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.
Approximation by simple functions is the way to go. Fix $\epsilon>0$ and let $f$ be such that $|f|\le1$ and $\left|\int_Ef\, d\nu\right|\ge\mu_3(E)-\frac\epsilon2$. Now choose a simple function $\varphi=\sum_{j=1}^na_j\mathbf1_{E_j}$ (with $E_j$ disjoint and $\bigcup E_j=E$) such that $\left|\int_E f\, d\nu-\int_E \varphi\, d\nu\right|\le\frac\epsilon2$. It follows that $$\mu_3(E)\le\left|\int_Ef\, d\nu\right|+\frac\epsilon2\le\left|\int_E\varphi\, d\nu\right|+\epsilon=\left|\sum_{j=1}^na_j\nu(E_j)\right|+\epsilon\le\sum_{j=1}^n|\nu(E_j)|+\epsilon\le\mu_1(E)+\epsilon,$$ where the second last inequality follows since necessarily $|a_j|\le1$. Since $\epsilon>0$ was arbitrary we are done.