[proofplan] We prove the equivalence directly from the definition of [uniform integrability](/page/Uniform%20Integrability). Tail decay implies uniform integrability by first obtaining a uniform $L^1$ bound from the decomposition into the bounded part $\{|f|\le M\}$ and the tail $\{|f|>M\}$, and then using the same decomposition over an arbitrary measurable set $A$. Conversely, uniform integrability gives a uniform $L^1$ bound; the set where $|f|>M$ has measure at most the uniform $L^1$ bound divided by $M$, so for large $M$ it is small enough for the uniform absolute continuity condition to control the tail integral. [/proofplan]

[step:Derive uniform integrability from uniform tail decay]Assume that \begin{align*} \lim_{M\to\infty}\sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M\}} |f(x)|\,d\mu(x) &= 0. \end{align*} If $\mathcal F=\varnothing$, the conclusion is immediate, with the usual convention that the supremum over the empty family is $0$. Assume henceforth in this step that $\mathcal F\neq\varnothing$. Choose $M_0>0$ such that \begin{align*} \sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M_0\}} |f(x)|\,d\mu(x) &\le 1. \end{align*} For each $f\in\mathcal F$, split $E$ into the measurable sets $\{x\in E:|f(x)|\le M_0\}$ and $\{x\in E:|f(x)|>M_0\}$. Since $\mu(E)<\infty$, we obtain \begin{align*} \int_E |f(x)|\,d\mu(x) &\le \int_{\{x\in E:|f(x)|\le M_0\}} M_0\,d\mu(x)+\int_{\{x\in E:|f(x)|>M_0\}} |f(x)|\,d\mu(x). \end{align*} Therefore \begin{align*} \int_E |f(x)|\,d\mu(x) &\le M_0\mu(E)+1. \end{align*} Taking the supremum over $f\in\mathcal F$ gives \begin{align*} \sup_{f\in\mathcal F}\int_E |f(x)|\,d\mu(x) &\le M_0\mu(E)+1<\infty. \end{align*} It remains to verify uniform absolute continuity of the integrals. Let $\varepsilon>0$. By the tail decay assumption, choose $M_1>0$ such that \begin{align*} \sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x) &< \frac{\varepsilon}{2}. \end{align*} Define \begin{align*} \delta &:= \frac{\varepsilon}{2M_1}. \end{align*} Let $A\in\mathcal E$ satisfy $\mu(A)<\delta$. For each $f\in\mathcal F$, decompose $A$ into $A\cap\{x\in E:|f(x)|\le M_1\}$ and $A\cap\{x\in E:|f(x)|>M_1\}$. Then \begin{align*} \int_A |f(x)|\,d\mu(x) &\le \int_{A\cap\{x\in E:|f(x)|\le M_1\}} M_1\,d\mu(x)+\int_{\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x). \end{align*} Thus \begin{align*} \int_A |f(x)|\,d\mu(x) &< M_1\delta+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Taking the supremum over $f\in\mathcal F$ proves uniform absolute continuity. Hence $\mathcal F$ is uniformly integrable.[/step]

