and $f \in L^p$ such that > 1. $M = \max\braks{\sup_{n \in \nat}\norm{f_n}_p, \norm{f}_p} < \infty$. > 2. $f_n \to f$ [[Almost Everywhere|a.e.]] > > then $f_n \to f$ [[Weak Topology|weakly]]. > > *Proof*. Identify $(L^p)^*$ with $L^q$[^1]. Let $g \in L^q$, then the measure $\abs{g}^qd\mu \ll \mu$. Hence for any $\eps > 0$, there exists $\delta > 0$ such that $\norm{g \cdot \one_E}_q^q < \eps$ for all $E \in \cm$ with $\mu(E) < \delta$. Let $A \subset X$ such that $\mu(A) < \infty$ and $\norm{g \cdot \one_A}_q^q < \eps$. By [[Egoroff's Theorem]], there exists $B \subset A$ such that $\mu(A \setminus B) < \delta$ and $f_n \to f$ [[Uniform Convergence|uniformly]] on $B$. Combining the two yields that $\norm{g \cdot \one_{X \setminus B}}_q^q < 2\eps$. > > In this case, > $ > \begin{align*} > \abs{\angles{f_n, g} - \angles{f, g}} &\le \angles{\abs{f_n - f}, g} \\ > &= \angles{\abs{f_n - f}, g \cdot \one_B} +\angles{\abs{f_n - f}, g \cdot \one_{X \setminus B}} > \end{align*} > $ > By uniform convergence, $\angles{\abs{f_n - f}, g \cdot \one_B} \to 0$. By absolute continuity, > $ > \begin{align*} > \angles{\abs{f_n - f}, g \cdot \one_{A \setminus B} + g\cdot \one_{X \setminus A}} &\le \norm{f_n - f}_p \cdot \norm{g \cdot \one_{X \setminus B}} _q \\ > &\le 2M \cdot (2\eps)^{1/q} > \end{align*} > $ > as $\eps$ was arbitrary, the limit must go to zero. [^1]: [[Lp Representation Theorem]]