Uniform Integrability and Vitali Convergence Theorem Confusion

measure-theoryreal-analysis

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?

Best Answer

$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$ 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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