The sum of mutually singular measures I have been trying to make an exercise related to mutually singular measures. Namely the following:

Exercise Let $\mu$ be a positive measure and $\nu_1, \nu_2$ be arbitrary measures all defined on the same measurable space. If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu$, then $\nu_1 + 
\nu_2 \perp \mu$.

Here I use the following definition of mutually singular

Definition (Mutually Singular) We call $\mu$ and $\nu$ mutually singular if there exists disjoint measurable sets on $E$ and $F$ such that $\nu(A) = \nu(A \cap E)$ and $\mu(A) = \mu(A \cap F)$.

Note: This phenomena of a set $E$ having the property that $\nu(A \cap E) = \nu(A)$ for all measurable $A$'s is dubbed as the measure $\nu$ being concentrated on $E$.
Naturally, I would begin by writing out what I have:
$\exists E,F, G,H \in \Sigma$ such that
\begin{align}
\nu_1(A) = \nu_1(A \cap E)\  \text{ and }\  \mu(A) = \mu(A \cap F)\ \text{ for all $A \in \Sigma$}.\\
\nu_2(B) = \nu_2(B \cap G)\  \text{ and }\  \mu(B) = \mu(B \cap H)\ \text{ for all $B \in \Sigma$}.\\
\end{align}
I have tried writing out various combinations of $E$ and $G$, but to no avail. From here I have tried to find some hints on the internet and this StackExchange post seems to have a solution to this problem. But the answer uses another definition, so from here I was wondering two things:

*

*Is there a way to directly prove it using this definition?

*Is there a reference to see how the definition provided in the answer of the StackExchange post and this definition are the same. Maybe after seeing how those definitions are equivalent, it might much more clear how to directly prove it.

Extra Attempt:
I have been searching for a reference on point 2. During this search, I found that mathworld claimed that the notion of $\nu$ being concentrated on a certain set $E$ is equivalent to saying $\nu(A) = 0$ whenever $A \cap E = \emptyset$ (Source).
It is easy to see how $\nu(A \cap E) = \nu(A)$ implies this, but the other way around I have not been able to prove. With this other definition of concentration, I might be able to more easily prove the original question.
 A: 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 $\nu_2$.
