10/10/24
Reference: Lecture notes 3.1, 3.2
Proposition (Properties of measurable sets)
(Intervals) Any interval $I\subset\R$ is measurable.
(Translation invariance) If$A\subset\R$ is measurable, then so is $A+c$ for any $c\subset \R.$
(Complements) If $A\subset\R$ is measurable, then so is $A^c =\R\setminus A$.
(Countable unions and intersections) If $A_1,A_2, \ldots\subset\R$ are measurable, then so are
$\bigcup_{n=1}^\infty A_n$ and $\bigcap_{n=1}^\infty A_n$
Proof. (1) Let $I=(a,b)$. Given $E$, cover it with open intervals $I_1,I_2,\ldots$. This gives us coverings
$E\setminus I \subset \bigcup_{n=1}^\infty I_n \setminus I$ and $E\cap I\subset \bigcup_{n=1}^\infty I\cap I_n$
The difference and intersection of two intervals are disjoint sums of intervals. They are not necessarily open but by the previous lemma, we can use intervals of any kind to define the Lebesgue outer measure. Therefore, using basic facts about length of intervals,
$\mu^(E\setminus I) + \mu^(E\cap I) \leq \sum_{n=1}^\infty l(I_n \setminus I)+\sum_{n=1}^\infty l(I_n \cap I) = \sum_{n=1}^\infty l(I_n)$
Taking infimum over all coverings we get
$\mu^(E\setminus I) + \mu^(E\cap I) \leq \mu(E)$
which shows that $I$ is measurable. For other types of intervals the statement will follow from (3).
(2) This is clear because $\mu^*$ is translation invariant.
(3) This is clear from the definition of a measurable set.
(4) First we show that the union of finitely many measurable sets is measurable. By induction, it suffices to consider the union of two such sets $A,B.$ We want to show that
$\mu^(E) = \mu^(E \cap (A\cup B)) + \mu^*(E \cap (A\cup B)^c)$