De la Vallée Poussin criterion on uniform integrability for infinite measures

Question

The de la Vallée Poussin criterion (which is often used in combination with the Dunford-Pettis theorem) is usually formulated for probability measures/finite measures, for example in [Bogachev: Measure Theory. (2007), Thm. 4.5.9.] the following version is given:

Let $(X, \mathfrak{A}, \mu)$ be a finite measure space. A family of functions $\Phi \subset L^1(X, \mu)$ is uniformly integrable if and only if there exists a nonnegative increasing function $G$ on $[0, + \infty)$ such that \begin{equation*} \lim\limits_{t \to +\infty} \frac{G(t)}{t} = \infty \quad \text{and} \quad \sup\limits_{f \in \Phi} \int_{X} G(|f(x)|) \, \mathrm{d}\mu (x) < \infty. \end{equation*} In such a case, one can choose a convex increasing function $G$.

The common definition of uniform integrability in the general case is:

The set $\Phi \subset L^1(X, \mu)$ is called uniformly integrable if $$ \forall \varepsilon > 0 \, \exists g \in L^1(X, \mu): \, \int_{\{|f| > g\}} |f| \, \mathrm{d} \mu \leq \varepsilon, \, \forall f \in \Phi.$$

But is there a version for infinite measures too? At least under certain conditions like $\sigma$-finite measures or atomless measures or at least for the (weighted) Lebesgue measure? Or is there an easy way to generalize it?

Would be very grateful for any references and help!

Answer 1

This result holds for any measure space.

Indeed, let $G$ be a Borel-measurable nonnegative function $G$ on $[0,\infty)$ such that $\lim_{t\to\infty}G(t)/t=\infty$ and hence $$\lim_{c\to\infty}h(c)=\infty,\text{ where }h(c):=\inf_{t>c}G(t)/t \tag{1}\label{1}$$ for real $c>0$. Assume also that $$S:=\sup_{f\in\Phi}\int_X d\mu\,G(|f|)<\infty. \tag{2}\label{2}$$ Then $\Phi$ is uniformly integrable.

Indeed, for each real $c>0$ $$\sup_{f\in\Phi}\int_A d\mu\,|f|\,1(|f|\le c) \le c\sup_{f\in\Phi}\int_A d\mu\to0 \tag{3}\label{3}$$ if $\mu(A)\to0$. Also, in view of \eqref{1} and \eqref{2}, for real $c>0$, $$\sup_{f\in\Phi}\int_A d\mu\,|f|\,1(|f|>c) \le\frac1{h(c)}\sup_{f\in\Phi}\int_X d\mu\,G(|f|)=\frac S{h(c)}\to0 \tag{4}\label{4}$$ as $c\to\infty$.

The desired result follows from \eqref{3} and \eqref{4}. $\quad\Box$