[proofplan] The proof is a direct Hölder estimate on arbitrary measurable subsets of $E$. The finite-measure hypothesis first converts the uniform $L^p$ bound into a uniform $L^1$ bound by applying Hölder to $E$ itself. Applying the same estimate to an arbitrary measurable set $A$ gives a factor $\mu(A)^{1/q}$, where $q$ is the Hölder conjugate of $p$, and this factor can be made uniformly small by choosing $\mu(A)$ small enough. [/proofplan] [step:Fix the uniform $L^p$ bound and the conjugate exponent] If $\mathcal F=\varnothing$, both asserted suprema over $\mathcal F$ are $0$ under the usual convention for non-negative quantities, and the conclusion is immediate. Assume henceforth that $\mathcal F\neq\varnothing$. Define the constant $C\in[0,\infty)$ by \begin{align*} C:=\sup_{f\in\mathcal F}\|f\|_{L^p(E)}. \end{align*} By hypothesis, $C<\infty$. Define the Hölder conjugate exponent $q\in(1,\infty)$ by \begin{align*} q:=\frac{p}{p-1}. \end{align*} Then \begin{align*} \frac{1}{p}+\frac{1}{q}=1. \end{align*} [/step] [step:Use Hölder's inequality to control integrals over measurable sets] Let $A\in\mathcal E$ be arbitrary, and define the indicator function $\mathbb{1}_A:E\to\{0,1\}$ by $\mathbb{1}_A(x)=1$ for $x\in A$ and $\mathbb{1}_A(x)=0$ for $x\in E\setminus A$. For each $f\in\mathcal F$, the function represented by $|f|\mathbb{1}_A$ is measurable, and the Hölder inequality applied to the product $|f|\mathbb{1}_A$ with exponents $p$ and $q$ gives \begin{align*} \int_A |f|\,d\mu(x)\le \|f\|_{L^p(E)}\|\mathbb{1}_A\|_{L^q(E)}. \end{align*} Since \begin{align*} \|\mathbb{1}_A\|_{L^q(E)}=\mu(A)^{1/q}, \end{align*} we obtain \begin{align*} \int_A |f|\,d\mu(x)\le C\mu(A)^{1/q}. \end{align*} Taking the supremum over $f\in\mathcal F$ yields \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)\le C\mu(A)^{1/q}. \end{align*} [guided] The point of the argument is that Hölder's inequality turns the size of the set $A$ into a quantitative small factor. Fix an arbitrary measurable set $A\in\mathcal E$ and define the indicator function $\mathbb{1}_A:E\to\{0,1\}$ by $\mathbb{1}_A(x)=1$ for $x\in A$ and $\mathbb{1}_A(x)=0$ for $x\in E\setminus A$. Fix also an arbitrary function $f\in\mathcal F$. We apply the Hölder inequality to the two [measurable functions](/page/Measurable%20Functions) $|f|:E\to[0,\infty]$ and $\mathbb{1}_A:E\to\{0,1\}$, using the conjugate exponents $p$ and $q$, where \begin{align*} \frac{1}{p}+\frac{1}{q}=1. \end{align*} This gives \begin{align*} \int_E |f|\mathbb{1}_A\,d\mu(x)\le \|f\|_{L^p(E)}\|\mathbb{1}_A\|_{L^q(E)}. \end{align*} The left-hand side is exactly the integral over $A$: \begin{align*} \int_E |f|\mathbb{1}_A\,d\mu(x)=\int_A |f|\,d\mu(x). \end{align*} The $L^q$ norm of the indicator function is computed directly from the definition of the $L^q$ norm: \begin{align*} \|\mathbb{1}_A\|_{L^q(E)}=\left(\int_E \mathbb{1}_A\,d\mu(x)\right)^{1/q}=\mu(A)^{1/q}. \end{align*} Since $C$ was defined as the supremum of the $L^p$ norms over $\mathcal F$, we have \begin{align*} \|f\|_{L^p(E)}\le C. \end{align*} Combining these three facts gives \begin{align*} \int_A |f|\,d\mu(x)\le C\mu(A)^{1/q}. \end{align*} Because the right-hand side does not depend on the particular choice of $f\in\mathcal F$, taking the supremum over all $f\in\mathcal F$ gives \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)\le C\mu(A)^{1/q}. \end{align*} [/guided] [/step] [step:Deduce the uniform $L^1$ bound from finite measure] Apply the preceding estimate with $A=E$. Since $\mu(E)<\infty$, we get \begin{align*} \sup_{f\in\mathcal F}\|f\|_{L^1(E)}=\sup_{f\in\mathcal F}\int_E |f|\,d\mu(x)\le C\mu(E)^{1/q}<\infty. \end{align*} Thus every element of $\mathcal F$ belongs to $L^1(E,\mathcal E,\mu)$, and the family is uniformly bounded in $L^1(E)$. [/step] [step:Choose a small-measure threshold to prove uniform integrability] Let $\varepsilon>0$ be arbitrary. If $C=0$, then the estimate from the previous step gives \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)=0 \end{align*} for every $A\in\mathcal E$, so any $\delta>0$ works. Assume now that $C>0$. Define \begin{align*} \delta:=\left(\frac{\varepsilon}{C}\right)^q. \end{align*} If $A\in\mathcal E$ satisfies $\mu(A)<\delta$, then the measurable-set estimate gives \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)\le C\mu(A)^{1/q}<C\delta^{1/q}=\varepsilon. \end{align*} Therefore, for every $\varepsilon>0$, there exists $\delta>0$ such that $\mu(A)<\delta$ implies \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\varepsilon. \end{align*} Together with the uniform $L^1$ bound, this is precisely [uniform integrability](/page/Uniform%20Integrability) of $\mathcal F$ in $L^1(E,\mathcal E,\mu)$. [/step]