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Let $\nu$ is a complex measure then its total variation $|\nu|$ is

(1) determined by the property that if $d\nu= f d\mu$ where $\mu$ is a positive measure, then $d|\nu|=|f|d\mu$.

(2) $|\nu|=|\nu_r|+|\nu_i|=({\nu}_r^+ + {\nu}_r^-)+ ({\nu}_i^+ + {\nu}_i^-)$.

First definition is given in Real Analysis by Folland. I want to ask if the second definition is true

Thank you!

Aarna
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1 Answers1

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$|\nu|(E)$ is defined to be $\int_E|f|d\mu$ whenever $\nu(E)=\int_E fd\mu$. You can talk about the whether the definition is well-defined. It doesn't make sense to say that $|\nu|=|\nu_r|+|\nu_i|$ a priori. What you can say is that, if $\lambda=|\nu_r|+|\nu_i|$, then $\nu <<\lambda$. By Radon-Nikodym theorem, we have $\nu(E)=\int_Efd\lambda$ and therefore $|\nu|(E)=\int_E|f|d\lambda$.

For Welldefinedness: Suppose that $d\nu=f_1d\mu_1=f_2d\mu_2$, then take $\phi=\mu_1+\mu_2$. See that $\mu_1<<\phi$ and $\mu_2<<\phi$. Again by Radon Nikodym Theorem, we have $$\frac{d\nu}{d\phi}=\frac{d\nu}{d\mu_1}\frac{d\mu_1}{d\phi}=f_1\frac{d\mu_1}{d\phi} \implies d\nu=f_1\frac{d\mu_1}{d\phi}d\phi$$ Similar argument gives that $$ d\nu=f_2\frac{d\mu_2}{d\phi}d\phi$$

Thus, $$|f_1|d\mu_1=|f_1|\frac{d\mu_1}{d\phi}d\phi=|f_2|\frac{d\mu_2}{d\phi}d\phi=|f_2|d\mu_2$$

tattwamasi amrutam
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  • Hello. So, we’re only assuming $\mu_1$ and $\mu_2$ are positive measures? If not $\sigma$-finite, though, then Radon-Nikodym Theorem is not applicable since this requires $\mu_1$ and $\mu_2$ to be $\sigma$-finite. Unless it’s not possible to define complex measures in terms of an integral wrt non-$\sigma$-finite measure? (p.s. This argument looks identical to that given on pg.93 of Folland…which, as far as I can tell, is not correct for the reasons already stated.) Could you help? Surely I’m missing something… – J.G.131 Oct 26 '24 at 15:13