I believe I have a proof for the following problem, though I do not rely on the finite measure space criterion. Any advice would be appreciated, especially if there is some connection with this problem to the $L^p$ version of Arzela-Ascoli and the Dunford-Pettis theorem.

Let $\Omega$ be a finite measure space. Prove that if $K \subset L^1(\Omega)$ is compact, then $K$ is equi-integrable.

Fix $\epsilon > 0$. Since $K$ is a compact metric space, it is totally bounded, so there exist $f_1,\dots,f_n$ such that $B(f_1,\epsilon), \dots, B(f_n,\epsilon/3)$ cover $K$. From standard measure theory, any given $f \in K$ is equi-integrable on its own (i.e., absolute continuity of the Lebesgue integral), so any finite set of functions is equi-integrable. Thus there exists $\delta > 0$ such that $h < \delta$ implies $\lVert{\tau_h f_i - f_i\rVert}_1 < \epsilon/3$ for each $i = 1,\dots,n$.

We claim that this choice of $\delta$ works. Indeed, for any $f \in K$, there exists $N$ such that $\lVert{f - f_N\rVert}_1 < \epsilon/3$. Then for $h < \delta$, $$ \lVert{\tau_h f - f\rVert} \leq \lVert{\tau_h f - \tau_h f_1\rVert} + \lVert{\tau_h f_1 - f_1\rVert} + \lVert{f_1- f\rVert} = 2\lVert{f - f_1\rVert} + \lVert{\tau_h f_1 - f_1\rVert} < \epsilon, $$ as desired.

EDIT: $\tau_h(f) = f(x+h)$. I also realized my definition of equi-integrability is a bit different: for $\epsilon > 0$ there exists $\delta $ such that $\int_A |f| < \epsilon$ whenever $|A|<\delta$ for all $f \in K$. Does what I prove suffice though?

  • $\begingroup$ What are $\tau$ and $\tau_h$? $\endgroup$ Mar 16, 2017 at 7:01
  • $\begingroup$ The $\tau$ was a typo. The $\tau_h$ is the shift operator (I thought this was standard notation) $\endgroup$
    – user369210
    Mar 16, 2017 at 7:06
  • $\begingroup$ How can you shift on an arbitrary measure space? Moreover, I don't understand why you shift at all? How is the shifting connected with the equi-integrability? $\endgroup$ Mar 16, 2017 at 7:11
  • $\begingroup$ I think everything is happening on Lebesgue space. Also, that was my other question -- can I connect shifting with equi-integrability? $\endgroup$
    – user369210
    Mar 16, 2017 at 7:12
  • $\begingroup$ What is a Lebesgue space? You wrote "Let $\Omega$ be a finite measure space". So, it can be anything. And even if, e.g., $\Omega = [0,1]$, how can you shift a function? Moreover, I don't see a connection between shifting and equi-integrability. $\endgroup$ Mar 16, 2017 at 7:16

1 Answer 1


I don't see how the shift operators come into play. You don't need them.

As you said, there exist $f_1,\ldots,f_n$ such that the open balls $B(f_j,\epsilon/2)$, $j=1,\ldots,n$, cover the compact set $K$. Since the finite set $\{f_1,\ldots,f_n\}$ is uniformly integrable, we can choose for given $\epsilon>0$ a constant $\delta>0$ such that

$$\int_{A} |f_j| < \frac{\epsilon}{2}$$

for all measurable sets $A$ with $|A|<\delta$ and $j=1,\ldots,n$. Now if $f \in K$, then we can find $j \in \{1,\ldots,n\}$ such that $\int |f-f_j| < \epsilon/2$. Hence, by the triangle inequality

$$\int_A |f| \leq \int_A |f-f_j| + \int_A |f_j| \leq \int |f-f_j| + \int_A |f_j| < \epsilon$$

for any measurable set $A$ with $|A|<\delta$. This proves that $K$ is equi-integrable.

  • $\begingroup$ Ah sweet. So this theorem holds for any totally bounded set? Also where does finite measure come into play? $\endgroup$
    – user369210
    Mar 16, 2017 at 7:27
  • $\begingroup$ Wait I see where you used the finite measure space condition, though my first question in my previous comment still holds. $\endgroup$
    – user369210
    Mar 16, 2017 at 7:43
  • 1
    $\begingroup$ @user321210 Yes, total boundedness is enough. The fact that the measure is finite is actually hidden in the very definition of uniform integrability; for non-finite measures uniform integrability is defined in a different way... this is nicely summarized in the book "Measures, integrals and martingales" by René Schilling where you can find a lot of different ways to define uniform integrability (for finite measure spaces they are all equivalent). $\endgroup$
    – saz
    Mar 16, 2017 at 7:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .