$\mu^* \left( \bigcup_{n=1}^{\infty} A_n\right) = 0$

Question

Let us have a fixed interval $I_0=[a,b]$ and let $A$ be a subset of $I_0.$ Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of subsets of $I_0$ s.t $\mu^* (A_n)$ (outer measure) is 0 for all natural $n$. How to prove that $$\mu^* \left( \bigcup_{n=1}^{\infty} A_n\right) = 0$$

without using the fact that $\bigcup_{n=1}^{\infty} A_n$ is measurable?

Thank you!

Answer 1

Hint: Use the $\sigma-$subadditivy and you obtain: $$ 0 \leq \mu^*(\bigcup A_n) \leq \sum \mu^*(A_n) $$