[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.
[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]
[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]
