Complex measure

Remarks

Taking all $E_n = \emptyset$ gives $\nu(\emptyset) = 0$. The left side does not depend on the order of the $E_n$, so every rearrangement of the series converges to the same value, and the series converges absolutely.

Example: let $\mu$ be a measure and $f \in L^1(\mu)$ complex-valued. Then $\nu(E) = \int_E f \, d\mu$ is a complex measure; countable additivity follows from the dominated convergence theorem with dominating function $\abs{f}$.

Non-example: Lebesgue measure on $\R$ is not a complex measure, because $\lambda(\R) = \infty$ is not a complex number.

Writing $\nu = \nu_r + i \nu_i$, the real and imaginary parts $\nu_r$ and $\nu_i$ are finite signed measures. Integration against $\nu$ is defined by $\int f \, d\nu = \int f \, d\nu_r + i \int f \, d\nu_i$.