Total variation of a measure

Remarks

The map $\abs{\nu}$ is a measure, and $\abs{\nu(E)} \le \abs{\nu}(E)$ for all $E \in \Sigma$. For a complex measure, $\abs{\nu}$ is a finite measure.

Example: let $\nu = \delta_0 - \delta_1$ on the subsets of $\R$, where $\delta_a$ is the Dirac measure at $a$. For $E = \set{0, 1}$, one has $\nu(E) = 0$, but the partition $\set{0}, \set{1}$ gives $\abs{1} + \abs{-1} = 2$. No partition gives more, so $\abs{\nu}(E) = 2$ and $\norm{\nu} = 2$.

Non-example: the map $E \mapsto \abs{\nu(E)}$ is not the total variation in general. For the same $\nu$ it gives $0$ on $\set{0, 1}$, and it is not even additive: $\abs{\nu(\set{0})} + \abs{\nu(\set{1})} = 2 \ne 0$.

For a signed measure with Jordan decomposition $\nu = \nu^+ - \nu^-$, one has $\abs{\nu} = \nu^+ + \nu^-$. If $\nu(E) = \int_E f \, d\mu$ with $f \in L^1(\mu)$, then $\abs{\nu}(E) = \int_E \abs{f} \, d\mu$.