I know similar questions have been asked before, but I'm looking for clarification of a proof. In Royden's book on real analysis, he proves that every set of positive measure contains a non-measurable set.
To prove that outer measures are not finitely additive, he proves the following claim: there are disjoint sets $A,B \subset \mathbb{R}$ for which $m^{*}(A \cup B) < m^{*}(A) + m^{*}(B)$.
His proof goes as follows: Assume $m^{*}(A \cup B)= m^{*}(A) + m^{*}(B)$ for every disjoint pair of sets $A$ and $B$. Then, by the definition of measurable set, every set must be measurable. This contradicts the preceding theorem (that every set of positive measure contains a non-measurable set).
I'm not exactly seeing the contradiction. Are the details of the proof that every set of positive measure contains a non-measurable set relevant?
Best Answer
I haven't seen Royden but one of the equivalent definitions of Lebesgue measurability is the following: $E$ is Lebesgue measurable if for every $X$,
$$m^{\star}(X) = m^{\star}(E \cap X)+ m^{\star}(E^c \cap X)$$
So, if you assume that $m^{\star}(A \cup B) = m^{\star}(A) + m^{\star}(B)$ whenever $A, B$ are disjoint, then it follows that every set is measurable - Since $X$ is a disjoint union of $E \cap X$ and $E^c \cap X$.
Luzin improved this to the following: For every $X$, there are disjoint $A, B \subseteq X$ such that $m^{\star}(X) = m^{\star}(A) = m^{\star}(B)$.