Integration with respect to sum of signed measures. Consider two finite signed measures $\mu$ and $\nu$ on a measurable space and a measurable mapping $f$ integrable with respect to both measure. The Jordan-Hahn decomposition yields
$$
\mu= \mu^+ - \mu^-, \quad \quad\text{and} \quad \quad  \nu=\nu^+-\nu^-.
$$
 We know that in general $(\mu+\nu)^+ \not = \mu^+ +\nu^+$ and $(\mu+\nu)^-\not= \mu^-+\nu^-$, but can I say that
$$
\int f d(\mu+\nu) \stackrel{?}{=} \int fd(\mu^++\nu^+) -\int f d(\mu^-+\nu^-).
$$
I have found two counterexamples for $\theta^+  = \mu^+ +\nu^+$ and $\theta^-= \mu^-+\nu^-$, but these examples still satisfy the above equality:


*

*$\mu=\delta_a-\delta_b$, $\nu=\delta_b -\delta_c$, such that $(\mu+\nu)^+ =\delta_a \not = \delta_a+\delta_b$ and $(\mu+\nu)^- = \delta_c \not = \delta_b+\delta_c$, but in this case we see that
$$
\int f d(\mu+\nu)= \int f d\delta_a - \int f d\delta_c = \int f d\delta_a +\int f d\delta_b - \int f d\delta_b- \int f d\delta_c
$$
$$
= \int fd(\mu^++\nu^+) - \int f d(\mu^-+\nu^-)
$$
so the equality still holds.

*Non-zero measures with $\mu=-\nu$, such that $(\mu+\nu)^+=0\not = \mu^++\nu^+$  and $(\mu+\nu)^-=0\not = \mu^-+\nu^-$, but
$$
\int f d(\mu+\nu) =0
$$
and we have that $\mu^+=\nu^-$ and $\mu^-=\nu^+$, yielding that
$$
\int fd\mu^+ -\int f d\mu^-+\int fd\nu^+ -\int fd\nu^- = \int fd\mu^+ -\int f d\mu^-+\int fd\mu^- -\int fd\mu^+ =0,
$$
so equality still holds.
This gives me reason to to thinks that the equality holds.
Can anyone please help me understand why the equality holds, or at least give me a counterexample for which it does not hold?
 A: Okay, It turned out to be rather easy to prove using an approximation argument.
\begin{align*}
 \int 1_E \, d(a\mu+b\nu) = (a\mu+b\nu)(E) = a\mu(E)+b\nu(E) = a\int 1_E\, d\mu +b\int 1_E \, d\nu.
 \end{align*}
     For any simple function $g=\sum_{i=1}^n x_i 1_{E_i}$ with $x_i\in \mathbb{R}$ and $E_i$ measurable, we get by similar reasoning that $ \int g \, d(a\mu+b\nu)= a \int g \, d\mu + b\int g\, d\nu$, using the definition that $\int g^+ \, d\mu^\pm:= \sum_{i=1}^n (x_i  \lor  0)\mu^\pm(E_i)$.
    Since $f$ is measurable, there exists a sequences of simple functions $(f_n)$ all bounded by $|f|$ such that  $\lim_{n\to\infty} f_n= f$ pointwise . Thus by Lebesgues dominated convergence theorem (assuming that $f$ is integrable with respect to all of the relevant measures), which obviously remains true for signed measures, we get that
    \begin{align*}
\int f \, d(a\mu+b\nu)=&  \lim_{n\to\infty} \int f_n \, d(a\mu+a\nu) \\
=& a\lim_{n\to\infty} \int f_n \, d\mu  + b\lim_{n\to\infty}\int f_n \, d\nu \\
=& a\int f \, d\mu + b\int f\, d\nu 
\end{align*}
