Pettis integral

Remarks

The vector $x_A$ is unique because the continuous linear functionals separate the points of $E$, by the Hahn–Banach theorem.

Example: let $E = \R^n$ and let each component $f_k$ of $f$ be integrable. Then $x_A = (\int_A f_1 \, d\mu, \dots, \int_A f_n \, d\mu)$ works, because every $\varphi \in E^*$ is a linear combination of the coordinates.

Non-example: let $X = \N$ with the counting measure, $E = c_0$, the space of sequences tending to $0$, and $f(n) = e_n$, the $n$-th unit vector. For each $\varphi \in E^* = \ell^1$, the sum $\sum_n \abs{\varphi(e_n)}$ is finite, so $\varphi \circ f$ is integrable. But for $A = \N$ the only candidate is the constant sequence $(1, 1, \dots)$, which is not in $c_0$, so $f$ is not Pettis integrable.

Every Bochner integrable function is Pettis integrable, with the same integral. In infinite-dimensional spaces the converse fails.