Question

Prove that if $$$ E \cup \{x\} $$$ is mesurable then E is mesurable when

and $$$ x \in E^c $$$ where is the complement of E

Answer 1

Let E C R and x elementof E^c suppose E union {x} is measurable. i.e., mu (A) = mu (A intersection (E union {x})) + mu (A intersection (E union {x})^c), (*) for any subset A C R we have to show E is measurable. let Bbe any subset of R. since mu is countable sub additive, therefore mu (B) lessthanorqualto mu (B intersection E) + mu (B intersection E^c) Let x elementof B. we take A = B intersection {x}^c in (*) and we get mu (B intersection {x}^c) = mu ((B intersection {x}^c) intersection (E union {x})) + mu ((B intersection {x}^c) intersection (E union {x}^c), = mu ((B intersection {x}^c intersection E) union (B intersection {x}^c intersection {x})) + mu ((B intersection {x}^c) intersection (E^c intersection {x}^c)) = mu (B intersection E) + mu (B intersection E^c intersection {x}^c) Rightarrow mu (B)