> [!theorem] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $\seq{f_n} \subset L^p$, then $\seq{f_n}$ is Cauchy in [$L^p$](Lp%20Space) if and only if: > 1. $\seq{f_n}$ is Cauchy [[Convergence in Measure|in measure]]. > 2. $\seq{\abs{f_n}^p}$ is [[Uniformly Integrable|uniformly integrable]]. > 3. For every $\eps > 0$, there exists $E \subset X$ such that $\mu(E) < \infty$ and $\int_{E^c} \abs{f_n}^p < \eps$ for all $n \in \nat$. > > *Proof*. Suppose that all three conditions holds. Let $\eps > 0$, $m, n \in \nat$, $E \subset X$ satisfying condition $(3)$. For any $\delta > 0$, let > $ > A_{mn, \delta} = \bracs{\abs{f_m - f_n} \ge \delta} \cap E > $ > This allows breaking the integral as > $ > \norm{f_m - f_n}_p \le \norm{f_m - f_n}_{L^p(E^c)} + \norm{f_m - f_n}_{L^p(E \setminus A_{mn})} +\norm{f_m - f_n}_{L^p(A_{mn})} > $ > Firstly, > $ > \begin{align*} > \norm{f_m - f_n}_{L^p(E \setminus A_{mn})}^p &\le \delta \mu(E) > \end{align*} > $ > Using condition $(2)$, $M \ge 0$ such that $\norm{f_k}_{L^p(\bracs{f_k} \ge M)} < \eps$, then > $ > \begin{align*} > \norm{f_m - f_n}_{L^p(A_{mn})} &\le \norm{f_m}_{L^p(A_{mn})} + \norm{f_n}_{L^p(A_{mn})} \\ > &\le \norm{f_m}_{L^p(\bracs{f_m \ge M})} + \norm{f_n}_{L^p(\bracs{f_n \ge M})} + 2M\mu(A_{mn})^{1/p} \\ > &< 2\eps + 2M\mu(A_{mn})^{1/p} > \end{align*} > $ > and > $ > \norm{f_m - f_n}_{L^p(E^c)} \le \norm{f_m}_{L^p(E^c)} + \norm{f_n}_{L^p(E^c)} < 2\eps^{1/p} > $ > by condition $(3)$. Therefore > $ > \norm{f_m - f_n}_p \le 2\eps + 2\eps^{1/p} + \delta \mu(E) + 2M\mu(A_{mn})^{1/p} > $ > By condition $(1)$, $\mu(A_{mn}) \to 0$ as $m, n \to \infty$. As $\delta$ and $\eps$ are both arbitrary, $\norm{f_m - f_n}_p \to 0$ as $m,n \to \infty$.