Finitely addivite finite regular set function on real Borel $\sigma$ algebra is already a measure. Let $\mu$ be a finitely additive set function on the Borel $\sigma$-algebra $B_{\mathbb{R}}$ with $\mu(\mathbb{R})&lt;\infty$ and $\mu (A) = \sup\{\mu(K) \mid K \subseteq A, \ K \text{ compact} \} $ for every $A \in B_{\mathbb{R}}$. Then $\mu$ is $\sigma$-additive and therefore a measure.

So far I did not find any way to tackle the above problem. Thanks for any inspiration!
We may assume that each $A_i$ is bounded, for $K$ is bounded. Now it suffices to show that for each $i$, we can find open set $O_i \supset A_i$ and having measure not "slightly" greater than that of $A_i$. For this, first enclose $A_i$ by an open interval $I_i$ and find a "big" compact $L_i$ inside $I_i \setminus A_i$. Finally put $O_i = I_i \setminus L_i$.
We may assume that each $A_i$ is bounded, for $K$ is bounded. Now it suffices to show that for each $i$, we can find open set $O_i \supset A_i$ and having measure not "slightly" greater than that of $A_i$. For this, first enclose $A_i$ by an open interval $I_i$ and find a "big" compact $L_i$ inside $I_i \setminus A_i$. Finally put $O_i = I_i \setminus L_i$.