I came across this posting where a definition of uniform integrability (found in Tao, T., Introduction Measure Theory. AMS, GTM vol 126, 2011) is given as follows:

Definition T: Suppose $(X,\mathscr{B},\mu)$ is a measure space (not necessarily finite). A sequence $f_n:X \rightarrow \mathbb{C}$ of absolutely Integrable functions is said to be uniformly Integrable if the following three statements hold

  1. (Uniform bounded on $L^1$) One has $\sup_n\|f_n\|_{L^1(\mu)}=\sup_n\int_{X}|f_n|d\mu <+\infty$.
  2. (No escape to vertical infinity) One has $\sup_n\int_{|f_n|\ge M}|f_n|d\mu\xrightarrow{M\rightarrow\infty} 0$.
  3. (No escape to width infinity) One has $\sup_n\int_{|f_n|\le\delta}|f_n|d\mu\xrightarrow{\delta\rightarrow0} 0$.

This is in contrast to the common used definition of uniform integrability by Hunt (Hunt, G.A., Martingales et Precessus de Markov, Paris: Dunod, 1966 pp. 254) which I state as follows:

Definition H: A family $\mathcal{F}\subset L_1$ is uniformly integrable iff $$\begin{align} \inf_{g\in L^+_1}\sup_{f\in\mathcal{F}}\int_{\{|f|>g\}}|f|\,d\mu=0 \tag{1}\label{one} \end{align}$$

Definition H is widely used on Probability theory and in functional analysis, for example, it provides an extension of Pettis-Dunford's theorem for $\sigma$-finite spaces. There are several known equivalences to Hunt's definition, some of which I will state below.

Observation: For a finite measure space, condition (3) Definition T is rather superfluous and, as it can be easily seen, Definitions H and T are equivalent in this setting.


  • (a): Are Definitions H and T equivalent for general measure spaces (or at least for infinite $\sigma$-finite measures, or only for countable families $\mathcal{F}\subset L_1$)?
  • (b): If not, are there any applications (problems) where Definition H does a job that Definition T can't deliver (and vice versa).

I spent some considerable amount trying to work through (a). I was unsuccessful to show either of the following statements:

Prop 1: Definition H implies Definition T.

Definition H does imply condition (1) and (2) in general measure spaces. Whether (3) also holds, escapes me.

Prop 2: Definition T implies Definition H.

Did not go far at all.

If all this is well known, a reference would be appreciated.

Here are some well known equivalencies of Definition H.

Notation: Suppose $(X,\mathscr{B},\mu)$ is a measure space.

  • When the the ambient space is clear from the context, we use $L_1$ as a shorthand for $L_1(X,\mathscr{B},\mu)$, and $\|f\|_1:=\int|f|\,d\mu$ (the $L_1$-norm). We also use the notation $L^+_1=\{f\in L_1: f\geq0\}$.
  • Given $g,h\in L_1$ with $g\leq h$ $\mu$-a.s., we denote $$[g,h]=\{f\in L_1: g\leq f\leq h\}$$
  • For any $\mathcal{A}\subset L_1$ and $f\in L_1$, $$d(f,\mathcal{A}):=\inf\{\|f-\phi\|_1: \phi\in\mathcal{A}\}$$

The following is from K. Bichteler, Integration: A functional approach, Birkhäuser Verlag, 1998. p.p. 57.

Definition B: A family $\mathcal{F}\subset L_1$ is uniformly integrable if for any $\varepsilon>0$, there are $g,h\in L_1$ with $g\leq h$ such that \begin{align}\sup_{f\in\mathcal{F}}d(f,[g,h])<\varepsilon\tag{2}\label{2} \end{align}

Comments: For any $x\in\mathbb{R}$ and $a<b$, define $x^b_a:=(a\vee x)\wedge b$. It is easy to check that if $c\leq a\leq b\leq d$, then $$|x-x^d_c|\leq |x-x^b_a|$$ Thus, Definition B may be rewritten as:

Definition B': $\mathcal{F}\subset L_1$ is uniformly integrable iff $$\inf_{g\in L^+_1}\sup_{f\in \mathcal{F}}d(f,[-g,g])=0$$

Comments: Definition B means that a family $\mathcal{F}\subset L_1$ is uniformly integrable if their elements are close to being dominated by an integrable function.

