equivalent conditions for uniform integrability

Let $(\mathrm{\Omega },ℱ,\mu )$ be a measure space and $S$ be a bounded subset of $L^1(\mathrm{\Omega },ℱ,\mu )$. That is, $\int |f|𝑑\mu$ is bounded over $f\in S$. Then, the following are equivalent.

1. For every $ϵ>0$ there is a $\delta >0$ so that

   for all $f\in S$ and .

2. For every $ϵ>0$ there is a $K>0$ satisfying

   for all $f\in S$.

3. There is a measurable function $\mathrm{\Phi }:ℝ\to [0,\mathrm{\infty })$ such that $\mathrm{\Phi }(x)/|x|\to \mathrm{\infty }$ as $|x|\to \mathrm{\infty }$ and

   $$\int \mathrm{\Phi }(f)𝑑\mu$$

   is bounded over all $f\in S$. Moreover, the function $\mathrm{\Phi }$ can always be chosen to be symmetric and convex.

So, for bounded subsets of $L^1$, either of the above properties can be used to define uniform integrability. Conversely, when the measure space is finite, then conditions (2) and (3) are easily shown to imply that $S$ is bounded in $L^1$.

To show the equivalence of these statements, let us suppose that for $f\in S$.

(1) implies (2)

For $ϵ>0$, property (1) gives a $\delta >0$ so that whenever $f\in S$ and . Choosing $K>L/\delta$, Markov’s inequality gives

and, therefore, .

(2) implies (3)

For each $n=1,2,\mathrm{\dots }$, property (2) gives a $K_n$ satisfying

$$\int {(|f|-K_n)}_{+}𝑑\mu \le {\int }_{|f|>K_n}|f|𝑑\mu \le 2^{-n}.$$

Without loss of generality, the $K_n$ can be chosen to be increasing to infinity, so we can define $\mathrm{\Phi }(x)={\sum }_n{(|x|-K_n)}_{+}$. Then,

$$\int \mathrm{\Phi }(f)𝑑\mu =\sum _n\int {(|f|-K_n)}_{+}𝑑\mu \le \sum _n{2}^{-n}=1.$$

(3) implies (1)

First, suppose that for $f\in S$.For $ϵ>0$, the condition that $\mathrm{\Phi }(x)/|x|\to \mathrm{\infty }$ as $|x|\to \mathrm{\infty }$ gives a $K>0$ such that $\mathrm{\Phi }(x)/|x|\ge 2M/ϵ$ whenever $|x|>K$.Setting $\delta =ϵ/2K$,

whenever and $f\in S$.