Jordan Decomposition Theorem Double Implication I'm working through the proof of the Jordan Decomposition Theorem in Zygmund and Wheeden and am stuck.
We define $\overline V(E) = \sup_{A \subseteq E, A \in \mathcal A} \varphi(A)$ and $\underline V(E) = -\inf_{A \subseteq E, A \in \mathcal A} \varphi(A)$.
Theorem: If $\varphi$ is an additive set function on $\mathcal A$, then $$\varphi (E) = \overline V(E) - \underline V(E)$$ for $E \in \mathcal A$.
Proof: If $A \subseteq E$ and $A \in \mathcal A$, then $\varphi(E) = \varphi(A) + \varphi(E \setminus A) = \varphi(A) - \big[ -\varphi(E \setminus A) \big]$. Since $\varphi(E)$ is fixed, we have $\varphi (A_k) \to \overline V(E)$ for a sequence of sets $A_k \in \mathcal A$ if and only if $-\varphi(E \setminus A_k) \to \underline V (E)$. Hence $\varphi(E) = \overline V(E) - \underline V(E)$, as claimed. $\square$
I don't understand why $\varphi(A_k) \to \overline V (E)$ if and only if $-\varphi(E \setminus A_k) \to \underline V(E)$. Can someone fill in the details.
 A: The two terms $\phi(A_k)$ and $-\phi(E\setminus A_k)$ only differ by a constant. 
Thus, any one of the two tends towards its supremum value iff the other does the same thing. 
EDIT : here is a more detailed explanation : 
There are three elementary properties you need to know :
(1) The minus sign reverses order, so in abstract nonsense
terms $-({\sf inf})$ is the same thing as ${\sf sup}(-)$.
(2) $\sf sup$ (or $\sf inf$) behaves well with respect to translation : in 
abstract nonsense terms again, ${\sf sup}(c+)$ is the same thing as
$c+{\sf sup}()$.
(3) $\cal A$ is closed under taking complements, so that
$$
\bigg\lbrace E\setminus A \ \bigg| \ A\subseteq E, A\in {\cal A} \bigg\rbrace=
\bigg\lbrace A \ \bigg| \ A\subseteq E, A\in {\cal A} \bigg\rbrace
$$
Then, you have
$$
\begin{array}{lcl}
\underline{V}(E) &=& -{\sf inf}_{A\subseteq E, A\in {\cal A}} \phi(A) \\
&=& {\sf sup}_{A\subseteq E, A\in {\cal A}} -\phi(A) \ (\text{by} \ (1))\\
&=& {\sf sup}_{A\subseteq E, A\in {\cal A}} -\phi(E \setminus A) \ (\text{by} \ (3))\\
&=& {\sf sup}_{A\subseteq E, A\in {\cal A}} -(\phi(E)-\phi(A)) \\
&=& {\sf sup}_{A\subseteq E, A\in {\cal A}} (-\phi(E))+\phi(A) \\
&=& (-\phi(E))+{\sf sup}_{A\subseteq E, A\in {\cal A}} \phi(A) \ (\text{by} \ (2))\\
&=& (-\phi(E))+\overline{V}(E)
\end{array}
$$
Thus, if a $\phi(A_k)$ converges to $x$, $-\phi(E\setminus A_k)$ converges to $x-\phi(E)$. Taking $x=\overline{V}(E)$, your double implication follows.