Now, for $g\geq0$, $|f-f^g_{-g}|=(|f|-g)_+$; hence, as $|x-x^a_{-a}|\leq |x-b|$ for all $|b|\leq a$, Definition B can also be rewritten as (see, for example (Klenke, A., Probability Theory, Springer 2006. pp. 134)

Definition H': A family $\mathcal{F}\subset L_1$ is uniformly integrable iff $$\begin{align} \inf_{g\in L^+_1}\sup_{f\in\mathcal{F}}\int_X(|f|-g)_+\,d\mu=0 \tag{1'}\label{onep} \end{align}$$

Comments: Observe that if $g\geq0$, then $$|f|\mathbb{1}_{\{|f|>2g\}}\leq (|f|-g)_+\mathbb{1}_{\{|f|>2g\}}+g\mathbb{1}_{\{|f|>2g\}}\leq 2(|f|-g)_+\mathbb{1}_{|f|>2g\}}$$ Hence, if $\mathcal{F}$ is uniformly integrable in the sense of Definition H' then, $$\begin{align} \inf_{g\in L^+_1}\sup_{f\in\mathcal{F}}\int_{\{|f|>g\}}|f|\,d\mu=0, \end{align}$$ that is, $\mathcal{F}$ is uniformly integrable in the sense of Definition H. Conversely, notice that for any $g\geq0$ $$(|f|-g)_+\leq|f|\mathbb{1}_{\{|f|>g\}}.$$ Hence, if $\mathcal{F}\subset L_1$ satisfies \eqref{one} then, it is uniformly integrable in the sense of Definition H'. This establishes the equivalence of Definitions B, H and H'.

Notice that for any $A\in\mathscr{B}$ and $g\geq0$ $$|f|\mathbb{1}_A\leq|f|\mathbb{1}_{\{|f|>g\}}+g\mathbb{1}_A$$ This leads to another equivalency

Theorem 1: A family $\mathcal{F}\subset L_1$ is uniformly integrable iff $a:=\sup_{f\in\mathcal{F}}\|f\|_1<\infty$, and for any $\varepsilon>0$ there exists $g_\varepsilon\in L^+_1$ and $\delta>0$ such that for any $A\in\mathscr{B}$, $$\int_Ag_\varepsilon<\delta\qquad\text{implies}\quad \sup_{f\in\mathcal{F}}\int_A |f|\,d\mu<\varepsilon$$

Proof: Necessity is obvious. For sufficiency, suppose $\mathcal{F}$ has the property described in the Theorem above. For $\varepsilon>0$ choose let $a$, $g_\varepsilon\in L^+_1$ and $\delta_\varepsilon$ as in the statement of the Theorem. For any $c>0$ $$\int_{\{|f|>cg\}}g\,d\mu \leq\frac{1}{c}\int_{\{|f|>cg_\varepsilon\}}|f|\,d\mu\leq\frac{a}{c}$$ Thus, for $c>\frac{a}{\delta_\varepsilon}$, we obtain that $\int_{\{|f|>cg_\varepsilon\}}g_\varepsilon\,d\mu<\delta_\varepsilon$ and so, $$\int_{\{|f|>cg_\varepsilon\}}|f|\,d\mu<\varepsilon$$ The uniform integrability follows.

