[proofplan] We prove the small-set form of [uniform integrability](/page/Uniform%20Integrability) directly. First we record the absolute continuity of the integral for a single $L^1$ function, proving it from truncation so that no external theorem is needed. Given $\varepsilon>0$, $L^1$ convergence makes all sufficiently late $f_n$ close to $f$ in $L^1$, so their integrals over small sets are controlled by the corresponding integral of $|f|$ plus a small error. The remaining finitely many functions are controlled simultaneously by taking the minimum of finitely many small-set parameters. [/proofplan]

[step:Prove absolute continuity of the integral for one $L^1$ function]Let $u\in L^1(E,\mathcal E,\mu)$, and let $\eta>0$. We claim that there exists $\delta_u>0$ such that every $A\in\mathcal E$ with $\mu(A)<\delta_u$ satisfies \begin{align*}\int_A |u|\,d\mu(x)<\eta.\end{align*} For each $M>0$, define the measurable set \begin{align*}B_M:=\{x\in E:|u(x)|>M\}.\end{align*} For $M>0$, let $h_M:E\to[0,\infty]$ be the [measurable function](/page/Measurable%20Function) defined by $h_M(x):=|u(x)|\mathbb{1}_{B_M}(x)$. If $M_1\le M_2$, then $B_{M_2}\subset B_{M_1}$, so $h_{M_2}\le h_{M_1}$ pointwise. Also $h_M(x)\to0$ for every $x\in E$ with $|u(x)|<\infty$, and $0\le h_M\le |u|$. Since $u\in L^1(E,\mathcal E,\mu)$, the [Dominated Convergence Theorem](/theorems/4) applied to $(h_M)$ along any sequence $M\to\infty$ gives \begin{align*}\lim_{M\to\infty}\int_{B_M}|u|\,d\mu(x)=0.\end{align*} Choose $M_u>0$ such that \begin{align*}\int_{B_{M_u}}|u|\,d\mu(x)<\frac{\eta}{2}.\end{align*} Define \begin{align*}\delta_u:=\frac{\eta}{2M_u}.\end{align*} If $A\in\mathcal E$ and $\mu(A)<\delta_u$, then splitting $A$ into $A\cap B_{M_u}$ and $A\setminus B_{M_u}$ gives \begin{align*}\int_A |u|\,d\mu(x)\le M_u\mu(A)+\int_{B_{M_u}}|u|\,d\mu(x)<\eta.\end{align*}[/step]

[guided]Fix $u\in L^1(E,\mathcal E,\mu)$ and $\eta>0$. We want to show that the integral of $|u|$ over a set $A$ becomes small whenever $\mu(A)$ is small. The obstruction is that $u$ need not be bounded, so we first separate the bounded part of $u$ from its large-value tail. For $M>0$, define \begin{align*}B_M:=\{x\in E:|u(x)|>M\}.\end{align*} Because $u\in L^1(E,\mathcal E,\mu)$, the nonnegative function $|u|$ has finite integral. We now justify that the large-value tail has vanishing integral. For $M>0$, let $h_M:E\to[0,\infty]$ be the measurable function defined by $h_M(x):=|u(x)|\mathbb{1}_{B_M}(x)$. As $M\to\infty$, the sets $B_M$ decrease and eventually exclude each point $x\in E$ with finite $|u(x)|$, so $h_M(x)\to0$ for $\mu$-a.e. $x\in E$. Moreover $0\le h_M\le |u|$, and $|u|\in L^1(E,\mathcal E,\mu)$. The [Dominated Convergence Theorem](/theorems/7529) therefore gives \begin{align*}\lim_{M\to\infty}\int_{B_M}|u|\,d\mu(x)=0.\end{align*} Therefore we may choose $M_u>0$ such that \begin{align*}\int_{B_{M_u}}|u|\,d\mu(x)<\frac{\eta}{2}.\end{align*} Now define \begin{align*}\delta_u:=\frac{\eta}{2M_u}.\end{align*} Let $A\in\mathcal E$ satisfy $\mu(A)<\delta_u$. On the set $A\setminus B_{M_u}$ we have $|u|\le M_u$, while on $A\cap B_{M_u}$ we use the already small tail bound. Thus \begin{align*}\int_A |u|\,d\mu(x)\le \int_{A\setminus B_{M_u}}|u|\,d\mu(x)+\int_{A\cap B_{M_u}}|u|\,d\mu(x).\end{align*} The first integral is bounded by $M_u\mu(A)$, and the second is bounded by the full tail integral over $B_{M_u}$. Hence \begin{align*}\int_A |u|\,d\mu(x)\le M_u\mu(A)+\int_{B_{M_u}}|u|\,d\mu(x)<M_u\delta_u+\frac{\eta}{2}=\eta.\end{align*} This proves the absolute continuity of the integral for the single function $u$.[/guided]

