Proving that $\nu(E\cap(A_{1}\cup A_{2}))=\nu(E\cap A_{1})+\nu(E\cap A_{2})$ I am going over my tutorials in my real analysis course and there
is an unproved statement that I am having difficulty to verify, and
I could use some help with.

Definition:
$\nu:\, P(X)\to[0,\infty]$ s.t $\nu(\emptyset)=0$,
  $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$ and
  s.t if $A\subseteq B$ then $\nu(A)\leq\nu(B)$ is called an outer
  measure.
Definition:
A set $A$ is called $\nu$ measurable if $\forall E\subseteq
 X:\,\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$.
Statement:
If $\{A_{i}\}_{i=1}^{n}$ are disjoint $\nu$ measurable sets (where $\nu$ is an outer measure) then
  $$\nu(E\cap(\cup_{i=1}^{n}A_{i}))=\sum_{i=1}^{n}\nu(E\cap A_{i})$$ The
  proof is by induction.

I have tried proving the last statement, the case $n=1$ is trivial
since the expression on the LHS is exactly the expression on the LHS,
after this I only tried to prove the claim for $n=2$, just to get
the idea, but I didn't manage to prove that.
If $A_{1},A_{2}$ are $\nu$ measurable then so is $A_{1}\cap A_{2}$
and $A_{1}\cup A_{2}$ and I hoped that I could find some good $E$
and a choise of $\nu$ measurable sets (like the two examples I gave)
that will yield the desired result, but I couldn't find any good choice
for $E$ and the $\nu$ measurable set.
Can someone please help me out ?
 A: For any two $A$ and $B$ and some measurable $M$ we have
$$\nu(A \cup B) = \nu\big((A \cup B) \cap M\big) + \nu\big((A \cup B) \cap M^c\big).$$
Let $A \cap B = \varnothing$, and suppose there exists $M$ that separates $A$ and $B$, that is, $M \cap A = A$ and $M \cap B = \varnothing$. It follows that
\begin{align}
\nu(A \cup B) &= \nu\big((A \cap M) \cup (B \cap M)\big) + \nu\big((A \cap M^c)\cup(B \cap M^c)\big) \\
&= \nu(A \cup \varnothing) + \nu(\varnothing \cup B) \\
&= \nu(A) + \nu(B).
\end{align}
The above is enough to conclude
\begin{align}
\nu\big(E \cap (A \cup B)\big) &= \nu\big((E \cap A) \cup (E \cap B)\big) \\
&= \nu(E \cap A) + \nu(E \cap B)
\end{align}
for any measurable disjoint $A$ and $B$ (just set $M = A$).
I hope it helps ;-)
A: Since $A_1$ is measurable, for all $S\subseteq X$ we have $\nu(S)=\nu(S\cap A_1)+\nu(S\cap {A_1}^\complement)$.
Use it for $S=E\cap(A_1\cup A_2)$. Then, $S\cap A_1=E\cap A_1$ and, since $A_2\subseteq {A_1}^\complement$, we also have  $S\cap {A_1}^\complement =E\cap A_2$.
