[proofplan] We prove the contrapositive obstruction directly from the small-set definition of [uniform integrability](/page/Uniform%20Integrability). If $\mathcal F$ were uniformly integrable, then measurable sets of sufficiently small $\mu$-measure would force the integrals of $|f|$ over those sets to be uniformly small for all $f\in\mathcal F$. Since $\mu(A_n)\to0$, one of the sets $A_n$ must be small enough to trigger this bound, contradicting the assumed lower bound for the corresponding function $f_n$. [/proofplan] [step:Assume uniform integrability and obtain a small-set bound] Assume, for contradiction, that $\mathcal F$ is uniformly integrable. By the definition of uniform integrability, applied with the positive number $\varepsilon_0/2$, there exists a number $\delta>0$ such that for every set $A\in\mathcal E$ satisfying $\mu(A)<\delta$, \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\frac{\varepsilon_0}{2}. \end{align*} [guided] We argue by contradiction, so suppose that $\mathcal F$ is uniformly integrable. The relevant definition says that the integrals of $|f|$ over small measurable sets are small uniformly in $f\in\mathcal F$: for every positive tolerance $\varepsilon>0$, there is a number $\delta>0$ such that every $A\in\mathcal E$ with $\mu(A)<\delta$ satisfies \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\varepsilon. \end{align*} We choose the tolerance to be $\varepsilon=\varepsilon_0/2$. This strict smaller number avoids any endpoint issue with the assumed lower bound $\varepsilon_0$. Therefore there exists $\delta>0$ such that whenever $A\in\mathcal E$ and $\mu(A)<\delta$, one has \begin{align*} \sup_{f\in\mathcal F}\int_A |f|\,d\mu(x)<\frac{\varepsilon_0}{2}. \end{align*} [/guided] [/step] [step:Choose one obstructing set below the uniform-integrability threshold] Since $\mu(A_n)\to0$, there exists an index $n_0\in\mathbb N$ such that \begin{align*} \mu(A_{n_0})<\delta. \end{align*} Applying the preceding small-set bound with $A=A_{n_0}$ gives \begin{align*} \sup_{f\in\mathcal F}\int_{A_{n_0}} |f|\,d\mu(x)<\frac{\varepsilon_0}{2}. \end{align*} [/step] [step:Contradict the assumed lower bound] Because $f_{n_0}\in\mathcal F$, the integral for this particular function is bounded above by the supremum over $\mathcal F$: \begin{align*} \int_{A_{n_0}} |f_{n_0}|\,d\mu(x)\le \sup_{f\in\mathcal F}\int_{A_{n_0}} |f|\,d\mu(x). \end{align*} Hence \begin{align*} \int_{A_{n_0}} |f_{n_0}|\,d\mu(x)<\frac{\varepsilon_0}{2}<\varepsilon_0. \end{align*} This contradicts the hypothesis that \begin{align*} \int_{A_{n_0}} |f_{n_0}|\,d\mu(x)\ge \varepsilon_0. \end{align*} Therefore $\mathcal F$ is not uniformly integrable. [/step]