Extension of sum of premeasures is sum of extensions of premeasures

Let $ \mu_1, \mu_2$ be two premeasures on a ring $\mathcal{R}$. Prove that$$(\mu_1 + \mu_2)^* = \mu_1^* + \mu_2^*$$ where $\mu^*$ is the Caratheodory extension of $\mu$, that is $$\mu^*(A) = \inf\left\{\sum_n \mu(A_n) : A\subseteq\bigcup_n A_n, (A_n)_n\subseteq \mathcal{R}\right\}.$$

Attempt: $(\mu_1 + \mu_2)^* \ge \mu_1^* + \mu_2^*$ because $(\mu_1 + \mu_2)^*(A)=\inf\{\sum_{n=1}^{\infty} \mu_1(A_n) + \mu_2(A_n) : A \subset \bigcup_{n=1}^{\infty} A_n \in \mathcal{R}\} \ge \inf\{\sum_{n=1}^{\infty} \mu_1(A_n) : A \subset \bigcup_{n=1}^{\infty} A_n \in \mathcal{R}\} + \inf\{\sum_{n=1}^{\infty} \mu_2(A_n) : A \subset \bigcup_{n=1}^{\infty} A_n \in \mathcal{R}\}=\mu_1^*(A) + \mu_2^*(A)$.

Is this true?

For the other direction, take a covering $\{E_n\}\subseteq \mathcal{R}$ of $A$ so that $\mu_i^*(A)> \sum_{n-1}^\infty \mu_i(E_n) - \frac{\epsilon}{2}$ for $i=1,2$ and hence $(\mu_1+ \mu_2)^*(A) \leq \sum_{n=1}^\infty \mu_1(E_n) + \mu_2(E_n) < \mu_1^*(A) + \mu_2^*(A)+\epsilon$ for all $\epsilon >0$.

My question is, does the existence of the covering $\{E_n\}$ guaranteed? How to construct such one? Do you have other idea?