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3 added 2 characters in body

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 $L^1$.

2 added 900 characters in body

Not a proof yet.

To do the general case along Bill's lines, perhaps someRadon-Nikodym argument. I used to know that material quitewell, so maybe I can come up with something.

Here is what we need. Let $X$ be a reflexive Banach spaceAssume $\int|\langle f(t),y\rangle|\,d\mu(t)<\infty$for all $y \in Y$.so that this equation (1) holds for all $y \in X^*$.

Perhaps this

Assume $X$ is separable, so that we also know$t \mapsto \|f(t)\|$ is measurable. For $k \in \mathbb N$,let $J_k = { t \in \Omega : k-1 \le \|f(t)\| < k}$.Thus $J_1, J_2, \dots$ is a Radon-Nikodym derivative argument measurable partition of $\Omega$.Restricting to a set $J_k$ we have the same problem, butnow $f$ is bounded, and therefore certainly Pettis integrable(even Bochner integrable) and since the dense subspace$Y$ separates points of $X$, we conclude (1) holds formeasure all $m$ with respect 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}$and$E_k = J_k \cap E$for$k \in \mathbb N$.(Suppose we have real scalars.) Apply Bill's sequentialargument to the vectors$\mu$.m(E_k)$ to concludethat 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 itis positive. Similarly it is integrable on the setwhere it is negative. So $\langle f(t),y\rangle$is in $L^1$.

1

Not a proof yet.

To do the general case along Bill's lines, perhaps some Radon-Nikodym argument. I used to know that material quite well, so maybe I can come up with something.

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 $X^*$. For each $E \in \mathcal F$, suppose there is $m(E) \in X$ such that $$\langle m(E),y\rangle = \int_E\langle f(t),y\rangle\,d \mu(t)$$ for all $y \in Y$. Then we want to conclude that $f$ is Pettis integrable so that this equation holds for all $y \in X^*$.

Perhaps this is a Radon-Nikodym derivative argument for vector measure $m$ with respect to $\mu$.