Here is what we need. Let $X$ be a reflexive Banach space (the graph of $T$ in the original), let $(\Omega,\mathcal F, \mu)$ be a probability space, let $f : \Omega \to X$ be scalarly measurable, let $Y$ be a dense subspace of $X^*$. Assume $\int|\langle f(t),y\rangle|\,d\mu(t)<\infty$ for all $y \in Y$. For each $E \in \mathcal F$, suppose there is $m(E) \in X$ such that $$ \langle m(E),y\rangle = \int_E\langle f(t),y\rangle\,d \mu(t) \qquad\qquad\hbox{(1)} $$ for all $y \in Y$. Then we want to conclude that $f$ is Pettis integrable so that (1) holds for all $y \in X^*$.
Assume $X$ is separable, so that we also know
$t \mapsto \|f(t)\|$ is measurable. For $k \in \mathbb N$,
let $J_k = \{ t \in \Omega : k-1 \le \|f(t)\| < k}$k\}$.
Thus $J_1, J_2, \dots$ is a measurable partition of $\Omega$.
Restricting to a set $J_k$ we have the same problem, but
now $f$ is bounded, and therefore certainly Pettis integrable
(even Bochner integrable) and since the dense subspace
$Y$ separates points of $X$, we conclude (1) holds for
all $E \subseteq J_k$ and all $y \in X^*$.
Now fix $y \in X^*$ possibly not in $Y$. Let
$E = \{ t \in \Omega : \langle f(t), y\rangle > 0}$
> 0\}$
and $E_k = J_k \cap E$ for $k \in \mathbb N$.
(Suppose we have real scalars.) Apply Bill's sequential
argument to the vectors $m(E_k)$ to conclude
that series $\sum_k m(E_k)$ converges unconditionally,
so in particular for this particular $y$ the series
$\sum_k \langle m(E_k),y\rangle$ converges and thus
$\langle f(t),y\rangle$ is integrable on the set where it
is positive. Similarly it is integrable on the set
where it is negative. So $\langle f(t),y\rangle$
is in $L^1$.
