# Equivalence condition to uniform integrability

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 $$f \in \mathcal{F}$$,

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

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

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