Let $X$ a Banach Space and $(\Omega, \Sigma, \mu)$ a measure space. A function $F:\Omega\rightarrow X$ is Dunford integrable if $x^\ast\circ F$ is $\mu$-integrable for every $x^\ast\in X^\ast$. The space of functions that are Dunford integrable, denoted by $\mathbb{D}(\mu,X)$, is a normed space with
$$
 \|F\|:=\sup\left\{\int\_\Omega|x^\ast\circ F|d\mu:x^\ast\in X^\ast,\,\|x^\ast\|\leq1\right\}
$$
Does anybody knows if the space $\mathbb{D}(\mu,X)$ is complete?



Is the space of vectorial functions that are Dunford integrable complete?
*This a revised and expanded version of my post.*


**No**, it is not complete. For simplicity suppose that $X$ is reflexive and separable in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite and non-atomic. In this case the space of Pettis integrable functions is complete if and only if $X$ is finite-dimensional as proved by Thomas



> 
> G.E.F. Thomas, *Totally Summable Functions with Values in Locally Convex Spaces*,
>  Measure Theory (Oberwolfach 1975), Lecture Notes in Math. Vol. **541** (1976), 117–131.
> 
> 
> 


This is based on a result o his which asserts that there exist an absolutely sequence summable sequence $(x\_n)\_{n=1}^\infty$ in $X$ (reflexivity is not needed) and a sequence $(f\_n)\_{n=1}^\infty$ in the unit ball of $L\_1(\mu)$ such that the vector measure


$$\nu(A) = \sum\_{n=1}^\infty \int\limits\_A f\_n(t)\,{\rm dt}\cdot x\_n$$


does not have a Pettis intregrable density.



> 
> G.E.F. Thomas, The Lebesgue-Nikodym Theorem for Vector Valued Radon Measures, *Memoirs. of AMS*, **139**, American Mathematical Society, Providence, 1974.
> 
> 
> 


The sequence $(\sum\_{k=1}^n f\_k\cdot x\_k)$ is Cauchy in $\mathcal{P}$ because $(x\_n)\_{n=1}^\infty$ is absolutely summable, yet it is not convergent as there is no function $F$ such that


$$\lim\_{n\to \infty}\int\limits\_A \sum\_{k=1}^n f\_k(t)\cdot x\_k\,{\rm d}t\to \int\_A F(t)\,{\rm d}t.$$


It seems to me that Pettis integrable functions form a closed subspace of the space of Dunford integrable functions, hence you may extend the above result as in the case where $X$ is infinite-dimensional, you have an incomplete, closed subspace of a normed space, so the space itself cannot be complete.


*Addendum*. Let me point out that if you want to do some functional analysis with the space of Pettis integrable functions, even though incomplete, it is [barrelled](http://www.ams.org/journals/proc/1992-114-03/S0002-9939-1992-1107271-2/S0002-9939-1992-1107271-2.pdf).


