Measures, as a supreme, proof Let $\Sigma$ be a $σ$-algebra over a set $X$, and $μ_1$ and $μ_2$ finite measures in it. It can be shown that the function $μ:\Sigma \to [0,\infty ]$ defined by
$$E\mapsto \sup_{F∈\Sigma}\{μ_1(E∩F)+μ_2(E\setminus F)\}$$
is a measure.
could someone help me see is a measure
I got stuck when i have to show 
$$\mu(\bigcup_{n=1}^\infty E_n) \geqslant \sum_{n=1}^\infty\sup_{F\in\Sigma} \left\{\mu_1(E_n\cap F) + \mu_2(E_n\setminus F) \right\} = \sum_{n=1}^\infty \nu(E_n).$$
$E_n $ a secuence of disjoints set
Know you can also use Radon-Nicodym, since $\mu_1$ and $\mu_2$ finite, but i'm more interested in the general case, when you have two measures
 A: We only need to prove 
$$
\nu(E)=\nu(\bigcup_{n=1}^\infty E_n) \geqslant \sum_{n=1}^\infty \nu(E_n)$$
as another case is proved in @Math1000's post. 
First we prove
$$
\nu(E_1\cup E_2)\geqslant \nu(E_1)+\nu(E_2)
$$
for disjoint $E_1$ and $E_2$.
By definition of $\nu$, there exist $F_1,F_2\in\Sigma$ such that
$$
\mu_1(E_1\cap F_1) + \mu_2(E_1\setminus F_1)\geqslant \nu(E_1)-\epsilon\quad\text{and}\quad \mu_1(E_2\cap F_2) + \mu_2(E_2\setminus F_2)\geqslant \nu(E_2)-\epsilon
$$
So 
\begin{align}
\nu(E_1\cup E_2)&\geqslant \mu_1((E_1\cup E_2)\cap (F_1\cup F_2)) + \mu_2(E_1\cup E_2\setminus F_1\cap F_2)
\\
&=\mu_1((E_1\cap (F_1\cup F_2))\cup (E_2\cap (F_1\cup F_2)) + \mu_2((E_1\setminus F_1\cap F_2)\cup (E_2\setminus (F_1\cap F_2))
\\
&=\mu_1(E_1\cap (F_1\cup F_2))+\mu_1( E_2\cap (F_1\cup F_2))+\\&\hspace{10 mm}\mu_2(E_1\setminus F_1\cap F_2)+\mu_2(E_2\setminus F_1\cap F_2)
\\
&\geqslant \mu_1(E_1\cap F_1)+\mu_1( E_2\cap F_2)+\mu_2(E_1\setminus F_1)+\mu_2(E_2\setminus F_2)
\\
&=\nu(E_1)+\nu(E_2)-2\epsilon
\end{align}
Since $\epsilon$ is arbitrary small, we have
$$
\nu(E_1\cup E_2)\geqslant \nu(E_1)+\nu(E_2)
$$
Thus
$$
\nu(E_1\cup E_2)=\nu(E_1)+\nu(E_2)\tag{1}
$$
Next we prove 
$$
E_1\subset E_2\implies \nu(E_1)\leqslant \nu(E_2)\tag{2}
$$
Since for any $F\in\Sigma$
$$
\mu_1(E_1\cap F) + \mu_2(E_1\setminus F)\leqslant \mu_1(E_2\cap F) + \mu_2(E_2\setminus F)
$$
So
$$
\nu(E_1)=\sup\{\mu_1(E_1\cap F) + \mu_2(E_1\setminus F)\}\leqslant \sup\{\mu_1(E_2\cap F) + \mu_2(E_2\setminus F)\}=\nu(E_2)
$$
Finally for any disjoint $E_n$, since 
$$\bigcup_{n\leqslant m}E_n \subset \bigcup_{n=1}^{\infty} E_n$$
By (1) and (2), there is
$$
\sum_{n\leqslant m}\nu(E_n)=\nu(\bigcup_{n\leqslant m} E_n)\leqslant \nu(\bigcup_{n=1}^{\infty}E_n)
$$
Let $m\to\infty$, it is proved.
A: Let $$\nu(E) = \sup_{F\in\Sigma}\{\mu_1(E\cap F) + \mu_2(E\setminus F)\}.$$
Then for any $F\in\Sigma$, $$\mu_1(\varnothing\cap F)+\mu_2(\varnothing\setminus F) = \mu_1(\varnothing) + \mu_2(\varnothing) = 0.$$
If $E_n$ is a sequence of disjoint sets in $\Sigma$, let $E=\bigcup_{n=1}^\infty E_n$. Then for any $F\in\Sigma$,
$$\mu_1(E\cap F)=\mu_1\left(\bigcup_{n=1}^\infty (E_n\cap F)\right) = \sum_{n=1}^\infty \mu_1(E_n\cap F), $$
and similarly
$$\mu_2(E\setminus F)=\mu_2\left(\bigcup_{n=1}^\infty (E_n\setminus F)\right) = \sum_{n=1}^\infty \mu_2(E_n\setminus F).$$
Hence
$$
\nu(E) = \sup_{F\in\Sigma}\left\{\sum_{n=1}^\infty \mu_1(E_n\cap F) + \mu_2(E_n\setminus F) \right\}$$
and so $$\nu(E) \leqslant \sum_{n=1}^\infty\sup_{F\in\Sigma} \left\{\mu_1(E_n\cap F) + \mu_2(E_n\setminus F) \right\} = \sum_{n=1}^\infty \nu(E_n).$$
It remains to show the reverse inequality. (I got stuck here :/ )
