Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, and let $V$ be a topological vector space with a dual space that separates points. Let $v_n : \Omega \to V$ be a sequence of Pettis integrable random variables, and write $\mathbb E[v_n]$ for the Pettis integral of $v_n$ (over $X$). Note that $\mathbb E[v_n]$ is a (non-random) vector in $V$, and is not a scalar value.
Let $\bar v_N := \frac{1}{N} \sum_{n=1}^N v_n$ denote the sample average. By linearity, $\bar v_N$ is Pettis integrable, and $\mathbb E[\bar v_N] = \frac{1}{N} \sum_{n=1}^N \mathbb E[v_n]$ in $V$.
Suppose that the partial sums $\frac{1}{N} \sum_{n=1}^N \mathbb E[\bar v_n]$ converge absolutely in the topology of $V$, in the sense that all rearrangements of the sum converge to a single vector $\lambda \in V$. The Weak Law of Large Numbers implies that $\langle \varphi, \mathbb E[\bar v_N] - \lambda \rangle \to 0$ for every functional $\varphi \in V^*$. Consequently, $\mathbb E[\bar v_N] \to \lambda$ in the weak topology on $V$.
Does it follow that $\mathbb E[\bar v_N] \to \lambda$ in the original topology on $V$? My intuition says no: weak convergence does not imply convergence. Still, I'm having trouble coming up with a counterexample. Can you provide one?