Let $F$ be a Sigma algebra on a set $X$, and let $\mu$ and $\nu$ be probability measures on $(X,F)$, i.e. $\mu(X)=\nu(X)=1$. Finally let $\eta=\mu-\nu$. Show that $$|\eta|(X)=2\sup_{E\in F}|\eta(E)|,$$ where $|\eta|$ is the total variation measure of $\eta$.
Now by definition $|\eta|(X)=\eta^+(X)+\eta^-(X)$ where $\eta=\eta^+-\eta^-$ is a Jordan decomposition, but I'm not sure how to find a Jordan decomposition of $\mu-\nu$. The other thought I had is to use the fact that $$|\eta|(X)=\sup\sum_{n=1}^k|\eta(E_n)|,$$ where the supremum is taken over all measurable partitions $\{E_1,...,E_n\}$, but I'm not sure how to show that $2\sup_{E\in F}|\eta(E)|=\sup\Sigma_{n=1}^k|\eta(E_n)|$, or equivalently $$2\sup_{E\in F}|\mu(E)-\nu(E)|=\sup\sum_{n=1}^k|\mu(E_n)-\nu(E_n)|.$$
Since $\mu$ and $\nu$ are finite, $\eta$ is this too, and we have also $\eta^{+}(X) \le 1$ and $\eta^{-}(X) \le 1$. Because $\mu$ and $\nu$ are probability measures, we can say more: In fact, we find that $$0= \mu(X)-\nu(X)=\eta(X)= \eta^{+}(X) - \eta^{-}(X).$$ Thus, if $X= P \cup H$ is a Hahn-decomposition of $\eta$, we have $\eta^{+}(P) =\eta^{-}(N).$ Take $E= P$ in order to get $$2 |\eta(E)| = 2\eta^{+}(P) = \eta^{+}(P) + \eta^{-}(N) = |\eta|(X).$$ On the other hand for any $A \in \mathcal{F}$, we have \begin{align} 2 |\eta(A)| &= 2|\eta^{+}(A)-\eta^{-}(A)| \\ &\le 2 \max \{ \eta^{+}(X),\eta^{-}(X)\} = \eta^{+}(X)+\eta^{-}(X) = |\eta|(X). \end{align}