Folland Exercise 3.3: Stuck + Possibly missing $\sigma$-finite hypothesis? This is Exercise 3.3 from Folland's Real Analysis, Second Edition stated exactly as it appears in the text:
''Let $\nu$ be a signed measure $(X, \mathcal{M})$.
$(a)$  $L^1(\nu) = L^1(|\nu|)$.
$(b)$  If $f \in L^1(\nu)$, $|\int f d\nu| \leq \int |f|d|\nu|$.
$(c)$  If $E \in \mathcal{M}$, $|\nu|(E) = \sup \{ |\int_E f d\nu| \colon |f| \leq 1 \}$.''
Parts $(a)$ and $(b)$ are easy to show.  My question concerns part $(c)$.  First off, I immediately assumed that Folland meant
$$ |\nu|(E) = \sup A := \sup \bigg\{ \bigg| \int_E f d\nu \bigg| \colon f \in L^1(\nu) \text{ and } |f| \leq 1 \bigg\}, $$
although, as we shall see soon, this interpretation may be up for debate.  
It is quick to show using $(b)$ that $|\nu|(E)$ is an upper bound for $A$.  For the other direction, what seems close to the right method is to first set a Hahn decomposition $P \cup N$ for $\nu = \nu^+ + \nu^-$, and then to let
$$f(x) = 1 \text{ if } x \in P \text{ and } f(x) = -1 \text{ if } x \in N.$$
Then, $|f| \leq 1$ and 
$$ \int_E f d\nu = \int_{P \cap E} f d\nu^+ - \int_{N \cap E} f d\nu^- = |\nu|(E) \geq 0, \tag{*}$$
so that $|\nu|(E) \in A$ and is thus the supremum.  
However, upon closer look, we have
$$ \int_X|f| \chi_E d\nu^+ = |\nu|(E \cap P) \;\;\; \text{ and } \;\;\; \int_X|f| \chi_E d\nu^- = |\nu|(E \cap N), $$
and if either of these quantities are $\infty$, then $f\chi_E \notin L^1(\nu)$ and equation $(*)$ becomes nonsense.  (Recall by definition that $f \in L^1(\nu)$ iff $f \in L^1(\nu^+) \cap L^1(\nu^-)$.)  
Since we are not given any finiteness conditions on $\nu$, we are thus forced to consider this possibility.  I have not been able to get around this difficulty, and I can only make headway by assuming that $\nu$ is at least $\sigma$-finite.
My question:  Can anyone see a way to prove the result without assuming $\nu$ is $\sigma$-finite?  (This would be the most desirable answer.)  Alternatively, is the $\sigma$-finite condition necessary (and/or sufficient)?  
Thanks.
 A: The problem as stated in Folland's book is correct as stated (no $L^1$ requirement): If $|\nu|(E)=\infty$ then, WLOG, $\nu^+(E)=\infty$ and then $f=\chi_{P\cap E}$ does the trick. If $|\nu|(E)<\infty$ then your argument takes care of things.
If you insist on requiring that $f\in L^1(\nu)$ then you're right that something must be said about $\nu$: Consider the measure on $\mathbb{R}$ given by $\mu(A)= \lambda(A)$ if $A\subset (0,1)$ (where $\lambda$ is Lebesgue measure) and $\mu(A)=\infty$ if $A\cap (0,1)^c\neq \emptyset$. Then for $E=(0,1)^c$ we have that $\mu(E)$ can't be approximated via $|\int_Ef d\mu|$ with $L^1(\mu)$ functions.
I'm pretty sure in this case the following condition is enough to guarantee what you want: Assume $\nu^+$ is infinite (recall only one of $\nu^\pm$ can be infinite by definition). Then for every $K>0$ there exists a measurable set $A_K$ such that $K<\nu^+(A_K)<\infty$ (i.e. $\nu^+$ is semifinite).
A: Let $\phi \in L^+$ be a simple function, and write $\phi = \sum_{i = 1}^n a_i \chi_{E_i}.$
Then
\begin{equation}
\label{eqn:simple integral sum}
\int \phi \,d|\nu| = \sum_{i = 1}^n a_i |\nu|(E_i) = \sum_{i = 1}^n a_i (\nu^+(E_i) + \nu^-(E_i)) = \sum_{i = 1}^n a_i \nu^+(E_i) + \sum_{i = 1}^n a_i \nu^-(E_i) = \int \phi \,d\nu^+ + \int \phi \,d\nu^-.
