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I want to prove that a finitely additive function that is upper-semicontinuous is a pre-measure, when considered on a semi-ring.

That is, when $A_i\in R,$ $\mu(A_1 \cup A_2\cup...\cup A_n)=\mu(A_1)+...+\mu(A_n)$ (Note that the unions are disjoint).

And if: $E_1\supset E_2...$ and $E_1\cap E_2...=\emptyset$, then $\text{lim}_{n\rightarrow \infty} \mu(E_i)=0$, whenever $E_i=\cup_{j=1}^nA_j$

Then: $\mu(\cup_{i=1}^\infty A_i)=\sum_n^\infty\mu(A_i)$ and $\mu(\emptyset)=0$

Any tips on how to do this?

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Taking $E_n=\emptyset$ for all $n$ we get $\mu(\emptyset)=0$. For countable additivity you have to assume that $\cup_n A_n$ is also in the semi-ring. Let $E_k=\cup_n A_n \setminus \cup_{n=1}^{k} A_n$, Verify that this sequence is decreasing and their intersection is empty. Hence $\mu (E_k) \to 0$ as $k \to \infty$. Now use additivity of $\mu$ to conclude that $\mu (\cup_n A_n) \leq lim \mu( \cup_{n=1}^{k} A_n)=\lim \sum_{n=1}^{k} \mu (A_n)=\sum_{n=1}^{\infty} \mu (A_n)$. The revesrse inequality is trivial.

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