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Our textbooks says "We infer from Proposition 23 of the preceding chapter that if $F$ is a family of functions on E that is uniformly integrable and tight over E, then each function in F is integrable over E."

Proposition 23:

Let f be a measurable funciton on E. If f is integrable over E, then for each $\epsilon < 0$, there is a $\delta < 0$ for which

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

Conversely, in the case $m(E)<\infty$, if for each $\epsilon < 0$, there is a $\delta < 0$ for which (26) [the above] holds, then f is integrable over E.

What I don't understand is...if there is a family of functions that is uniformly integrable, then how is it possible for one of the functions in that family to $*not*$ be integrable? Why do we need tightness to conclude that?

Thanks in advance

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    Sounds like a not so well written paragraph in a text. – Maesumi Mar 03 '13 at 23:01
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    I agree that it is not very well written. The definition of "uniformly integrable family" should require that every single member of the family is integrable. Otherwise the language gets confusing. – Giuseppe Negro Mar 03 '13 at 23:07
  • It actually seems to follow from Proposition 25, not 23. – jodag Nov 04 '18 at 18:09

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Royden & Fitzpatrick defines uniform integrability for a collection of measurable functions. A uniformly integrable function (i.e. the collection of measurable functions that consist of a single function) fails to be integrable precisely when its domain has unbounded measure. Also see this discussion.

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Alp Uzman
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