Proving that the set of $\nu$ measurable sets $M\subseteq P(X)$ is an algebra I am not sure I am using the standard definitions so I will open by
defining what I need:

  
*
  
*Let $X$ be a set, $\nu:\, P(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any
  $\{A_{i}\}_{i=1}^{\infty}\subseteq P(x)$ (not neccaseraly disjoint) it
  holds that
  $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$
  
*Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$

The exercise asks to prove that the set of $\nu$ measurable sets
$M\subseteq P(X)$ is an algebra.
I have proved $\emptyset,X\in M$ and that $A\in M\implies A^{c}\in M$
but I am having problems proving closer under union and intersection.
I assume that $A_{1},A_{2}$ are $\nu$ measurable so I get that for
any $E$: $$\nu(E)=\nu(E\cap A_{1})+\nu(E\cap A_{1}^{c})$$ 
$$\nu(E)=\nu(E\cap A_{2})+\nu(E\cap A_{2}^{c})$$
And I need to prove that for any $E'$: $$\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu(E'\cap(A_{1}\cap A_{2})^{c})$$
which is the same as $$\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu((E'\cap A_{1}^{c})\cup(E'\cap A_{2}^{c}))$$
and a similar result to prove closer under union.
I guess that it all have to do with choosing the right $E$'s from
knowing that $A_{i}$ are $\nu$ measurable, but I tried different
options for an hour now and I don't see this going anywhere.
I need some help in showing closer under union and intersection
 A: If $A\in \mathcal{P}(X)$, then $E=(E\cap A)\cup (E\cap A^c)$ for alle $E\in\mathcal{P}(X)$. By 1. we have that
$$
\nu(E)\leq \nu(E\cap A)+\nu(E\cap A^c),
$$
and hence $A\in M$ if and only if 
$$
\nu(E)\geq \nu(E\cap A)+\nu(E\cap A^c),\quad E\in\mathcal{P}(X).\qquad (*)
$$
Let $A,B\in M$ be given and let us show that $A\cup B\in M$ by showing that $A\cup B$ satisfies $(*)$. If $E\in\mathcal{P}(X)$, then
$$
\begin{align*}
\nu(E)&=\nu(E\cap A)+\nu(E\cap A^c)\\
&=\nu(E\cap A \cap B)+\nu(E\cap A\cap B^c)+\nu(E\cap A^c\cap B)+\nu(E\cap A^c\cap B^c)\\
&\geq \nu\big((E\cap A\cap B)\cup(E\cap A\cap B^c)\cup(E\cap A^c\cap B)\big)+\nu(E\cap A^c\cap B^c)\\
&=\nu(E\cap (A\cup B))+\nu(E\cap (A\cup B)^c)
\end{align*}
$$
and hence $A\cup B\in M$. To show that also $A\cap B\in M$, we just use that $(A\cap B)^c=A^c\cup B^c$ which is in $M$ and hence $A\cap B\in M$.
In the last equality we used that
$$
(E\cap A\cap B)\cup (E\cap A\cap B^c)\cup (E\cap A^c\cap B)=E\cap \big((A\cap B)\cup (A\cap B^c)\cup (A^c\cap B)\big) \\
=E\cap\big(A\cup (A^c\cap B)\big)=E\cap (A\cup B).
$$
