Showing existence of the Pettis integral Suppose $\xi\sim \mathbb{P}_\xi$ is a random element in a separable reflexive Banach space of weak order one. I was able to show the map
$$T_\xi :X^*\to L^1(\Omega,\mathbb{R}),\quad T_\xi f=f\circ \xi$$
is continuous. I am trying to show the existence of the Pettis integral, i.e. the element $m_\xi \in X$, such that
$$f(m)=\int_\Omega f(\xi (\omega))\mathbb{P}(d \omega),\quad \forall f\in X^*$$
If we define the functional $$g_\xi(f)=\int f\circ \xi d\mathbb{P}$$
and show it's continuous, then its in $X^{**}$ and the result follows from reflexivity. Let $f_n\to f$ in $X^*$.
$$\lim g_\xi (f_n)= \lim \int T_\xi f_n =^!\int\lim T_\xi f_n =\int T_\xi f =g_{\xi}(f)$$
But I don't see why the integral itself needs to be continuous on $T_\xi(X^*)$, as it needs not be bounded.
 A: I'm a little confused by your setup, since you seem to claim that $f\circ \xi$ takes values in $X$, but to me this looks like a scalar valued function if $\xi\in X$? In this case, it is a completely standard application of the convergence theorem.
Edit: This was supposed to be a comment, but this is what happens when working from a mobile device. I should probably mention before I give any answer, I am not an expert on this subject, so please correct me if I make an error.
The function $f\circ \xi$ seems to me to be a scalar-valued function on $\Omega$ and, moreover, this is an $L^1$ function with respect to $\Bbb P$. Now the sequence $f_n$ converges to $f$ in $X$ and, since you have already shown $T_\xi$ is continuous, $f_n\circ \xi$ is an $L^1$-convergent sequence whose limit is $f\circ \xi$. Now we need two standard facts: 1) convergent sequences are bounded; and 2) $L^1$-convergence implies pointwise convergence almost everywhere of a subsequence. We may instead consider this subsequence. Thus there exists a constant $M > 0$ for which $|f_n\circ \xi| \leq M$ a.e. on $\Omega$, with respect to $\Bbb P$. and $f_n(\xi(\omega)) \to f(\xi(\omega))$ for almost all $\omega\in \Omega$. (I am assuming $\Bbb P$ is a probability measure, given the context.) Thus $g_\xi(f_n) \to g_\xi(f)$ as you claimed, using the dominated convergence theorem.
