Uniform integrability

Remarks

Example: on $[0, 1]$ with Lebesgue measure, let $\mathcal{F} = \set{f : \abs{f} \le 2}$. For $M \ge 2$ each set $\set{\abs{f} > M}$ is empty, so the supremum is $0$.

Non-example: on $[0, 1]$, let $f_n = n \mathbf{1}_{(0, 1/n)}$. For $n > M$ the integral of $f_n$ over $\set{f_n > M}$ equals $1$, so the supremum is $1$ for every $M$.

On a finite measure space the condition is equivalent to two conditions together. First, $\sup_{f \in \mathcal{F}} \int \abs{f} \, d\mu < \infty$. Second, for every $\varepsilon > 0$ there is $\delta > 0$ with $\int_A \abs{f} \, d\mu < \varepsilon$ for all $f \in \mathcal{F}$ whenever $\mu(A) < \delta$. Any family dominated by one integrable function is uniformly integrable. Bounded families in $L^p$ for some $p > 1$ are uniformly integrable. On infinite measure spaces other versions of the definition are in use.