According to my Real Analysis textbook, a family $\scr{F}$ of measurable functions on $E$ is said to be uniformly integrable over $E$ provided for each $\epsilon$ $>$ $0$, there is a $\delta$ $>$ $0$ such that for each $f$ $\in$ $\scr{F}$, if $A$ $\subseteq$ $E$ is measurable and $m(A)$ $<$ $\delta$, then $\int_{A}$$|f|$ $<$ $\epsilon$. That is fine, but my textbook doesn't really give any good examples or counterexamples. I am studying for a final exam in Real Analysis, and would like some input in regards to any examples or counterexamples for uniform integrability of Lebesgue integrable functions. Thanks!


You might be able to find more examples of UI and non-UI functions from probability books. But the definition there is different, and not equivalent to yours (similar ideas though).

Here are some examples that you can find/adapt from Wikipedia. You still need to prove them because the definition there is different.

  1. Any finite set of Lebesgue integrable functions is U.I. (because of absolute continuity of integral.)
  2. Any family bounded by an $L^1$ function: $\{f\in L^1: |f|\le g\}$ where $g\in L^1$ is U.I.
  3. $\{f\in L^p:\|f\|_{L^p}\le C\}$ with $p>1$ is U.I. (because of Holder's inequality)

Examples of non-U.I. families should not be too hard. You can construct a sequence of functions with the "mass" more and more concentrated. For example, consider the family of functions $\{f_n(x)=n \mathbf{1}_{[0,\frac{1}{n}]}:n=1,2,\cdots\}$. It's easy to use your definition to prove that this family is not U.I. (But it is U.I. in the sense defined in probability theory, which you can find from Wikipedia.)

  • 1
    $\begingroup$ There seems to be an error here: the given family $f_n = 1_{[0,\frac{1}{n}]}$ is uniformly integrable in every sense, and it converges to $0$ in $L^1$. Perhaps you meant to take $f_n = n 1_{[0,\frac{1}{n}]}$? That fails to be uniformly integrable in every sense. $\endgroup$ – Nate Eldredge Mar 11 '12 at 4:33

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