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Please help me with this exercise:

Let $\mathcal{F}$ be a family of functions, each of which is integrable over $E$. Show that $\mathcal{F}$ is uniformly integrable over $E$ if and only if for each $\epsilon > 0$, there is $\delta > 0$ such that for each $f \in \mathcal{F}$,

$$\textit{if } U \textit{ is open and } m(E \cap U) < \delta, \textit{ then } \int_{E \cap U} |f| < \epsilon.$$

The forward direction is rather easy but I am having a hard time with the reverse direction. Thanks in advance.

Just in case, The Royden's book define uniformly integrablility as follows:

A family $\mathcal{F}$ of measurable functions on $E$ is said to be uniformly integrablility over $E$ provided for each $\epsilon > 0$, there is $ \delta > 0$ such that for each $f \in \mathcal{F}$,

$$\textit{if } A \subset E \textit{ is measurable and } m(A) < \delta, \textit{ then } \int_{A} |f| < \epsilon.$$

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  • $\begingroup$ How did you define uniform integrability? $\endgroup$ – Ian Oct 9 at 2:37
  • $\begingroup$ I just edited the question... $\endgroup$ – Aldebaran Oct 9 at 2:47
  • $\begingroup$ Then the problem is just approximating measurable sets by open sets. $\endgroup$ – Ian Oct 9 at 2:52
  • $\begingroup$ I really cannot figure out the details, would you be nice enough to help me?? $\endgroup$ – Aldebaran Oct 9 at 2:56
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Let $\epsilon>0$

Then exists $\delta>0$ such that, for each $f\in \mathcal{F}$:

if $U$ open and $m(U \cap E)<\delta$ then $\int_{E \cap U}|f| <\epsilon$

Let $f \in \mathcal{F}$ and $A \subseteq E$ such that $m(A)<\delta$

By regularity of the Lebesgue measure exists $O \supseteq A$ open such that $m(O)<\delta$, thus $m(O \cap E)\ \leq m(O)<\delta$

So $\int_{O \cap E}|f|<\epsilon.$

But $A \subseteq E \cap O$ and $\int_{A}|f| \leq \int_{O \cap E}|f|<\epsilon$

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