It's helpful to keep in mind the uniqueness condition for the Jordan decomposition theorem:
If $\lambda$ is a signed measure and there are positive measures $\lambda_1,\lambda_2$ with $\lambda = \lambda_1-\lambda_2$ and $\lambda_1\perp\lambda_2,$ then $\lambda_1=\lambda^+$ and $\lambda_2 = \lambda^-.$
This is where the assumption $\mu\perp\nu$ comes into play; setting $\lambda = \mu-\nu$ then implies $\mu = \lambda^+ = (\mu-\nu)^+.$
For the other direction, one can use the Hahn Decomposition Theorem (which is a precursor of Jordan decomposition). Since $\lambda = \mu-\nu$ is a signed measure, this theorem provides a $\lambda$-positive set $P$ and $\lambda$-negative $N$ such that $X = P\cup N$ and $P\cap N=\emptyset.$ These sets further have the property $\lambda^+(A)=\lambda(A\cap P)$ and similarly for $\lambda^-$ in the Jordan decomposition $\lambda = \lambda^+-\lambda^-.$
Taking $P$ and $N$ as above, I claim $E_1=P$ and $E_2=N$ work in showing $\mu\perp\nu.$ Indeed, $\mu = \lambda^+$ implies $\mu(A\cap N) = \lambda^+(A\cap N) = \lambda((A\cap N)\cap P)= 0$ for any measurable set $A.$ Since $\mu = \lambda^+$ implies $\nu = \lambda^-,$ we also have $\nu(A\cap P) = \lambda^-(A\cap P) = \lambda((A\cap P)\cap N) = 0.$ This completes the proof.
Let me give you part hint, part thought process. There are some details to work out here, but please give it a try.
Trying to connect $E$ and $G$ is hard, because we don't have much direct connection between $\nu_1$ and $\nu_2$. I think you also have too many sets floating around, which makes things into a symbol soup. Let's see if we can clear it up.
In principal, the idea here is this: we know that all of the mass of the measure $\mu$ is isolated from the masses of the measures $\nu_1$ and $\nu_2$. In your existing breakdown of sets ($E$, $F$, $G$, and $H$), the real difficulty you're grappling with is the fact that these were produced by some theorem factory and it is hard to see how they can relate to eachother. So, how about this: can we prove that it is possible to find just two disjoint sets, call them $U$ and $I$, such that we can write the following?
$$
\nu_1(A)=\nu_1(A\cap U),\qquad\nu_2(A)=\nu_2(A\cap U), \qquad\mu(A)=\mu(A\cap I)\tag{1}
$$
You can think of this as a sort of simultaneous decomposition of the space into a part that hosts $\mu$ and a part that hosts both $\nu_1$ and $\nu_2$, whereas you can think of what you are given as the ability to decompose the space once for $\nu_1$/$\mu$ and once for $\nu_2$/$\mu$.
IF you can prove this, then your result is easy to come by, because
$$
(\nu_1+\nu_2)(A)=\nu_1(A)+\nu_2(A)=\nu_1(A\cap U)+\nu_2(A\cap U)=(\nu_1+\nu_2)(A\cap U).
$$
So, how do we prove (1)?
Take the sets $E$, $F$, $G$, and $H$ as you defined them above. We know that all of the mass of $\mu$ is concentrated inside $F$; but, we also know that all the mass of $\mu$ is concentrated inside of $H$. Can you show that this implies that for any $A$,
$$
\mu(A)=\mu(A\cap(F\cap H))?
$$
If you can prove that, then $F\cap H$ is a great candidate for $I$ above.
What, then, should $U$ be? The masses of $\nu_1$ and $\nu_2$ are restricted to $E$ and $G$, respectively; could we take $U:=E\cup G$? You need to show that this is disjoint from $I:=F\cap H$, and you need to show that the intersection property holds for both $\nu_1$ and $\nu_2$.
Best Answer
Suppose $\nu \perp \mu$. There exist disjoint measurable sets $E_1,E_2$ with union $$X such that $\mu(A \cap E_1)=0$ and $\nu (A \cap E_2)=0$ for all $A$. This implies that $|\nu| (A \cap E_2)=0$ for all $A$ [because $|\nu| (B) =\sup \sum_n |\nu (C_n)|$ where the supremum is over all partitions $(C_n)$ of $B$]. Hence $|\nu| \perp \mu$.
The remaining implications are all immediate from the following: $\nu^ {\pm } \leq |\nu|$ and $|\nu| =\nu ^{+}+\nu ^{+}$.
It is not clear to me as to what definition of $|\nu|$ you are using. I have used the definitition from Rudin's book.