Q:

Show the equality of total variation of measure

If $\mu$ is a complex measure on a $\sigma$-algebra $M$, and if $E \in M$, define $$\lambda(E) = \sup\sum \mu|E_i|,$$ the supremum being taken over all finite partitions $\{E_i\}$ of $E$. Does it follow that $\lambda = |\mu|?$
I know that this result hold if $\mu$ is charge. But in this case?

A:

The answer is affirmative.
It is easy to see that $\lambda\leq |\mu|$. Let's prove the reverse inequality.
Let $\epsilon>0$ and let $\{E_i\}_{i=1}^\infty$ a partition of $E$. Since the series $\sum_{i=1}^\infty |\mu(E_i)|$ converges, there exists $n_0\in\mathbb{N}$ such that $$\sum_{i=n_0+1}^\infty|\mu(E_i)|<\epsilon.$$ Define $\{F_i\}_{i=1}^{n_0+1}$ by $F_i=E_i$ for $i\in\{1,...,n_0\}$ and $F_{n_0+1}=\bigcup_{i=n_0+1}^\infty E_i$. Then $\{F_i\}_{i=1}^{n_0+1}$ is a finite partition of $E$ and $$\sum_{i=1}^\infty|\mu(E_i)|= \sum_{i=1}^{n_0}|\mu(E_i)| + \sum_{i=n_0+1}^{\infty}|\mu(E_i)| < \sum_{i=1}^{n_0}|\mu(F_i)| + \epsilon \leq \lambda(E)+\epsilon.$$ Since $\{E_i\}_{i=1}^\infty$ is arbitrary, we have $$|\mu|(E)\leq\lambda(E)+\epsilon.$$ Since $\epsilon>0$ is arbitrary, $|\mu|\leq \lambda$.

