We have the outer measure
$$\mu^{*}(A) = \inf \left\{ \sum_{n=1}^{\infty} \mu_0 (A_n) \: | \: A_n \in \mathcal{A}_0, A \subset \bigcup_{n=1}^{\infty} A_n \right\}$$
The inner measure is:
$$\mu_{*} (A) = \mu_0 (X) - \mu^{*} (X \backslash A)$$
Where $\mu_0 (A) = \sum_{n=1}^{\infty} \mu_0 (A_n)$ and $\mathcal{A}_0$ is a $\sigma$ algebra.
But why is $\mu_{*} \le \mu^{*}$?
I always end up with a contradiction.
My proof would be:
$$\mu_{*} (A) - \mu^{*} (A) = \mu_0 (X) - \mu^{*} (X \backslash A) - \mu^{*} (A)$$
Because:
$$\mu^{*} (A) = \mu_0 (A) - \mu_{*} (\varnothing)$$
We get:
$$\mu_{*} (A) - \mu^{*} (A) = \mu_0 (X) - \mu^{*} (X \backslash A) - \mu_0 (A) + \mu_{*} (\varnothing)$$
$$\mu_{*} (A) - \mu^{*} (A) = \mu_0 (X \backslash A) - \mu^{*} (X \backslash A)$$
Because by definition $\mu^{*}(A) = \inf \sum_{n=1}^{\infty} \mu_0 (A_n) \le \sum_{n=1}^{\infty} \mu_0 (A_n)$ this means, that $\mu_0 (X \backslash A) - \mu^{*} (X \backslash A) \ge 0$
We get:
$$\mu_{*} (A) - \mu^{*} (A) \ge 0$$
$$\mu_{*} (A) \ge \mu^{*} (A)$$
So where's my mistake?
