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 $f \in \mathcal{F}$,

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.