Bartle–Dunford–Schwartz integral

Remarks

Example: every bounded measurable $f$ is $\mu$-integrable. Take simple $s_n$ with $\abs{s_n - f} \le 1/n$ everywhere. For a simple $s$ with $\abs{s} \le c$ one has $\norm{\int_B s \, d\mu} \le c \, \norm{\mu}(\Omega)$. Here $\norm{\mu}(\Omega)$ is the supremum of $\norm{\sum_i a_i \mu(A_i)}$ over finite measurable partitions $(A_i)$ of $\Omega$ and reals $\abs{a_i} \le 1$; it is finite when $\mu$ is countably additive. Apply this to $s = s_n - s_m$. So $\int_B s_n \, d\mu$ is a Cauchy sequence in $E$.

Non-example: let $\mu$ be Lebesgue measure on $(0, 1]$, viewed as a vector measure with values in $\R$, and $f(x) = 1/x$. For a real-valued finite measure, integrability in the sense of clause 2 is the same as Lebesgue integrability. Since $\int_0^1 dx / x = \infty$, no sequence as in clause 2 exists, and $f$ is not $\mu$-integrable.

When $E = \R$ and $\mu$ is a finite signed measure, the integral is the Lebesgue integral against $\mu$. For each continuous linear functional $x^*$ on $E$, $x^* \big( \int_B f \, d\mu \big) = \int_B f \, d(x^* \circ \mu)$.