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Motivation

The notion of uniform integrability is important for formulating the Vitali convergence theorem. Unfortunately, different authors define uniform integrability differently, which causes quite a lot of confusion, as evident in the number of questions about uniform integrability and Vitali convergence theorem in MSE. However, after reading many of those questions and the answers, I feel that many students are still confused. In this post, I want to clear all the confusion once and for all.

Definitions

Throughout this post, I would use the following definitions.

Let $(X, \mathcal{F}, \mu)$ be a measure space and $\Phi$ a collection of measurable functions on $(X, \mathcal{F})$, taking values in the extended real line.

  1. We say $\Phi$ is uniformly bounded in $L^1$ if $$ \sup_{f \in \Phi} \int_X |f| d\mu < \infty $$
  2. We say $\Phi$ does not escape to vertical infinity if for any $\epsilon > 0$, there exists $M > 0$ such that $$\sup_{f \in \Phi} \int_{|f| \geq M} |f| d\mu < \epsilon $$
  3. We say $\Phi$ does not escape to width infinity if for any $\epsilon > 0$, there exists $m > 0$ such that $$\sup_{f \in \Phi} \int_{|f| \leq m} |f| d\mu < \epsilon $$
  4. We call $\Phi$ equi-integrable if for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $A \in \mathcal{F}$ is a measurable set with $\mu(A) < \delta$, we have $$\sup_{f \in \Phi} \int_A |f| d\mu < \epsilon $$
  5. We call $\Phi$ tight if for any $\epsilon > 0$, there exists a measurable set $X_0 \in \mathcal{F}$ such that $\mu(X_0) < \infty$ and $$\sup_{f \in \Phi} \int_{X \setminus X_0} |f| d\mu < \epsilon $$

Confusion

Here comes the confusion about the definitions of uniform integrability:

  1. Measure theory textbooks usually define uniform integrability as being equi-integrable.
  2. Probability theory textbooks usually define uniform integrability as not escaping to vertical infinity.
  3. In Tao's blog post, uniform integrability was defined as being uniformly bounded in $L^1$, not escaping to vertical infinity and not escaping to width infinity.
  4. Yet some other authors define uniform integrability as being uniformly bounded in $L^1$ and equi-integrable.

What adds to more confusion is, the different definitions of uniform integrability are only equivalent under certain assumptions, while in general they are not equivalent. Moreover, different authors formulate the Vitali convergence theorem under different definitions of uniform integrability.

Claim

It is well-known that if $\mu$ is a finite measure, then $\Phi$ does not escape to vertical infinity if and only if it is uniformly bounded in $L^1$ and equi-integrable. For the general case, I would like to propose the following claim. To avoid confusion, I would avoid the term "uniformly integrable" altogether.

Let $(X, \mathcal{F}, \mu)$ be a measure space and $\Phi$ a collection of measurable functions on $(X, \mathcal{F})$, taking values in the extended real line. We do not make any other assumption on $\mu$ or $\Phi$. Then the following conditions are equivalent:

  1. $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity and does not escape to width infinity
  2. $\Phi$ does not escape to vertical infinity and is tight
  3. $\Phi$ is uniformly bounded in $L^1$, equi-integrable and tight

Now, let $(f_n)$ be a sequence of measurable functions and $f$ another measurable function on $(X, \mathcal{F})$. Suppose that:

  1. the collection $\Phi = \{f_n : n \in \mathbb{N}\}$ satisfies any one of the equivalent conditions above,
  2. the sequence $(f_n)$ converges to $f$ either almost everywhere or in measure,

then we have $f \in L^1(\mu)$, and $(f_n)$ converges to $f$ in $L^1(\mu)$.

Questions

  1. Is my claim correct?
  2. Is there any book or paper that makes a concerted effort to clear the confusion around uniform integrability and Vitali convergence theorem?

Edit: 3. Iosif Pinelis' answer gives an counter-example in which a sequence of functions is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity. However, the sequence is neither equi-integrable nor tight. So how to connect Tao's definition of uniform integrability with equi-integrability and tightness?

