Abstract

Let $\mathfrak {X}$ be a Banach space and $(\Omega ,\Sigma ,\mu )$ be a finite measure space. A strongly measurable $f:\Omega \to \mathfrak {X}$ is Pettis integrable if and only if there exists a Young’s function $\Phi$ with ${\lim _{t \to \infty }}\Phi (t)/t = \infty$ such that ${x^ \ast }f \in {L^\Phi }(\mu )$ for all ${x^ \ast } \in {\mathfrak {X}^ \ast }$.

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