Let $\lambda$ a charge and $\mu$ a measure in $(X,\mathcal{X})$, with $\lambda\,\bot\,\mu $. So, $\exists A,B\in\mathcal{X}$ such that $A\cap B=\emptyset$, $X=A\cup B$ and $\lambda(A)=\mu(B)=0.$
I'm trying to prove that $\lambda^{+}\bot\,\mu $.
Let $X=P\cup N$ the Hahn's decomposition of $X$ with respect with $\lambda$. I draw a picture to help me to find sets $C,D$ that leads me to $\lambda^{+}\bot\,\mu$.
By the picture, I taught in $C=B\cap N,D=A\cup(B\cap P)$. Here, $\mu(D)=\mu(B\cap N)\leq \mu (B)=0\implies \mu(D)=0,$ but I can't prove that $\lambda(C)=0$. What can I do?
This does not work. Consider $X =\{a,b,c,d\}$ and set $A = \{a\}$, $B = \{b\}$, $C = \{c\}$, and $D = \{d\}$. Define measures $\lambda^\pm$ and $\mu$ by
$\mu(A)=\mu(B)=1$, $\mu(C) = \mu(D) = 0$,
$\lambda^-(B) = \lambda^-(C) = \lambda^-(D) = 0$, and $\lambda^-(A) = 1$,
$\lambda^+(B) = \lambda^+(C) = \lambda^+(D) = 1$, and $\lambda^+(A) = 0$,
as well as $\lambda = \lambda^+-\lambda^-$. Then $\lambda(A\cup B) = 0$ and $\mu(C\cup D) = 0$. However, you will not find any set $E$ such that $\lambda^+(E) = 0$ and $\mu(X\setminus E) = 0$ because the first condition is only satisfied for $E = A$ and $E=\emptyset$, but $\mu(X\setminus A) = 1$ and $\mu(X\setminus\emptyset) = 2$.