Definition of Lebesgue Outer Measure: Given a set $E$ of $\mathbb R$, we define the Lebesgue Outer Measure of $E$ by, $$m^*(E) = \inf \left\{\sum_{n=1}^{+\infty} \ell(I_n): E \subset \bigcup_{n=1}^{+\infty}I_n \right\}$$ where $\ell(I_n)$ denotes the length of interval (bounded and nonempty interval).
Definition of measurable set: A set $E$ measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.
If $E \subset \mathbb R$ with $m^*(E) = 0$ and $\exists$ finite interval $I,$ such that $E \subset I, $, then is $E$ measurable?
If $E \subset \mathbb R$ with $m^*(E) = 0$, then is $E$ measurable?
Best Answer
Yes. Suppose that ${\frak m}^\ast E = 0$. On one hand, the inequality: $${\frak m}^\ast T \leq {\frak m}^\ast(T \cap E) + {\frak m}^\ast(T \cap E^c)$$always holds. On the other hand: $${\frak m}^\ast(T \cap E) \leq {\frak m}^\ast E = 0 \implies {\frak m}^\ast(T \cap E) = 0, \quad {\frak m}^\ast(T \cap E^c) \leq {\frak m}^\ast T$$gives the other inequality.