Integration in complex measure Let $v$ be a complex measure in $(X,M)$. Then $L^{1}(v)=L^{1}(|v|)$.
I have made:
$L^1(v)\subset L^1(|v|)$?.
Let $g\in L^1(v)$
As $v<<|v|$ and $|v|$ is finite measure, then for chain rule, $g.(\frac{dv}{d|v|})\in L^1(|v|)$.
As $|\frac{dv}{d|v|}|=1\;|v|-a.e$, then $|g|=|g||\frac{dv}{d|v|}|\;|v|-a.e$, then $\int |g|d|v|=\int |g||\frac{dv}{d|v|}|d|v|=\int |g.\frac{dv}{d|v|}|d|v|< \infty$.
Therefore $g\in L^1(|v|)$.

Let $g\in L^1(|v|).
For \;$j=r,i$,\; we \:have, v_{j} ^ {\pm} \leq v_j^{+}+v_j^{-}=|v_j| \leq |v|$, then 
$g\in L^1(v_r^{+})\cap L^1(v_i^{+})\cap L^1(v_r^{-})\cap L^1(v_i^{-})$ , then $g\in L^1(v)$
