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.
Best Answer
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