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2 Answers 2

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$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$ Your claim is not quite correct.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, n\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is not tight.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.


On the the other hand, if $\Phi$ does not escape to vertical infinity, then $\Phi$ is equi-integrable (the condition that $\Phi$ does not escape to width infinity is not needed here). Indeed, suppose that $\Phi$ does not escape to vertical infinity, and take any real $\ep>0$. Then there is a real $M>0$ such that $\sup_{f\in\Phi}\int_{|f|\ge M}|f|\,d\mu<\ep/2$. Let now $\de:=\ep/(2M)$. Then for any $f\in\Phi$ and any $A\in\mathcal F$ such that $\mu(A)<\de$ we have \begin{equation} \int_A|f|\,d\mu\le\int_{|f|\ge M}|f|\,d\mu+\int_A M\,d\mu<\ep/2+M\de=\ep. \end{equation} So, $\Phi$ is equi-integrable, as claimed.

Also, if $\Phi=\{f_n\colon\, n\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.

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  • $\begingroup$ Thank you for this counter-example. Knowing this, I'm now even more confused. How do I connect Tao's definition of uniform integrability with equi-integrability and tightness? $\endgroup$
    – user141240
    Commented Oct 3, 2021 at 1:06
  • $\begingroup$ Seems like this MSE post also asked a similar question but only got a partial answer: math.stackexchange.com/questions/4042711/… $\endgroup$
    – user141240
    Commented Oct 3, 2021 at 1:10
  • $\begingroup$ @Iosif Pinelis, It seems to me the sequence in your example is equi-integrable. You can simply take $\delta=\epsilon$. Indeed, equi-integrability will always hold for a family of uniformly bounded functions. $\endgroup$ Commented Nov 4, 2021 at 1:49
  • $\begingroup$ @YuvalPeres : Thank you for your comment. This is now fixed. $\endgroup$ Commented Nov 4, 2021 at 3:07
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This is a comment which I hope will be of interest to you, but it will be too long for this format. The type of conditions that you mention were analysed in some detail by Walter Schachermayer in the 70´s in the context of functional analysis. I am not sure if the term measure space has a universally accepted meaning but the basic setting he used was that of a positive Radon measure on a locally compact space which is bounded on compacta. One can also work with a $\sigma$-finite measure on a $\sigma$-algebra but beyond this one risks treading on "Here be dragons" territory.

The general setting is the scale of $L^p$ spaces. As is well known, the boundary case $p=\infty$ presents a number of disadvantages (bad density and duality properties, too rigorous a notion of convergence for many purposes, problems with tensor products and exponential type laws,...). It shares these with many spaces of bounded objects (continuous functions on a completely regular space, holomorphic functions, operator algebras, in particular von Neumann algebras, ...). This problem has been addressed by many mathematicians, in many contexts, but the basic idea goes back to Saks in his proof of what is now known as the Vitali-Hahn-Saks theorem. In our context, it consists of replacing the norm with the finest locally convex topology $\beta_1$ which agrees with that of $L^1_{\text{loc}}$ (the local $L^1$-topology) on the unit ball. This construction occurs frequently in the literature, under such monikers as "strict topology", "mixed topology",....

Then Schachermayer´s result is the equivalence of the following conditions on a subset $B$ of §L^1$;

  1. a) for every $\epsilon>0$ there is a $\delta>0$ so that if $A$ is an integrable subset of the space with $\mu(A)\leq \delta$, then $$ \int_A |x(t)|d\mu\leq \epsilon $$ for each $x$ in $B$;

b) for each $\epsilon >0$, there is a compact subset $K$ so that for each $x$ in $B$, $\int |x(t) |d\mu\leq \epsilon$, the integral being over the complement of $A$.

  1. $B$ is $\beta_1$ equicontinuous;

  2. $B$ is relatively weakly compact.

This means that $\beta_1$ is the Mackey topology.

These results are written up in the third chapter of the monograph "Saks Spaces and Applications to Functional Analysis".

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