# Can the total variation measure of a complex measure be related to its real and imaginary parts?

If $$\mu$$ is a signed measure with Jordan decomposition $$\mu=\mu^+-\mu^-$$, then the total variation measure of $$\mu$$ is equal to $$\mu^++\mu^-$$. My question is, is it similarly possible to express the total variation measure of a complex measure $$\mu$$ in terms of its real and imaginary parts $$\mu_r$$ and $$\mu_i$$?

Maybe $$|\mu_r|^2 + |\mu_i|^2$$ or the square root of that or something?

The short answer is no. To begin with, the total variation of a complex measure is defined as $$|\mu|(E):=\sup\left\{\sum_{j\in J}|\mu(A_j)|:\{A_j\}_{j\in J}\text{ is a disjoint measurable partition of }E\right\}.$$
Now, here's a counterexample to your question. In $$\mathbb{R}$$ with Borel sigma algebra, consider the measure $$\mu=\delta_1-(1+i)\delta_0$$. Thus, according to your notation, $$\mu_r=\delta_1-\delta_0$$ and $$\mu_i=-\delta_0$$. Then $$|\mu_r(\{0,1\})|=2$$ and $$|\mu_i(\{0,1\})|=1$$ but $$|\mu|(\{0,1\})=|\mu(\{0\})|+|\mu(\{1\})|=\sqrt{2}+1.$$