Let \((\Omega,\Sigma,\mu)\) be a measure space with finite measure and \(X\) a Banach space. A weakly \(\mu\)-measurable function \(f:\Omega\to X\) is Pettis integrable if \(\langle x^*,f(\centerdot)\rangle\in L_ 1(\mu)\) for each \(x^*\in X^*\) and if for every \(A\in\Sigma\) there is an element \(\int_ Afd\mu\) in \(X\) — the Pettis integral of \(f\) over \(A\) - – such that \(\langle x^*,\int_ Afd\mu\rangle=\int_ A\langle x^*,f(\centerdot)\rangle d\mu\) for all \(x^*\in X^*\). On the space \({\mathcal P}(\mu,X)\) of all Pettis integrable functions \(\Omega\to X\), a norm is given by \[ \| f\|:=\sup_{x^*\in B_{X^*}}\int_ \Omega|\langle x^*,f(\centerdot)\rangle | d\mu, \] but Banach space is far away. It is in fact known that if \(\mu\) doesn’t have any atoms then \({\mathcal P}(\mu,X)\) is only complete when \(X\) is finite dimensional.
However, the main result of this paper asserts that \({\mathcal P}(\mu,X)\) is always barrelled. It is obtained as a consequence of a theorem on certain Boolean algebras of projections in general locally convex spaces. There are also some interesting applications to spaces \(L_ p(X)\) of Bochner integrable functions in a normed space \(X\). If \(1\leq p<\infty\), then \(L_ p(X)\) is always barrelled, regardless of whether \(X\) has this property or not.