Let $m^\ast(E)$ < ∞.If for every interval (a, b) we have that $b-a$=$m^\ast((a,b)∩E) + m^\ast((a,b) ∩ E^c)$ then $E$ is lebesgue measurable

Question

Since we are considering only few one type if sets (i.e open intervals) , i don't know how to prove $m^\ast(A)$=$m^\ast((A∩E) + m^\ast(A∩ E^c)$ using only information that $b-a$=$m^\ast((a,b)∩E) + m^\ast((a,b) ∩ E^c)$.Is the question wrong ?

Answer 1

Note that the assumption is saying that for any interval $I = (a,b)$ we have that $$m^*(I) = m^*(I \cap E) + m^*( I \cap E^c)$$

Sub additivity gives us that $ m^*(A) \leq m^*(A \cap E) + m^*( A \cap E^c)$ so we will prove the reverse inequality

By the definition of outer measure we can choose a covering of $A$ by open intervals $I_k$ such that $\bigcup_k I_k = O$ and

$$ m^*(A) + \epsilon > \sum_k m^*(I_k)\\= \sum m^*(E \cap I_k)+ (E^c \cap I_k) \\ \ \ \ \ \ \ \ \ \ \ \geq m^*\big(\bigcup_k E \cap I_k \big) + m^*\big(\bigcup_k E^c \cap I_k \big) \\ = m^*(E \cap O ) + m^*(E^c \cap O ) \\ \ \ \geq m^*(E \cap A) + m^*(E^c \cap A) $$