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Transactions of the American Mathematical Society
Transactions of the American Mathematical Society
ISSN 1088-6850(online) ISSN 0002-9947(print)

 

Pettis integrability


Author: Gunnar F. Stefánsson
Journal: Trans. Amer. Math. Soc. 330 (1992), 401-418
MSC: Primary 46G10; Secondary 28B05, 46B20
MathSciNet review: 1070352
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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}}{\vert _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)$.


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DOI: http://dx.doi.org/10.1090/S0002-9947-1992-1070352-0
PII: S 0002-9947(1992)1070352-0
Keywords: Banach space, Pettis integral, weak measurability, $ {\text{weak}}^{\ast} $ integral
Article copyright: © Copyright 1992 American Mathematical Society