Let
- $\Omega$ be a set
- $\mathcal S\subseteq2^\Omega$ with $\emptyset\in\mathcal S$
- $E$ be a normed $\mathbb R$-vector space
- $\mu:\mathcal S\to E$ be $\sigma$-additive, i.e. if $(A_n)_{n\in\mathbb N}\subseteq\mathcal S$ is disjoint with $\biguplus_{n\in\mathbb N}A_n\in\mathcal S$, then $$\mu\left(\biguplus_{n\in\mathbb N}A_n\right)=\sum_{n\in\mathbb N}\mu(A_n)\tag1$$
Now, let $$|\mu|(S):=\sup\left\{\sum_{i=1}^n\left\|\mu(S_i)\right\|_E:n\in\mathbb N\text{ and }S_1,\ldots,S_n\in\mathcal S\text{ are disjoint with }\biguplus_{i=1}^nS_i\subseteq S\right\}$$ for $S\in\mathcal S$ and$^1$ $$\left\|\mu\right\|:=\sup_{\left\|\varphi\right\|_{X'}\:\le\:1}|\varphi\circ\mu|\;.$$
It's easy to see that $$\left\|\mu\right\|\le|\mu|\;.\tag2$$
Assume $|\mu|(\Omega)<\infty$. Let $\nu:\mathcal S\to\mathbb R$ be $\sigma$-additive. I want to show that $$|\mu|=\nu\tag3$$ if and only if $$\left\|\mu\right\|\le\nu\tag4$$ and $$\nu\le\lambda\;\;\;\text{for all }\sigma\text{-additive }\lambda:\mathcal S\to\mathbb R\text{ with }\left\|\mu\right\|\le\lambda\tag5\;.$$
How can we show that? For example, in the direction "$\Rightarrow$", $(4)$ holds by $(2)$, but how can we show $(5)$? If we fix such a $\lambda$ and assume $|\mu|>\lambda$, are we able to deduce a contradiction? Clearly, we obtain $|\mu|>\left\|\mu\right\|$ ... Maybe we can find a $S\in\mathcal S$ with $|\mu|(S)=\left\|\mu\right\|(S)$ and obtain the contradiction in that way ...
$^1$ Note that the used notation is common for the total variation $|\mu|$ and semivariation $\left\|\mu\right\|$ of $\mu$. However, be careful and, for example, don't confuse the mapping $\mathcal S\ni S\mapsto|(\varphi\circ\mu)(S)|$ with the total variation $|\varphi\circ\mu|$ of $\varphi\circ\mu$. The same care is necessary in the distinction of $\mathcal S\ni S\mapsto\left\|\mu(S)\right\|_E$ and the semivariation $\left\|\mu\right\|$ of $\mu$.