I know there are some similar questions which have already tackled the problem. But I need verification and help for a (hopefully) different approach I am thinking.
Let $X$ be a set and $\Sigma$ be a $\sigma$-algebra. Let $\mu_1$ and $\mu_2$ be two measures on $X$, both with domain $\Sigma$. Set $$\mu E=\inf\left\{\mu_1(E \cap F)+\mu_2(E\backslash F):F\in\Sigma\right\}$$ for each $E\in \Sigma$. Show that $\mu$ is a measure on X, and that it is the greatest measure, with domain $\Sigma$, such that $\mu E\leq \min(\mu_1 E, \mu_2 E)$ for every $E \in \Sigma$.
Claim 1 - It is possible to simply take infimum over $F\subseteq E$.
Since, we always have that $E\cap F=E \cap (E\cap F)$ and $E\backslash F=E\backslash(E\cap F)$. We can replace $F$ with $F\cap E$ as $F \cap E$ is always measurable in $\Sigma$. I think it's as if we are forming equivalence classes and taking the representative as exactly what happens inside $E$. Hence, I modify the expression and write: $$\mu E=\inf\left\{\mu_1(E \cap F)+\mu_2(E\backslash F):F\in\Sigma \; \land \; F\subseteq E \right\}$$
Claim 2 - I will take two disjoint sets $E_1$ and $E_2$ and define $F_i=E_i \cap F$. It follows very similarly for infinite case $(E_i)_{i\in\mathbb N}$.
$$\mu (E_1 \cup E_2)=\inf\left\{\mu_1(E_1 \cap F_1)+\mu_2(E_1\backslash F_1) + \mu_1(E_2 \cap F_2)+\mu_2(E_2\backslash F_2) \\ :F\in\Sigma \; \land \; F\subseteq E_1 \cup E_2 \; \land \; (F_1 = E_1 \cap F) \land (F_2=E_2\cap F) \right\}$$
I think this expression simply reduces to $$\mu (E_1 \cup E_2)=\inf\left\{\mu_1(E_1 \cap F_1)+\mu_2(E_1\backslash F_1) : F_1 \subseteq E_1 \;\land\; F_1 \in \Sigma \right\} \\ + \inf\left\{\mu_1(E_2 \cap F_2)+\mu_2(E_2\backslash F_2) : F_2 \subseteq E_2 \;\land\; F_2 \in \Sigma \right\}$$ because $F_1,E_1$ and $F_2,E_2$ are disjoint measurable sets and the terms of the expression are additive. Hence, we can minimize both additive terms simultaneously by selecting a favorable $F$ as $F=F_1 \cup F_2$. With this, it becomes obvious that $\mu (E_1 \cup E_2)=\mu E_1 + \mu E_2$ and a similar proof for countably infinite sequence.
But I am a bit doubtful for the argument in claim 2 even though visually with Venn diagrams it seems very obvious. Is there anyway to set it rigorously without using any $\epsilon/2^n$ argument?
I think I have got this.
Take any two disjoint $E_1,E_2$ and define $F_1'=E_1\cap F$, $F_2'=E_2\cap F$ for arbitrary $F\subseteq E_1\cup E_2$
$\mu (E=E_1\cup E_2)=\inf\left\{\mu_1(E \cap F)+\mu_2(E\backslash F):F\in\Sigma \; \land \; F\subseteq E \right\} \\= \inf\left\{\mu_1(E_1 \cap F_1')+\mu_2(E_1\backslash F_1') + \mu_1(E_2 \cap F_2')+\mu_2(E_2\backslash F_2') \\ :F\in\Sigma \; \land \; F\subseteq E_1 \cup E_2 \; \land \; (F_1' = E_1 \cap F) \land (F_2'=E_2\cap F) \right\} \\ \geq \inf\left\{\mu_1(E_1 \cap F_1)+\mu_2(E_1\backslash F_1):F_1\in\Sigma \; \land \; F_1\subseteq E_1 \right\} + \inf\left\{\mu_1(E_2 \cap F_2)+\mu_2(E_2\backslash F_2):F_2\in\Sigma \; \land \; F_2\subseteq E_2 \right\}\\=\mu(E_1)+\mu(E_2)$
just by using the properties of infimum. We have to prove the other inequality. Take any disjoint $F_1\subseteq E_1$ and $F_2\subseteq E_2$, define $F'=F_1 \cup F_2$ and $E=E_1\cup E_2$ then the expression
$$\mu_1(E_1 \cap F_1)+\mu_2(E_1\backslash F_1)+ \mu_1(E_2 \cap F_2)+\mu_2(E_2\backslash F_2)= \mu_1(E\cap F')+\mu_2(E\backslash F')\geq \inf\left\{\mu_1(E \cap F)+\mu_2(E\backslash F):F\in\Sigma \; \land \; F\subseteq E \right\}$$ holds for arbitrary $F_1$ and $F_2$. Hence, by taking infimum over $F_1$ and $F_2$. We recover the inequality that $$\mu(E_1)+\mu(E_2)\geq \mu(E)$$ And by combining both the results, we get for any two disjoint $E_1,E_2$ that $$\mu(E)=\mu(E_1)+\mu(E_2)$$