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