Equivalent definitions of total variation of a complex measure Let $(X,M,\mu)$ be a complex measure space.  
Then for $E \in M$ we have $|\mu|(E)=sup\{\Sigma_k| \mu(E_k)|: E = \sqcup_{k=1}^\infty E_k $ where $ E_k \in M\}$
Prove that for each $E\in M$
$$\begin{align*} 
|\mu|(E)&=\sup\left\{\sum_{k=1}^N |\mu(E_k)| : \{E_k\}_{k=1}^N \ \text{is a finite partition of} \ E\right\}\\
&=\sup\left\{\Big|\int_E f\,d\mu\Big| : f \ \text{is measurable and} \ |f|\leq 1\right\}\end{align*}$$
I've been having a real rough time trying to show these inequalities... I feel like I'm running into some sort of tunnel vision, I've been stuck on this for awhile now, I'd really appreciate it if someone could walk me through this one.. Thanks
 A: Try setting $f = \chi_{E_k} sgn(\mu(E_k))$, then you get that $$\left|\int f d\mu \right| = \sum|\mu(E_k)|$$ and conclude that
$$ \sup\left\{\sum_{k=1}^N |\mu(E_k)| : \{E_k\}_{k=1}^N \ \text{is a finite partition of} \ E\right\}\\
\leq \sup\left\{\Big|\int_E f\,d\mu\Big| : f \ \text{is measurable and} \ |f|\leq 1\right\}$$
For the reverse inequality use an approximation of $f$ by simple functions $\phi_j = \sum_{i=1}^{n_j} \lambda_i^j \chi_{E_i^j}$ such that $|\phi_j(x)| \leq |f(x)| \leq 1$ (this always exists). Now
\begin{align}
\left| \int_E f\,d\mu \right| & = \left| \int_E (\lim_j \phi_j)\,d\mu\right| = \lim_j \left| \int_E ( \sum_{i=1}^{n_j} \lambda_i^j \chi_{E_i^j})\,d\mu \right|\\
& \leq \lim_j  \sum_{i=1}^{n_j} \left| \int_E (  \chi_{E_i^j})\,d\mu \right|  \leq \lim_j  \sum_{i=1}^{n_j} \left| \mu(E_i^j)\right|\end{align}
where you use the dominated convergence theorem to pull the limit out, and the triagle inequality and linearity of the integral to pull the sum out. This shows the reverse inequality.
