Prove if $E_1$ and $E_2$ are measurable, so is $E_1 \cap E_2$ Prove if $E_1$ and $E_2$ are measurable, so is $E_1 \cap E_2$.  
Definition of measurable is as follows: A subset $E$ of $X$ is called measurable whenever $\mu(A)=\mu(A \cap E)+\mu(A \cap E^c)$ holds for all $A$ subset of $X$. 
To be a measure it must also satisfy the following properties:


*

*$\mu(\emptyset)=0$

*$\mu(A)\leq \mu(B)$ if $A \subset B$ (that is, $\mu$ is monotone)

*$\mu\left(\bigcup_{i=1}^\infty E_i \right) \leq \sum_{i=1}^\infty \mu(E_i)$ holds for every sequence of subsets $E_i$ of $X$ (that is, $\mu$ is subadditive).


This question just appears to be a matter of using definitions but I am a little confused on the definitions of $\mu$ and how to go about proving this.
 A: I think in the Chasky demonstration something is not right in the line where he asserts that $A \cap(E_2\cup E_1) \subseteq (A\cap E_2)\cup (A \cap E_1^c \cap E_2)$, or at least it seems false to me but I could be wrong.
I think an easier one could be the following demonstration, where I use m instead of mu for the outer measure.
So  to say that $E_1\cap E_2$ is measurable, you just need to show that for any set A subset of R
$$m(A)\geq m(A\cap (E_1\cap E_2)) + m(A\cap(E_1\cap E_2)^c)$$ 
Let's take A as any subset of R. Since $E_1$, is measurable, it means that:
$$ m(A)=m(A\cap E_1) + m(A\cap E_1^c)  \tag a $$
Now since $E_2$ is measurable too, you have that
$$ m(A\cap E_1)=m(A\cap E_1\cap E_2) + m(A\cap E_1\cap E_2^c) \tag b$$ and also that
$$ m(A\cap E_1^c)=m(A\cap E_1^c\cap E_2) + m(A\cap E_1^c\cap E_2^c) \tag c$$
Now substituting $(b)$ and $(c)$ in $(a)$ you get the following:
$$ m(A)=m(A\cap E_1\cap E_2) + m(A\cap E_1\cap E_2^c)+m(A\cap E_1^c\cap E_2) + m(A\cap E_1^c\cap E_2^c) \tag d$$
Now it is easy  to prove that the following holds:
$$\begin{align}(A\cap E_1\cap E_2^c)&\cup(A\cap E_1^c\cap E_2)\cup (A\cap E_1^c\cap E_2^c)\\&= A\cap((E_1\cap E_2^c)\cup(E_1^c\cap E_2)\cup(E_1^c\cap E_2^c))\\ &=A\cap(E_1^c \cup E_2^c)=A\cap (E_1\cap E_2)^c\end{align}$$
So by sub-additive you have that
$$ m(A\cap (E_1\cap E_2)^c) \leq m(A\cap E_1\cap E_2^c)+m(A\cap E_1^c\cap E_2) + m(A\cap E_1^c\cap E_2^c) \tag e$$
Substituting $(e)$ in $(d)$ you have the final result
$$\begin{align} m(A)&=m(A\cap E_1\cap E_2) + m(A\cap E_1\cap E_2^c)+m(A\cap E_1^c\cap E_2) + m(A\cap E_1^c\cap E_2^c) \\&\geq m(A\cap E_1\cap E_2) + m(A\cap(E_1\cap E_2)^c)\end{align}$$
I hope this helps
A: If $E_1$ and $E_2$ are measurable, then by definition $E_1^c $ and $E_2^c$ are measurable and $E_1^c \cup E_2^c$ is measurable (I will prove this below) and hence
$$ ( E_1^c \cup E_2^c )^c = E_1 \cap E_2 \; \text{must be measurable}$$
By using your definition, it is trivial to see that the complement of a measurable set is measurable.
Now, we show $E_1 \cup E_2 $ is measurable. By hypothesis, since $E_1$ is measurable, take $A \subseteq X$, such that
$$ \mu(A) = \mu(A \cap E_1) + \mu(A \cap E_1^c) $$
Now, since $E_2$ is measurable, then 
$$ \mu(A \cap E_1^c) = \mu(A \cap E_1^c \cap E_2) + \mu(A \cap E_1^c \cap E_2^c)$$
Now, combine these two equation to obtain
$$ \mu(A) = \mu(A \cap E_1) + \mu(A \cap E_1^c \cap E_2) + \mu(A \cap E_1^c \cap E_2^c)$$
Now, since $A \cap  (E_2 \cup E_1) \subseteq (A \cap E_2) \cup (A \cap E_1^c \cap E_2)$, then
$$ \mu(A) = \mu(A \cap E_1) + \mu(A \cap E_1^c \cap E_2) + \mu(A \cap E_1^c \cap E_2^c) \geq $$
$$ \geq \mu(A \cap (E_1 \cup E_2) + \mu(A \cap (E_1 \cup E_2)^c )$$
by monotonocity and subadditivty. Then, notice by definition $E_1 \cup E_2 $ must be measurable and your problem is solved.
