If U be a family ofopen P- nul setsl in Borel sigmafield in R^n then P(union of U)=0.

Question

I have seen this problem in my measuretheoretic probability class note but I can not do this . Let P be a probability measure on (R^n,BR^n) and U a family of open P null sets , then , P( ∪U )=0

Answer 1

$\mathbb R^{n}$ is separable metric space. Hence it is second countable and any union of open sets in it can be expressed as a countable union of some subfamily. From this the conclusion is obvious.