Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 4 down vote accepted

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.)

share|cite|improve this answer
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. – Nate Eldredge Mar 11 '12 at 4:33

At the risk of being insulting I'll rephrase it like they do in the intro-to-proofs class.

Just like the notion of 'uniform convergence' or 'uniform continuity' it means that as you move around your set, things only change in a bounded way.

If we know that a family $\mathscr{F}$ is uniformly integrable over $E$, that means that if you tell me a size, no matter how small, I can tell you how small a subset of $E$ you have to choose so that the integral over it is negligible for any function in $\mathscr{F}$.

It means that the functions in your family don't get too big anywhere as you go around your family. The family is 'uniform.'

With this intuition we can come up with an example of a non-uniformly integrable set:

Consider the family $\mathscr{F}$ that contains the sequence of functions $(f_n)$ over $[0,1]$ where: $$f_n(x):=\left\{\begin{array}{ll}n & x\in\left[0,\frac{1}{n}\right]\\ 0 & \text{otherwise}\end{array}\right. .$$

Fix some small $\varepsilon>0$. No matter how small a size $\delta>0$ we choose, we can always find a set $A$ where for some $h\in\mathscr{F},\ \int_A |h| \ dx >\varepsilon.$ In particular, choose $A=[0,\delta]$, and $h=f_N,$ for any $N>\frac{100}{\delta}.$ Now that integral is at least $100.$

So no matter how small (but positive) a size we choose, we can find some place in $\mathscr{F}$ where the function is big there.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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