In Real Analysis 2 by Terence Tao, the author claims that:
Lemma 7.4.5 and Proposition 7.3.3, when combined,
imply that there exist non-measurable sets
With $m^*$: outer measure, $m$: Lebesgue measure:
Lemma 7.4.5 (Finite additivity)
If $(E_j)_{j \in J}$ are a finite collection of disjoint measurable sets and any set A (not necessarily measurable), we have:
$m^*(A \, \cap \bigcup \limits_{j \in J} E_j) = \sum \limits_{j \in J} m^*(A \, \cap E_j)$
Furthermore, we have:
$m(\bigcup \limits_{j \in J} E_j) = \sum \limits_{j \in J} m(E_j)$
Proposition 7.3.3 (Failure of finite additivity)
There exists a finite collection $(A_j)_{j \in J}$ of disjoint subsets of $\mathbb{R}$, such that
$m^*(\bigcup \limits_{j \in J} A_j) \neq \sum \limits_{j \in J} m^*(A_j)$
Could you please give me some explanations why this claim is true ? (or at least some intuitive explanations).
P/s: I quote the definition of Lebesgue measure $m$ by Terence Tao for further information:
Definition 7.4.1 (Lebesgue measurability)
Let $E$ be a subset of $\mathbb{R^n}$.
We say that $E$ is Lebesgue measurable, or measurable for short, iff we have the identity:
$m^*(A) = m^*(A \cap E) + m^*(A$ \ $E)$
for every subset $A$ of $\mathbb{R^n}$.
If $E$ is measurable, we define Lebesgue measure of $E$ to be $m(E) = m^*(E)$.
Best Answer
By proposition 7.3.3 one of the sets $A_j$ of the collection must be non-measurable,because if all the sets $A_j$ where measurable,then the set $B=\bigcup_{j \in J}A_j$ would be measurable and we would have the equality $m^*(B)=\sum_{j \in J}m^*(A_j)$ byLemma 7.4.5