Extension of measure is less or equal to outer measure when restricted to semiring [closed]
Solution 1:
Given any $A\in\mathscr{A}$, we have \begin{align*} \nu(A) &\leq \inf \left \{\sum_{n = 1}^\infty \nu(A_n): \forall n \in \Bbb N, \ A_n \in \mathscr{H}, \ A \subseteq \bigcup_{n = 1}^\infty A_n \right\} = \\ & = \inf \left \{\sum_{n = 1}^\infty \mu(A_n): \forall n \in \Bbb N, \ A_n \in \mathscr{H}, \ A \subseteq \bigcup_{n = 1}^\infty A_n \right\} = \\ &= \mu^*(A) \end{align*} So, for all $A\in\mathscr{A}$, $\nu(A) \leq \mu^*(A)$.