I have proved $\implies$ direction i.e. if $\mu\perp\nu$, then total variation of $\mu-\nu$ is $\mu+\nu$. This post answer the same direction.
For the converse part,I define $\lambda=\mu+\nu$. Then by Radon Nikodym Theorem, there are $f,g\in L^1(\mu)$ such that $d\mu=f d\lambda$, $d\nu=gd\lambda$. Then $d(\mu-\nu)=(f-g)d\lambda\implies d|\mu-\nu|=|f-g|d\lambda\implies d(\mu+\nu)=|f-g|d\lambda\implies (f+g)d\lambda=|f-g|d\lambda$. Hence, $f+g=|f-g|$ $\lambda$-a.e.
Now I define $E=\{x:\ f(x)\ne0\}$ and $F=\{x:\ g(x)\ne0\}$. So $\mu$ and $\nu$ are concentrated on $E$ and $F$ respectively. But I cannot prove $\nu$ is concentrated outside $E$ (or $\mu$ is concentrated outside $F$). I need to somehow use $f+g=|f-g|$ to get the desired result.
Can anyone help complete the proof? Thanks for you help in advance.
Let me define $A=\{f>g\},B=\{g\le f\}$ (both are measurable). On $A$, the measure induced by $|f-g|$ is equal to that induced by $f-g$, that is:
$$\mu+\nu=\mu-\nu$$
Likewise, on $B$, we have - by comparing R-N derivatives: $$\mu+\nu=\nu-\mu$$Identically. We conclude that: $$\nu(U)=-\nu(U),\,\mu(V)=-\mu(V)$$For all measurable $U\subseteq A,V\subseteq B$. So, $\nu,\mu$ are nil on $A,B$. We also have $A\sqcup B=X$.
By my book's definition of the relation $\mu\perp\nu$, we are done! I'm not sure what you mean by 'concentrated' in this context.