Pettis integrability

Abstract:

A weakly measurable function $f:\Omega \to X$ is said to be determined by a subspace $D$ of $X$ if for each ${x^{\ast } } \in {X^{\ast } }$, ${x^{\ast }}{|_D} = 0$ implies that ${x^{\ast }}\;f= 0$ a.e. For a given Dunford integrable function $f:\Omega \to X$ with a countably additive indefinite integral we show that $f$ is Pettis integrable if and only if $f$ is determined by a weakly compactly generated subspace of $X$ if and only if $f$ is determined by a subspace which has Mazur’s property. We show that if $f:\Omega \to X$ is Pettis integrable then there exists a sequence (${\varphi _n}$) of $X$ valued simple functions such that for all ${x^{\ast }} \in {X^{\ast }}$, ${x^{\ast }}f= {\lim _n}{x^{\ast }} {\varphi _n}$ a.e. if and only if $f$ is determined by a separable subspace of $X$. For a bounded weakly measurable function $f:\Omega \to {X^{\ast } }$ into a dual of a weakly compactly generated space, we show that $f$ is Pettis integrable if and only if $f$ is determined by a separable subspace of ${X^{\ast }}$ if and only if $f$ is weakly equivalent to a Pettis integrable function that takes its range in ${\text {cor}}_f^{\ast } (\Omega )$.