Positive and negative variations of a charge, which is an integral. 
Let $f \in L(X, \mathbb X, \mu)$ and consider the charge $v: \mathbb X \to \mathbb R$ given by $v(A) = \int_A f \, d\mu$. Then the positive and negative variations of $v$ are given by $v^+(A) = \int_A f^+ \, d\mu$ and $v^-(A) = \int_A f^-(A) \, d\mu$.

Proof: Let $P$, $N$ be the Hahn decomposition for $v$. Then it is $v^+(A) = v(A \cap P)$ and $v^-(A) = -v(A \cap N)$. We have that
\begin{align*}
v^+(A) = v(A \cap P) = \int_{A \cap P} f \, d\mu = \int_A \chi_P f \, d\mu = \int_A \chi_Pf^+ \, d\mu - \int_A \chi_P f^- \, d\mu.
\end{align*}
$P$ is positive if $v(A \cap P) \ge 0$ for all $A \in \mathbb X$. So I know that
\begin{align*}
\int_A \chi_P f^- \, d\mu \le \int_A \chi_Pf^+ \, d\mu.
\end{align*}
But how can I show that $\int_A \chi_P f^- \, d\mu = 0$? If it would be greater than $0$ then $x \in P$ and $f < 0$. So $f^+ = \max\{f, 0\} = 0$. It follows
\begin{align*}
0 = \int_A \chi_Pf^+ \, d\mu \ge \int_A \chi_P f^- \, d\mu > 0,
\end{align*}
a contradiction. Can I do this?
 A: 
If it would be greater than $0$ then $x \in P$ and $f < 0$. So $f^+ = \max\{f, 0\} = 0$.

It's not entirely clear (to me, at least) what you mean here. From $\int_A \chi_Pf^-\,d\mu > 0$, you can deduce that $f < 0$ on a subset of $A$ with positive $\mu$-measure, and using that, you can reach the desired conclusion.
But let's start from a different point:
Let $G = \{ x : f(x) > 0\}$, $S = \{x : f(x) < 0\}$, and $E = \{x : f(x) = 0\}$. Then we have $f^+ = f\chi_G$ and $f^- = f\chi_S$.
Since $\nu^+(A) \geqslant 0$ for all $A$, we have in particular
$$0 \leqslant \nu^+(S) = \nu(S\cap P) = \int_{S\cap P} f\,d\mu = - \int_{S\cap P}f^-\,d\mu.$$
Since $f^- > 0$ on $S$, that means $\mu(S\cap P) = 0$. Similarly, we obtain $\mu(G\cap N) = 0$.
And thus we have
$$\begin{align}
\nu^+(A) &= \nu(A\cap P)\\
&= \int_{A\cap P} f\,d\mu\\
&= \int_{A\cap P} f^+\,d\mu - \int_{A\cap P} f^-\,d\mu\\
&= \int_{A\cap P} f^+\,d\mu - \int_{A\cap P \cap S} f^-\,d\mu\\
&= \int_{A\cap P} f^+\,d\mu\\
&= \int_A f^+\,d\mu.
\end{align}$$
The penultimate equality follows from $\mu(P\cap S) = 0$, and the last from $\mu(G\cap N) = 0$, which implies that $f^+ = 0$ almost everywhere [$\mu$] on $A\cap N$.
That $\nu^-(A) = \int_A f^-\,d\mu$ can be shown analogously, or by
$$\nu^-(A) = \nu^+(A) - \nu(A) = \int_A f^+ - f\,d\mu = \int_A f^-\,d\mu.$$
