Question
Suppose $\mathfrak M$ is a $\sigma$-ring on $\mathbb R^n$, and $\mu$ is a nonnegative additive function on $\mathfrak M$. Let $\mu^*(A)=\inf\,\left\{\,\mu(E)\,\big\vert\,E\in\mathfrak M\land E\supset A\,\right\}$ and $\mu_*(A)=\sup\,\left\{\,\mu(E)\,\big\vert\,E\in\mathfrak M\land E\subset A\,\right\}$. Given that $A_1\supset A_2\supset\dotsb$ are bounded, $\bigcap_nA_n=\emptyset$.
- What about $\lim_{n\to\infty}\mu^*(A_n)$ and $\lim_{n\to\infty}\mu_*(A_n)$? Are they zeros?
- Suppose $\mathcal E$ is the elementary sets of $\mathbb R^n$, and $\mu$ is extended from an additive, regular, nonnegative and finite set function on $\mathcal E$, just as Baby Rudin chapter 11 did, how about the answers to the preceding questions?
Background
In our calculus reference book, there's a theorem stated:
(Arzela's dominated convergence theorem) Suppose $\{f_n\}$ is Riemann-integrable on $[a,b]$, and uniformly bounded. If $f_n\to f$ (pointwise), and $f$ is Riemann-integrable on $[a,b]$, then $$\lim_{n\to\infty}\int_a^bf_n(x)dx=\int_a^bf(x)dx$$
In order to prove that, there's a lemma stated:
(Lewin's lemma) Suppose $A_1\supset A_2\supset\dotsb$ are bounded and $\bigcap_nA_n$ is empty. Put $$\alpha_n=\sup\,\left\{\,m(E)\,\big\vert\,E\in\mathcal E\land E\subset A_n\,\right\}$$ Then $$\lim_{n\to\infty}\alpha_n=0$$ Where $m(E)$ is the length of the elementary set $E$.
The proof to that lemma, which is not rather general, is a bit tricky. Since I learnt a bit about measure theory these days, I tried to generalize the lemma into measures. I don't know whether it's true. Thanks!
