Integral with respect to the sum of two complex measures Does the integral value w.r.t. the sum of two complex measures coincide with the sum of the integral values w.r.t. each of the complex measures?
I write the details below.
Let $(S, \mathscr{A})$ be a measurable space, and $\nu, \mu$ complex measures on $\mathscr{A}$.
We define a complex measure $\tau$ by $\tau := \nu + \mu$. 
Question 1.
Let $f: S \to \mathbb{C}$ be a measurable function. We consider the following two conditions:
(1) $f \in L^1(|\tau|)$
(2) $f \in L^1(|\nu|) \cap L^1(|\mu|)$.
I checked that if (2) holds then so does (1). But I don't know whether (2) holds under (1).
Can (2) be proven under (1)?
Question 2.
Let $f \in L^1(|\nu|) \cap L^1(|\mu|)$. Then does the following equation hold?:
\begin{equation}
\int_S f d\tau = \int_S f d\nu + \int_S f d\mu.
\end{equation}
(I understand the equation above holds if $\nu$ and $\mu$ are ordinary measures(i.e. $[0, \infty]$-valued measures).)
Thank you in advance.

Definitions:
We define an integral w.r.t. a complex measure($\mathbb{C}$-valued measure) as follows.  
(Remark: Complex measures take $\mathbb{C}$-values, so they are not allowed to take infinite values.)
Let $(S, \mathscr{A}, \mu)$ be a complex measure space. For $A \in \mathscr{A}$, we define 
$$
\begin{equation}
    |\mu|(A) := \sup\left\{ \sum_{B \in \mathscr{B}} |\mu(B)|\  \Bigl|\  \mathscr{B} \textrm{ is a finite partition of } A \textrm{ into } \mathscr{A}\textrm{-measurable sets} \right\}.
\end{equation}
$$
It is well-known that $|\mu|$ is a real finite measure. And we can write $\mu_1 (:= \mathrm{Re}\mu) = \mu_1^+ - \mu_1^-$ and $\mu_2 (:= \mathrm{Im}\mu) = \mu_2^+ - \mu_2^-$ by Jordan decomposition, where $\mu_1^\pm$ and $\mu_2^\pm$ are positive finite measures on $\mathscr{A}$.
For $f \in L^1(|\mu|)$, we define the integral of $f$ w.r.t. the complex measure $\mu$ by
\begin{align}
    \int_S f d\mu &:= \int_S f d\mu_1 + i \int_S f d\mu_2,
\end{align}
where
\begin{align}
    \int_S f d\mu_1 &:= \int_S f d\mu_1^+ - \int_S f d\mu_1^-, \\
    \int_S f d\mu_2 &:= \int_S f d\mu_2^+ - \int_S f d\mu_2^-.
\end{align}
($\int_S f d\mu_1^\pm$ and $\int_S f d\mu_2^\pm$ are the integrals w.r.t. $[0, \infty)$-valued measures, which we are familiar with.)
 A: Question 1.
Let $\mu = -\nu$ be the Lebesgue measure on $(0, 1).$ Then $\tau := \mu+\nu$ is the zero measure so every measurable function $f$ is in $L^1(|\tau|)$. But for example $f(x) = 1/x$ is neither in $L^1(|\mu|)$ nor $L^1(|\nu|)$.
Question 2.
I give a sketch of a possible proof.
For a complex measure $\tau$ we have by definition
$\int f \, d\tau = \int f \, d(\operatorname{Re}\tau) + i \int f \, d(\operatorname{Im}\tau)$
where $\operatorname{Re}\tau$ and $\operatorname{Im}\tau$ are realvalued measures.
Thus assume that $\tau = \mu + \nu$ are realvalued measures. Integration against such is defined by doing a Jordan decomposition into two ordinary (i.e. nonnegative) measures $\tau^+$ and $\tau^-$ living on disjoint sets, and then setting $\int f \, d\tau = \int f \, d\tau^+ - \int f \, d\tau^-.$
Idea: Let $S_\tau^+ \cup S_\tau^-$, $S_\mu^+ \cup S_\mu^-$, $S_\nu^+ \cup S_\nu^-$ be Hahn decompositions of $S$ for $\tau$, $\mu$ and $\nu$ respectively. Then split $S$ into the eight sets of the form $S_\tau^\pm \cap S_\mu^\pm \cap S_\nu^\pm$ and do integration on each of these. For example, on $S_\tau^+ \cap S_\mu^+ \cap S_\nu^-$ we have 
$$
\int f \, d\tau = \int f \, d\tau^+ = \int f \, d\mu^+ - \int f \, d\nu^- = \int f \, d\mu + \int f \, d\nu.
$$
Then we just add the integrals over all such sets and get the result $\int f \, d\tau = \int f \, d\mu + \int f \, d\nu$ for realvalued measures.
