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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, $\bigcap_nA_n=\emptyset$.

  1. What about $\lim_{n\to\infty}\mu^*(A_n)$ and $\lim_{n\to\infty}\mu_*(A_n)$? Are they zeros?
  2. 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!

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1. Well, the limits exist, since $\mu^*$ and $\mu_*$ are monotone set functions, but the limits are not necessarily $0$. There are standard counterexamples that apply for the case where $\mathfrak{M}$ is $\sigma$-algebra, $\mu$ is a measure and the $A_n$ are measurable. - 2. What do you mean by "regular"? How do you extend $\mu$ to the $\sigma$-ring generated by $\mathcal{E}$ (if that's what you want?)? –  Thomas Feb 22 '13 at 13:11
    
@Thomas From Baby Rudin: A nonnegative additive set function $\phi$ defined on $\mathcal E$ is said to be regular if the following is true: To every $A\in\mathcal E$ and to every $\epsilon>0$ there exists $F\in\mathcal E$, $G\in\mathcal E$ such that $F$ is closed, $G$ is open, $F\subset A\subset G$ and $\phi(G)-\epsilon\le\phi(A)\le\phi(F)+\epsilon$. –  Frank Science Feb 23 '13 at 2:51
    
@Thomas The extension process is referred to Baby Rudin 11.7 to 11.10, just like the Lebesgue measure extended from the length function of elementary sets. –  Frank Science Feb 23 '13 at 2:54
    
@Thomas If $\{A_n\}$ are measurable, it seems that $\mu(A_n)\to0$ since $A_n$ are bounded. We can write $A_1=\bigcup_n(A_n\backslash A_{n+1})$, since $\forall a\in A_1,\exists N: a\in A_N\land a\not\in A_{N+!}$, and $\mu(A_1)=\sum_n\mu(A_n\backslash A_{n+1})$. Since both sides are finite, we have $\mu(A_m)=\sum_{n\ge m}\mu(A_n\backslash A_{n+1})\to0$ as $m\to\infty$. –  Frank Science Feb 23 '13 at 2:58
    
Regular: ok. - Extension: But you already have the set function defined on elementary sets. I don't see how regularity on elementary sets lets you extend the additive set function to the $sigma$-ring generated by the elementary sets. - Monotone convergence: Ok, suppose your content extends to a Borel measure on $\mathbb{R}^n$, then it is finite on bounded sets and your argument for measurable sets works. –  Thomas Feb 25 '13 at 9:41

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