The case $\mu(\Omega)<\infty$ is of interest in the Theory of Probability. In this case, the infimum in Definition H (equivalently in Definition H') can be taken over nonnegative numbers:

Theorem 2: Suppose $(X,\mathscr{F},\mu)$ is a finite measure space. A family $\mathcal{F}\subset L_1$ is uniformly integrable iff $$\begin{align} \inf_{a\geq0}\sup_{f\in\mathcal{F}}\int_{\{|f|>a\}}|f|\,d\mu=0\tag{4}\label{four} \end{align}$$ or equivalently $$\begin{align} \inf_{a>0}\sup_{f\in\mathcal{F}}\int_X(|f|-a)_+\,d\mu=0\tag{4'}\label{fourp} \end{align}$$

Proof: Given $\varepsilon>0$ choose $g_\varepsilon\in L^+_1$ and $a_\varepsilon\in(0,\infty)$ such that $$\begin{align} \sup_{f\in \mathcal{F}}\int_{\{|f|>g_\varepsilon\}}|f|\,d\mu<\varepsilon/2, \qquad \int_{\{g_\varepsilon>a_\varepsilon\}}g\,d\mu<\varepsilon/2\end{align}$$ Then, for any $f\in \mathcal{F}$ $$\int_{\{|f|>a_\varepsilon\}}|f|\,d\mu\leq\int_{\{|f|>g_\varepsilon\}}|f|\,d\mu+\int_{\{g>a_\varepsilon\}}g_\varepsilon\,d\mu<\varepsilon$$ Conversely, when $\mu(X)<\infty$, the family of nonnegative constant functions is contained in $L^+_1$ and so, \eqref{four} implies \eqref{one}

Comment: Versions of this theorem are stated as a definition in many probability textbooks.

For $\sigma$-finite measures, here is another equivalency found in the literature.

Theorem 3: Suppose $(X,\mathscr{B},\mu)$ is $\sigma$-finite and let $h\in L^+_1$ with $h>0$ a.s. A family $\mathcal{F}\subset L_1$ is uniformly integrable iff

  • (i) $a:=\sup_{f\in\mathcal{F}}\|f\|_1<\infty$, and
  • (ii) for any $\varepsilon>0$ there is $\delta>0$ such that, for any $A\in\mathscr{B}$ $$\int_Ah\,d\mu<\delta,\qquad\text{then}\quad \sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon$$

Comment: Notice that if $\mu(X)<\infty$, one may take $h\equiv1$ to obtain another familiar definition of uniform integrability that appears in many probability textbooks.

Proof to Theorem 3: $\sigma$-finiteness implies the existence of functions $h\in L^+_1$ with $h>0$ a.s. If $\mathcal{F}$ is uniformly integrable then, choose $g\in\mathcal{L}^+_1$ such that $$\sup_{f\in\mathcal{F}}\int_{\{|f|>g\}}|f|\,d\mu\leq\varepsilon/3$$ Then, for any $f\in\mathcal{F}$ $$\int_X|f|\,d\mu\leq\int_{\{|f|>g\}}|f|\,d\mu+\int_{\{|f|\leq g\}}g\,d\mu<\varepsilon/3 +\|g\|_1$$ Since $\phi_n:=\mathbb{1}_{\{g>nh\}}\xrightarrow{n\rightarrow\infty}0$ a.s., dominated convergence implies that $\int\phi_n g\,d\mu\xrightarrow{n\rightarrow\infty}0$. Choose $N\in\mathbb{N}$ large enough so that $\int \phi_n g\,d\mu<\varepsilon/3$. For any $A\in\mathscr{B}$ $$ |f|\mathbb{1}_A\leq |f|\mathbb{1}_{\{|f|>g\}} + g\mathbb{1}_{\{g>Nh\}}+ Nh\mathbb{1}_A$$ Thus, for $\delta=\varepsilon/(3N)$, $\int_Ag\,d\mu<\delta$ implies $\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon$.

The converse follows as in the proof of Theorem 3.

  • $\begingroup$ Nice question, I also think it is weird that the definition in Tao's book doesn't prevent escape to horizontal infinity (which is the counterexample in the answer below). $\endgroup$ Jul 29, 2022 at 8:47

3 Answers 3


The definitions are not equivalent, even on the real line. For every integrable function $f \ge 0$, the family of translates $\{x \mapsto f(x-c)\}_{c \in {\mathbb R}}$ is uniformly integrable by definition T but not by definition H. In fact, definition T does not yield that an almost everywhere convergent, uniformly integrable sequence converges in $L^1$.

