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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.