[guided]Assume that the tails vanish uniformly: \begin{align*} \lim_{M\to\infty}\sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M\}} |f(x)|\,d\mu(x) &= 0. \end{align*} The definition of uniform integrability has two parts: a uniform $L^1$ bound and uniform smallness of integrals over sets of small measure. We prove both. First, choose $M_0>0$ so large that \begin{align*} \sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M_0\}} |f(x)|\,d\mu(x) &\le 1. \end{align*} For any fixed $f\in\mathcal F$, split the integral over $E$ into the region where $|f|$ is bounded by $M_0$ and the region where $|f|$ is larger than $M_0$. On the bounded region, the integrand is at most $M_0$; on the tail region, the uniform tail estimate applies. Therefore \begin{align*} \int_E |f(x)|\,d\mu(x) &\le \int_{\{x\in E:|f(x)|\le M_0\}} M_0\,d\mu(x)+\int_{\{x\in E:|f(x)|>M_0\}} |f(x)|\,d\mu(x). \end{align*} Since $(E,\mathcal E,\mu)$ is a finite [measure space](/page/Measure%20Space), $\mu(E)<\infty$, so the first term is bounded by $M_0\mu(E)$. The second term is bounded by $1$. Hence \begin{align*} \int_E |f(x)|\,d\mu(x) &\le M_0\mu(E)+1. \end{align*} This bound is independent of $f$, so \begin{align*} \sup_{f\in\mathcal F}\int_E |f(x)|\,d\mu(x) &\le M_0\mu(E)+1<\infty. \end{align*} Now we prove the small-set condition. Let $\varepsilon>0$. Choose $M_1>0$ so large that \begin{align*} \sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x) &< \frac{\varepsilon}{2}. \end{align*} The idea is to control the bounded part by making the measure of $A$ small, and to control the unbounded part by the tail estimate. Define \begin{align*} \delta &:= \frac{\varepsilon}{2M_1}. \end{align*} If $A\in\mathcal E$ and $\mu(A)<\delta$, then for each $f\in\mathcal F$, \begin{align*} \int_A |f(x)|\,d\mu(x) &\le \int_{A\cap\{x\in E:|f(x)|\le M_1\}} |f(x)|\,d\mu(x)+\int_{A\cap\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x). \end{align*} On $A\cap\{x\in E:|f(x)|\le M_1\}$, we have $|f(x)|\le M_1$, so \begin{align*} \int_{A\cap\{x\in E:|f(x)|\le M_1\}} |f(x)|\,d\mu(x) &\le M_1\mu(A)<M_1\delta=\frac{\varepsilon}{2}. \end{align*} Also, \begin{align*} \int_{A\cap\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x) &\le \int_{\{x\in E:|f(x)|>M_1\}} |f(x)|\,d\mu(x)<\frac{\varepsilon}{2}. \end{align*} Combining the two estimates gives \begin{align*} \int_A |f(x)|\,d\mu(x) &< \varepsilon. \end{align*} The choice of $\delta$ did not depend on $f$, so \begin{align*} \sup_{f\in\mathcal F}\int_A |f(x)|\,d\mu(x) &\le \varepsilon. \end{align*} This proves uniform absolute continuity, and together with the uniform $L^1$ bound proves that $\mathcal F$ is uniformly integrable.[/guided]

[step:Use uniform integrability to make high-level sets uniformly small] Assume that $\mathcal F$ is uniformly integrable. Define the uniform $L^1$ bound \begin{align*} C &:= \sup_{f\in\mathcal F}\int_E |f(x)|\,d\mu(x). \end{align*} By uniform integrability, $C<\infty$. Let $\varepsilon>0$. By uniform absolute continuity, choose $\delta>0$ such that for every $A\in\mathcal E$ with $\mu(A)<\delta$, \begin{align*} \sup_{f\in\mathcal F}\int_A |f(x)|\,d\mu(x) &<\varepsilon. \end{align*} For $M>0$ and $f\in\mathcal F$, define the measurable high-level set \begin{align*} A_{f,M} &:= \{x\in E:|f(x)|>M\}. \end{align*} Since $M\mathbb{1}_{A_{f,M}}(x)\le |f(x)|$ for $\mu$-almost every $x\in E$, integration gives \begin{align*} M\mu(A_{f,M}) &\le \int_E |f(x)|\,d\mu(x)\le C. \end{align*} Hence \begin{align*} \mu(A_{f,M}) &\le \frac{C}{M}. \end{align*} Choose $M_\varepsilon>0$ such that $C/M_\varepsilon<\delta$. Then for every $M\ge M_\varepsilon$ and every $f\in\mathcal F$, \begin{align*} \mu(A_{f,M}) &<\delta. \end{align*} Therefore the uniform absolute continuity estimate applied to $A=A_{f,M}$ gives \begin{align*} \int_{\{x\in E:|f(x)|>M\}} |f(x)|\,d\mu(x) &<\varepsilon. \end{align*} Taking the supremum over $f\in\mathcal F$ yields \begin{align*} \sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M\}} |f(x)|\,d\mu(x) &\le \varepsilon \end{align*} for all $M\ge M_\varepsilon$. Since $\varepsilon>0$ was arbitrary, the desired limit is $0$. [/step]

[step:Conclude the equivalence] The first step proves that uniform tail decay implies uniform integrability. The second step proves that uniform integrability implies uniform tail decay. Hence \begin{align*} \mathcal F \text{ is uniformly integrable} \end{align*} if and only if \begin{align*} \lim_{M\to\infty}\sup_{f\in\mathcal F}\int_{\{x\in E:|f(x)|>M\}} |f(x)|\,d\mu(x) &= 0. \end{align*} This proves the theorem. [/step]