Definition H does imply definition T. We will verify property (3.) in T, assuming a sequence $f_n$ satisfies H.

Given $\epsilon>0$, find $g \in L_1^+$ such that $$\sup_n \int_{|f_n|>g} |f_n| \, d\mu <\epsilon \,. \quad (*)$$

We have $$ \sup_n \int_{|f_n| \le g \wedge \delta} |f_n| \, d\mu \le \int g \wedge \delta \, d\mu \to 0 \quad \text{as} \; \delta \downarrow 0$$ by dominated convergence. Together with (*), this gives $$ \limsup_{\delta \downarrow 0} \sup_n \int_{|f_n| \le \delta} |f_n| \, d\mu \le \epsilon \,,$$ and property (3.) in definition T follows.


This is a comment for @YuvalPeres which a rather long to add in the comment section.

The last part (last edit in your posting) is what I failed to see immediately. Given $\varepsilon>0$ I chose $g\in L^+_1$ such that $$\sup_{f\in\mathcal{F}}\int_{\{|f|>g\}}|f|\,d\mu<\varepsilon$$ Then, for any $f\in \mathcal{F}$ \begin{align} \int_{\{|f|\leq\delta\}}|f|\,d\mu\leq \int_{\{|f|>g\}}|f|\,d\mu+\color{darkgreen}{\int_{\{|f|\leq\delta\}\cap\{ |f|\leq g\}}|f|\,d\mu} \end{align} Now I see that I could have continued as $$\begin{align} \color{darkgreen}{\int_{\{|f|\leq\delta\}\cap\{ |f|\leq g\}}|f|\,d\mu}&= \int_{\{|f|\leq g\wedge \delta\}}|f|\,d\mu\\ &\leq \int_{\{|f|\leq g\wedge \delta\}}g\wedge\delta\,d\mu \leq \int g\wedge\delta\,d\mu\xrightarrow{\delta\rightarrow0}0 \end{align}$$

So indeed, Hunt's definition implies Definition T in the general setting.

As your simple example shows, Definition T does not imply Hunt's (without additional assumptions). So it seems, Hunt's notion of uniform integrability stands again another test of time.

Still, I wonder why would Terry introduce such definition of uniform integrability (some specific application in Harmonic Analysis for which Hunt's notion was not useful, perhaps?).

In any event, thanks for your answer and for showing a detail (although rather simple in retrospect) that I missed in my previous attempts prop 1 (in my posting).

  • 1
    $\begingroup$ I also like def. H. I think that (in Euclidean space, at least) definition H is equivalent to Def. T plus tightness of the measures $A \mapsto \int_A f \, d\mu$. If you require tightness then part 3 of def T becomes redundant. so one can see requirement 3 in T as a weakening of tightness, and if you want to restore H you just need to check tightness in addition to T1 and T2. $\endgroup$ Apr 27, 2022 at 3:25
  • 1
    $\begingroup$ PS Perhaps you would like to improve en.wikipedia.org/wiki/Uniform_integrability ? The H definition is missing there. $\endgroup$ Apr 27, 2022 at 3:27
  • 1
    $\begingroup$ @YuvalPeres: just made an edit to the wiki article. I hope my changes get improved rather than reversed to the previous version. $\endgroup$
    – Mittens
    Apr 27, 2022 at 21:24
  • $\begingroup$ Thanks! I hope so too, that was a glaring omission. $\endgroup$ Apr 28, 2022 at 1:49

Tao states in his blog note a stronger definition of uniform integrability in place of the older one, with property $(3)$ properly changed to:

(No escape to width or horizontal infinity) For every ${\varepsilon>0}$, there is a finite measure subset ${E_\varepsilon}$ of ${X}$ such that ${\int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq \varepsilon}$ for all ${n}$.

Under this strengthened condition the two definitions indeed become equivalent.


