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
- (Uniform bounded on $L^1$) One has $\sup_n\|f_n\|_{L^1(\mu)}=\sup_n\int_{X}|f_n|d\mu <+\infty$.
- (No escape to vertical infinity) One has $\sup_n\int_{|f_n|\ge M}|f_n|d\mu\xrightarrow{M\rightarrow\infty} 0$.
- (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.
Problem(s):
- (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.
Best Answer
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.