Collection is uniformly integrable, but individual is not integrable

Question

Could you give me an example about:

"a collection of functions that is uniformly integrable but each (or some) function in the collection is not integrable."

This sounds counterintuitive? However according to Royden's "Real analysis":

• Definition of "uniform integrability": A family $\mathcal F$ of measurable functions on $E$ is said to be uniformly integrable over $E$ if for every $\epsilon >0$ there is a $\delta >0$ such that for each $f \in \mathcal F$, if $A\subset E$ is measurable and $m(A)<\delta$, then $\int_A|f|<\epsilon$.

• Property of integrability: When $m(E)<\infty$, $f$ is (Lebesgue) integrable over $E$ if and only if for every $\epsilon >0$ there is a $\delta >0$ such that if $A\subset E$ is measurable and $m(A)<\delta$, then $\int_A|f|<\epsilon$

When $m(E)=\infty$ then there is only one direction $\Rightarrow$ in the proposition about integrability right above. So I think my desired counter example should be about a collection of functions uniformly integrable on a set of infinite measure.

Thank you.

Answer 1

I see your point, and guest's comment provides an immediate example ($\mathbb{R}\to\mathbb{R},x\mapsto1$) for a nonintegrable but uniformly integrable function.

I believe the reason why Royden & Fitzpatrick defines uniform integrability for a collection of measurable functions, instead of integrable functions, is to emphasize the analogy of this concept and equicontinuity, at least at the formal level:

1. Let $\mathcal{F}$ be a collection of measurable functions from $E\subseteq\mathbb{R}$ to $\mathbb{R}$. Then

   • If $\mathcal{F}\subseteq L^1$, then $\forall f\in\mathcal{F},\forall\varepsilon>0,\exists\delta>0,\forall A\in\mathcal{M}:$

$$A\subseteq E, m(A)<\delta\implies \int_A |f|<\varepsilon.$$

• $\mathcal{F}$ is uniformly integrable (or equiintegrable, or uniformly absolutely continuous) if $\forall\varepsilon>0,\exists\delta>0,\forall f\in\mathcal{F},\forall A\in\mathcal{M}:$

$$A\subseteq E, m(A)<\delta\implies \int_A |f|<\varepsilon.$$

2. Let $(X,d)$ be a metric space, $x_0\in X$, $\mathcal{F}\subseteq\mathbb{R}^X$. Then
   • $\mathcal{F}$ is equicontinuous at $x_0$ if $\forall\varepsilon>0,\exists\delta>0,\forall f\in\mathcal{F},\forall x\in X:$

$$d(x,x_0)<\delta\implies |f(x)-f(x_0)|<\varepsilon.$$

Just as equicontinuity is generally not defined exclusively for continuous functions, Royden & Fitzpatrick chooses not to define uniform integrability exclusively for integrable functions.

Though admittedly this analogy is rather weak, and perhaps as a remedy to this all of the results in the corresponding section in the book relates either to collections of integrable functions, or collections of measurable functions with finite-measured domains (e.g. the Vitali Convergence Theorem).