First assume definition H.

To show uniform bound on $L_1$ norm, note that $\forall \varepsilon > 0$, $\exists h: X \rightarrow [0, +\infty]$ which is absolutely integrable and such that $\forall n \geq 1$, $\int_X |f_n| d\mu = \int_{|f_n|>h} |f_n|\ d\mu + \int_{|f_n|\leq h} |f_n|\ d\mu \leq \int_{X} h\ d\mu + \varepsilon < \infty$. Hence ${\sup_n \|f_n\|_{L^1(\mu)} < +\infty}$.

To show no escape to vertical infinity, note that since $h$ is absolutely integrable, it is finite a.e. So we can modify it on a null set and assume that $h \leq K$ on $X$ for some $K > 0$. Then $\forall M \geq K$, $\sup_n\int_{|f_n|\ge M}|f_n|\ d\mu\ \leq \sup_n \int_{|f_n| > h} |f_n|\ d\mu \leq \varepsilon$. That is, ${\sup_n \int_{|f_n| \geq M} |f_n|\ d\mu \rightarrow 0}$ as ${M \rightarrow +\infty}$.

Finally, to show no escape to width or horizontal infinity, let $\varepsilon > 0$. Use the fact that the constant sequence $h_n = h$ is uniformly integrable, we can find a finite measure subset ${F_\varepsilon}$ of ${X}$ such that ${\int_{X \backslash F_\varepsilon} h\ d\mu \leq \varepsilon}$. $\forall n \geq 1$, definition H essentially says that the bulk of the $L_1$ mass of $f_n$ is concentrated in the set $|f_n| \leq h$. Define the set $E^n_{\varepsilon} := \{x \in X: |f_n(x)| \leq h(x)\} \bigcap F_\varepsilon$. Then we have $\mu(E^n_{\varepsilon}) < \infty$ for every $n$. Furthermore, we have:

$\begin{multline*} {\int_{X \backslash E^n_\varepsilon} |f_n|\ = \int_{(|f_n| > h) \cap F_\varepsilon} |f_n| + \int_{(|f_n| \leq h) \cap F_\varepsilon^\complement} |f_n| + \int_{(|f_n| > h) \cap F_\varepsilon^\complement} |f_n| \leq 2\int_{|f_n| > h} |f_n| + \int_{X \backslash F_\varepsilon} h \leq 3\varepsilon} \end{multline*}$. Where we have used the identity $\displaystyle (A \bigcap B)^\complement = (A \bigcap B^\complement) \bigcup (A^\complement \bigcap B) \bigcup (A \bigcup B)^\complement.$ By taking $E_\varepsilon = \bigcup_{n=1}^\infty E_\varepsilon^n$, which again has finite measure since it is a subset of $F_\varepsilon$, we see that ${\int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq 3\varepsilon}$ for all ${n}$, as desired.

Therefore H $\rightarrow$ T.

Conversely, assume definition T. Let $\varepsilon > 0$. By property $(2)$ and $(3)$, there exist some $M > 0$, and a finite measure set $E_\varepsilon$ such that ${\int_{|f_n| \geq M} |f_n|\ d\mu} \leq \varepsilon$ and ${\int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq \varepsilon}$, $\forall n \geq 1$. Define $h := M1_{E_\varepsilon}$. Clearly, $h$ is absolutely integrable and $\forall n \geq 1$, we have $\displaystyle \int_{|f_n| > h} |f_n|\ d\mu = \int_{(|f_n| > h) \cap E_\varepsilon} |f_n|\ d\mu + \int_{(|f_n| > h) \cap E_\varepsilon^\complement} |f_n|\ d\mu \leq \int_{|f_n| \geq M} |f_n|\ d\mu + \int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq 2\varepsilon,$ which implies that $\sup_n \int_{|f_n| > h} |f_n|\ d\mu \leq 2\varepsilon.$

Thus T $\rightarrow$ H and the two definitions are equivalent.


You must log in to answer this question.

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