[step:Control the limiting function and the finitely many initial terms on small sets] Fix $\varepsilon>0$. Since $f_n\to f$ in $L^1(E)$, there exists $N\in\mathbb N$ such that, for every $n\ge N$, \begin{align*}\|f_n-f\|_{L^1(E)}<\frac{\varepsilon}{3}.\end{align*} Apply the preceding step to each member of the finite family \begin{align*}\mathcal G:=\{f\}\cup\{f_1,\dots,f_{N-1}\}.\end{align*} For every $g\in\mathcal G$, choose $\delta_g>0$ such that $\mu(A)<\delta_g$ implies \begin{align*}\int_A |g|\,d\mu(x)<\frac{\varepsilon}{3}.\end{align*} Define \begin{align*}\delta:=\min_{g\in\mathcal G}\delta_g.\end{align*} Then $\delta>0$, and for every $A\in\mathcal E$ with $\mu(A)<\delta$ and every $g\in\mathcal G$, \begin{align*}\int_A |g|\,d\mu(x)<\frac{\varepsilon}{3}.\end{align*} [/step]

[step:Control the tail of the sequence by $L^1$ closeness to the limit] Let $A\in\mathcal E$ satisfy $\mu(A)<\delta$. If $n\ge N$, then the triangle inequality for absolute values gives \begin{align*}|f_n(x)|\le |f(x)|+|f_n(x)-f(x)|\end{align*} for $\mu$-a.e. $x\in E$. Integrating this inequality over $A$ with respect to $\mu$ yields \begin{align*}\int_A |f_n|\,d\mu(x)\le \int_A |f|\,d\mu(x)+\int_A |f_n-f|\,d\mu(x).\end{align*} Since $f\in\mathcal G$ and $\mu(A)<\delta$, the first term is less than $\varepsilon/3$. Since $A\subset E$, the second term is bounded by the full $L^1$ norm: \begin{align*}\int_A |f_n-f|\,d\mu(x)\le \int_E |f_n-f|\,d\mu(x)=\|f_n-f\|_{L^1(E)}<\frac{\varepsilon}{3}.\end{align*} Therefore, for every $n\ge N$, \begin{align*}\int_A |f_n|\,d\mu(x)<\frac{2\varepsilon}{3}<\varepsilon.\end{align*} [/step]

[step:Conclude uniform integrability of the whole family] Let $A\in\mathcal E$ satisfy $\mu(A)<\delta$. If $g\in\mathcal F$, then either $g=f$, or $g=f_n$ for some $n<N$, or $g=f_n$ for some $n\ge N$. In the first two cases, $g\in\mathcal G$, so \begin{align*}\int_A |g|\,d\mu(x)<\frac{\varepsilon}{3}<\varepsilon.\end{align*} In the last case, the preceding step gives \begin{align*}\int_A |g|\,d\mu(x)<\varepsilon.\end{align*} Thus \begin{align*}\sup_{g\in\mathcal F}\int_A |g|\,d\mu(x)<\varepsilon.\end{align*} Since $\varepsilon>0$ was arbitrary, $\mathcal F$ is uniformly integrable. [/step]