Uniform integrability

A family $\mathcal F\subset L^1(X,\mu)$ on a measure space $(X,\mathcal A,\mu)$ is uniformly integrable if

\sup_{f\in\mathcal F}\|f\|_1<\infty \qquad\text{and}\qquad \lim_{M\to\infty}\ \sup_{f\in\mathcal{F}} \int_{\{|f|>M\}} |f|\,d\mu = 0.

When $\mu(X)<\infty$ , the tail condition alone implies the uniform $L^1$ bound.

A sequence $(f_n)$ is uniformly integrable if the family $\{f_n:n\ge 1\}$ is uniformly integrable.

Interpretation

The condition rules out increasingly tall tails that retain substantial $L^1$ -mass. Together with suitable convergence hypotheses, it permits passage to limits in the Lebesgue integral.

Examples

If $|f|\le g$ almost everywhere for every $f\in\mathcal F$ , where $g\in L^1(X,\mu)$ , then $\mathcal F$ is uniformly integrable.
On $([0,1],\mathcal B,\lambda)$ , the functions $f_n=n\mathbf 1_{[0,1/n]}$ are not uniformly integrable: for any $M>0$ , choose $n>M$ . Then $\int_{\{|f_n|>M\}} |f_n|\,d\lambda = \int_0^{1/n} n\,dx = 1,$ so the supremum of the tail integrals does not tend to $0$ .