I have the following question about the meaning of the total variation measure. If $(X,\mathcal{A})$ is a measurable space and $\nu$ is a signed measure, the total variation of $\nu$ is defined as $|\nu|=\nu^++\nu^-$ and this gives us an intuition that $|\nu|$ measures "how much charge" has been put on a set, no matter that some of it is positive and some of it is negative, excluding one another. Now if $\mu=\mu_1+i\mu_2$ is a complex measure with $\mu_1, \mu_2$ being signed finite measures, the total variation of $\mu$ is defined as follows:$$|\mu|(E)=\sup\{\sum_{j=1}^{n}|\mu(E_j)|: (E_j)_{j=1}^{n}\text{are disjoint and in }
\mathcal{A}, \bigcup_{j=1}^{n}E_j=E\}$$
My intuition would be to define $|\mu|$ as $|\mu_1|+|\mu_2|$, a measure that is able to count the total charge that has been assigned to a set by $\mu_1$ and $\mu_2$ both. If someone could explain the reasoning behind the complex total variation definition, I would be grateful.
Best Answer
The problem is that $|\cdot|$ is not additive. Consider the complex measure $$ \mu=(a+ib)\delta.$$ Whatever the definition of $|\mu|$ is, I expect that, in this case, it should boil down to $|\mu|=\sqrt{a^2+b^2}$. With your expected definition $|\mu|:=|\mu_1|+|\mu_2|$, however, we would get $|\mu|=|a|+|b|$.
This example might suggest the definition $|\mu|:= \sqrt{ |\mu_1|^2+|\mu_2|^2}.$ This is not good either. If $\mu=(a+ib)\delta_x + (c+id)\delta_y$, with $x\ne y$, then we expect $$ |\mu|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}.$$ With the above definition we would instead have $|\mu|=\sqrt{(a+c)^2+(b+d)^2}$.