I was reading the proof of the following Lemma:
If $E_1, E_2$ are two Lebesgue measurable sets then their union and intersection are also Lebesgue measurable.
The proof is as follows:
Denote by $m^*(E)$ the outer measure of a set $E$.
We are only going to prove the case of union of two sets, the case of intersection is proven similarly.
We know that a set $E$ is called measurable iff for every $A\subset \mathbb{R}$ it holds $$m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$$ where $E^c$ is the complement of $E$. In our case this relation holds for the sets $E_1$ and $E_2$.
Furthermore for every $A\subset \mathbb{R}$ $$A = (A\cap E) \cup (A\cap E^c)$$ and using the subadditivity of outer measure we have: $$m^*(A)\leq m^*(A\cap E)+m^*(A\cap E^c)$$
So to prove that a set is measurable we only need to prove the reverse inequality (because if they are both true then equality must hold): $$m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^c)$$
Thus, in our case we want to show: $$m^*(A)\geq m^*(A\cap (E_1\cup E_2))+m^*(A\cap (E_1\cup E_2)^c)$$
We know that $E_1$ is measurable so $$m^*(A)= m^*(A\cap E_1)+m^*(A\cap E_1^c)$$
Now, this is where my trouble begins. The book states: As $E_2$ is also measurable it holds $$m^*(A\cap E_1^c)= m^*(A\cap E_1^c\cap E_2)+m^*(A\cap E_1^c\cap E_2^c)$$
I do not seem to understand where this last relation is coming from. The definition states that the first relation must hold for every subset of $\mathbb{R}$ on the left hand side, but the set $A\cap E_1^c$ is always a subset of $E_1^c$ and cannot thus run through all subsets of $\mathbb{R}$. What am I missing? The proof finishes by manipulating the last two relations and the subadditivity of outer measure.
I have searched on SE, but couldn't find any similar questions or methods of proof. It would mean the world if somebody could give me a little insight. Thanks in advance!
Best Answer
You have already stated that $E_2$ is measurable iff for every $B\subset\mathbb{R}$
$m^*(B)=m^*(B\cap E_2)+m^*(B\cap E_2^c)$.
Set $B:=A\cap E_1^C$.