> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $\seqi{f} \subset L^1$ be a family of [[Integrable Function|integrable functions]]. The family $\seqi{f}$ is **uniformly absolutely continuous** if for every $\eps > 0$, there exists $\delta > 0$ such that > $ > \mu(E) < \delta \Rightarrow \sup_{i \in I}\int_E \abs{f_i} < \eps > $ > [!theorem] > > Let $\seqi{f} \subset L^1$ be a family of functions. > 1. If $\seqi{f}$ is [[Uniformly Integrable|uniformly integrable]], then $\seqi{f}$ is also uniformly absolutely continuous. > 2. If $\seqi{f}$ is uniformly bounded in $L^1$ and uniformly absolutely continuous, then $\seqi{f}$ is also uniformly integrable. > 3. If $\mu(X) < \infty$ and $\seqi{f}$ is uniformly absolutely continuous, then $\seqi{f}$ is also uniformly bounded in $L^1$. > > *Proof*. Suppose that the assumptions $(1)$ holds and let $\eps > 0$. Let $M \ge 1$ such that $\sup_{i \in I}\int_{\bracs{\abs{f_i} \ge M}}\abs{f_i} < \eps/2$, so > $ > \mu(\bracs{\abs{f_i} \ge M}) = \int \one_{\bracs{\abs{f_i} \ge M}} \le \int_{\bracs{\abs{f_i} \ge M}}\abs{f_i} \le \eps/2 > $ > for all $i \in I$. So for any $E \in \cm$, > $ > \begin{align*} > \int_E \abs{f_i} &\le \int_{E \cap \bracs{\abs{f_i} \ge M}}\abs{f_i} + \int_{E \cap \bracs{\abs{f_i} < M}}\abs{f_i} \\ > &\le \eps/2 + M\mu(E) > \end{align*} > $ > Let $\delta < \eps/(2M)$, then the desired result holds. > > Now suppose that the assumptions of $(2)$ holds and let $\eps > 0$. Let $M \ge 0$, then by [[Markov's Inequality]], > $ > \mu(\bracs{\abs{f_i} \ge M}) \le \frac{\norm{f_i}_1}{M} \le \frac{\sup_{i \in I}\norm{f_i}_1}{M} > $ > If $\sup_{i \in I}\norm{f_i}_1/{M} < \delta$, then $\sup_{i \in I}\int_{\bracs{\abs{f_i} \ge M}}\abs{f_i} < \eps$. > > Lastly suppose that the assumptions of $(3)$ holds. Let $\delta > 0$ such that $\sup_{i \in I}\int_E \abs{f_i} \le 1$ for all $E \in \cm$ with $\mu(E) < \delta$. Firstly, there exists a finite covering $\seqf{E_j}$ of $X$ such that $\mu(E_j) < \delta$ for each $j \in [n]$. Therefore $\sup_{i \in I}\norm{f_i}_1 \le n$.