\end{equation}
Hence, if $f \in L^1(\nu)$ then
$$\int |f| \,d|\nu| = \sup\left\{ \int \phi \,d\nu^+ + \int \phi \,d\nu^-\ \middle|\ \phi \in L^+ \text{ simple with } \phi \le |f| \right\} \le \int |f| \,d\nu^+ + \int |f| \,d\nu^- < \infty$$ because $L^1(\nu) = L^1(\nu^+) \cap L^1(\nu^-).$
Therefore $L^1(\nu) \subseteq L^1(|\nu|).$
Conversely, if $f \in L^1(|\nu|)$ it is clear from \eqref{eqn:simple integral sum} that $\int \phi \,d\nu^+ \le \int \phi \,d|\nu| \le \int |f| \,d|\nu|$ for all simple $\phi \in L^+$ with $\phi \le |f|,$ so $\int |f| \,d\nu^+ \le \int |f| \,d|\nu| < \infty.$
This shows that $L^1(|\nu|) \subseteq L^1(\nu^+),$ and a similar argument shows that $L^1(|\nu|) \subseteq L^1(\nu^-).$
Therefore $L^1(|\nu|) \subseteq L^1(\nu).$
Let $A$ and $B$ be disjoint measurable sets covering $X$ such that $A$ is $\nu^+$-null and $B$ is $\nu^-$-null.
Then $g \chi_A = 0$ $\nu^+$-a.e.\ and $g \chi_B = 0$ $\nu^-$-a.e. for every measurable function $g.$
In particular, from \eqref{eqn:simple integral sum} $\int \phi \,d|\nu| = \int \phi \,d\nu^+$ for all simple functions $\phi \in L^+$ with $\phi \le f^+ \chi_B$ or $\phi \le f^- \chi_B.$
This implies that
$$\int f \chi_B \,d|\nu| = \int f^+ \chi_B \,d|\nu| - \int f^- \chi_B \,d|\nu| = \int f^+ \chi_B \,d\nu^+ - \int f^- \chi_B \,d\nu^+ = \int f \chi_B \,d\nu^+,$$ and similarly $\int f \chi_A \,d|\nu| = \int f \chi_A \,d\nu^-.$
Moreover $|\chi_A - \chi_B| = \chi_X$ because $A \cup B = X$ and $A \cap B = \varnothing.$
Therefore
\begin{align*}
\left|\int f \,d\nu\right|
&= \left|\int f (\chi_A + \chi_B) \,d\nu^+ - \int f (\chi_A + \chi_B) \,d\nu^-\right| \\
&= \left|\int f \chi_B \,d\nu^+ - \int f \chi_A \,d\nu^-\right| \\
&= \left|\int f \chi_B \,d|\nu| - \int f \chi_A \,d|\nu|\right| \\
&= \left|\int f (\chi_B - \chi_A) \,d|\nu|\right| \\
&\le \int |f (\chi_B - \chi_A)| \,d|\nu| \\
&= \int |f| \,d|\nu|.
\end{align*}
Define $g = \chi_B - \chi_A.$
Then $|g| \le 1$ and hence $|\nu|(E) = |\int_E g \,d\nu| \le \sup\{ |\int_E f \,d\nu| \mid |f| \le 1 \}$ because
\begin{align*}
\int_E g \,d\nu
&= \int (\chi_B - \chi_A) \chi_E \,d\nu \\
&= \int \chi_B \chi_E \,d\nu^+ - \int -\chi_A \chi_E \,d\nu^- \\
&= \int \chi_{(B \cap E)} \,d\nu^+ + \int \chi_{(A \cap E)} \,d\nu^- \\
&= \nu^+(B \cap E) + \nu^-(A \cap E) \\
&= \nu^+(A \cap E) + \nu^+(B \cap E) + \nu^-(A \cap E) + \nu^-(B \cap E) \\
&= \nu^+(E) + \nu^-(E) \\
&= |\nu|(E) \ge 0.
\end{align*}
This completes the proof if $|\nu|(E) = \infty.$
Assume that $|\nu|(E) < \infty$ and let $f$ be a measurable function with $|f| \le 1.$
Then $f \chi_E \in L^1(|\nu|) = L^1(\nu)$ because $\int_E |f| \,d|\nu| \le \int \chi_E \,d|\nu| = |\nu|(E) < \infty.$
Hence, by the previous exercise $|\int_E f \,d\nu| \le \int |f \chi_E| \,d|\nu| \le \int \chi_E \,d|\nu| = |\nu|(E).$
Therefore $\sup\{ |\int_E f \,d\nu| \mid |f| \le 1 \} \le |\nu|(E).$
