In page 129 of Rudin's real and complex analysis, it assumes that $\mu$ is a complex Borel measure and defines integration by $\int f d\mu = \int fh\ d|\mu|$, where $d\mu = h\ d|\mu|$ and $|h|=1$.
Then it claims that $\int \chi_E d\mu = \mu(E)$ is a special case, but I'm having trouble to see why $\int_E h d|\mu|$ is equal to $\mu(E)$. I tried to prove it using the definitions, but it didn't seem to work. I'm not sure what I'm missing...
We can decompose $\mu$ as $\mu_1^+-\mu_1^-+i(\mu_2^+-\mu_2^-)$, where $\mu_j^{\pm}$ are positive measures. One can see that $$\frac{\mathrm d\mu_1^+}{\mathrm d|\mu|}=(\Re h)^+,\frac{\mathrm d\mu_1^-}{\mathrm d|\mu|}=(\Re h)^-,\frac{\mathrm d\mu_2^+}{\mathrm d|\mu|}=(\Im h)^+,\frac{\mathrm d\mu_2^-}{\mathrm d|\mu|}=(\Im h)^-,$$ where $x^+$ denote $\max\{0,x\}$ and $x^+-x^-=x$. We thus get that \begin{align} \mu(E)&=\mu_1^+(E)-\mu_2^+(E)+i(\mu_2^+(E)-\mu_2^-(E))\\ &=\int \chi_E((\Re h)^+-(\Re h)^-+i((\Im h)^+)-(\Im h)^-)\mathrm d|\mu|\\ &=\int \chi_Eh\mathrm d|\mu|. \end{align}