Let $\nu$ be a complex measure on $(X, \mathcal{M})$. If $E \in \mathcal{M}$, define:
- $\mu_1(E) = \sup\{\sum_1^n{|v(E_j)|}:n \in N, E_1, ..., E_n$ disjoint$, E = \bigcup_1^n{E_j}\}$
- $\mu_2(E) = \sup\{\sum_1^{\infty}{|v(E_j)|}:E_1, E_2,...$ disjoint$, E = \bigcup_1^{\infty}{E_j}\}$
- $\mu_3(E) = \sup\{|\int_E{fdv}|: |f| \le 1\}$.
Prove that $\mu_1 = \mu_2 = \mu_3 = |\nu|$
This problem is exercise 21, chapter 3 from Real Analysis of Folland. I followed the structure in his book, and up to now, I can prove that $\mu_1 \le \mu_2 \le \mu_3$ and $\mu_3 = |\nu|$. So the only point I stuck right now is to prove that $\mu_3 \le \mu_1$. I think about approximating $f$ using simple functions. But $f$ is a complex function, so the approximation is quite complicated. Anyone has any suggestion? Thanks so much for your help, I really appreciate.