Given a Banach space $X$ and a measure space $(\Omega ,\mu )$, one says that a function $$ f:\Omega \to X $$ is Dunford integrable, or scalarly integrable if, for every $\varphi $ in the topological dual $X'$, the function $$ x\in \Omega \mapsto \varphi \big (f(x)\big )\in \mathbb C $$ belongs to $L^1(\Omega , \mu )$.
If, in addition, for every measurable subset $E\subseteq \Omega $, there exists $x_E\in X$ such that $$ \int_E\varphi \big (f(x)\big )\, d\mu (x) = \varphi (x_E), \quad\forall \varphi \in X', $$ then $f$ is said to be Pettis integrable and each $x_E$ is called the Pettis integral of $f$ over $E$.
If $X$ is reflexive, every Dunford integrable function is Pettis integrable (this follows easily from the Closed Graph and the Riesz-Fischer Theorems).
On the downside, there are elementary examples of Dunford integrable $c_0$-valued functions which are not Pettis integrable.
My questions are thus:
Question 1. Is there a Dunford integrable function taking values in $B(H)$ (bounded operators on Hilbert's space) which is not Pettis integrable?
Question 2. Given a non-reflexive space $X$, can one prove that there exists a Dunford integrable $X$-valued function which is not Pettis integrable?
PS: Besides the result mentioned above regarding reflexive spaces, there is another important sufficient condition for Pettis integrability due to Dimitrov and Diestel: If $X$ does not contain any isomorphic copy of $c_0$, then each strongly measurable (meaning a.e. limit of simple functions) and Dunford integrable $X$-valued function is Pettis integrable.
