If $K$ is compact, then $m_*(K) = m^*(K)$

Question

I was reading http://people.math.harvard.edu/~shlomo/212a/11.pdf

In page 25, it mentions that:

If $K$ is a comapct set, then $m_*(K) = m^*(K)$ since $K$ is a comapct set contained in itself.

I don't quite understand how the inner measure equal the outer measure here.

Answer 1

I misread the definition of inner measure.

The inner measure is given as: $$m_\ast(A)=\sup\lbrace m^*(K): K\subseteq A \text{ is compact}\rbrace.$$

Hence, when $K$ compact, by definition $$m_*(K) = m^*(K)$$