Definition 1.3.1 An outer measure is an extended real-valued set function $\mu^*$ having the following properties:
(i) The domain of $\mu^*$ consists of all the subsets of $X$.
(ii) $\mu^*$ is nonnegative.
(iii) $\mu^*$ is countably subadditive.
(iv) $\mu^*$ is monotone.
(v) $\mu^*(\emptyset)=0$.

Definition 1.8.1 An outer measure $\mu^*$ on $(X,\rho)$ is called a metric outer measure if it satisfies the folowing property:
(vi) If $\rho(A,B)>0$, then $\mu^*(A\cup B)=\mu^*(A)+\mu^*(B)$.

Lemma 8.1.1 Let $\mu^*$ be a metric outer measure and let $A,B$ be any sets such that $A\subset B$, $B$ open. For any positive integer $n$, let $$A_n=\left\{x;x\in A,\rho(x,B^c)\geq\frac{1}{n}\right\}\mbox{.}$$ Then $$\lim_{n\rightarrow\infty}\mu^*(A_{2n})=\mu^*(A)\mbox{.}$$ Proof. $\quad$ Since $A_n\subset A_{n+1}$ and $\mu^*$ is monotone, the sequence $\{\mu^*(A_n)\}$ is an increasing sequence. Also, $\mu^*(A_n)\leq\mu^*(A)$. It therefore suffices to show that $$\tag{1.8.1}\lim_{n\rightarrow\infty}\mu^*(A_{2n})\geq\mu^*(A)\mbox{.}$$ $\quad$ Each point $y$ of $A$ lies in the open set $B$. Hence there is some $\epsilon$-neighborhood of $y$ that is contained in $B$. It follows that $\rho(y,B^c)>\epsilon$. This shows that $$A\subset\bigcup^\infty_{n=1}A_n\mbox{.}$$ Since $A_n\subset A$ for all $n$, we have $$\tag{1.8.2}A=\bigcup^\infty_{n=1}A_n\mbox{.}$$ Let $G_n=A_{n+1}-A_n$ for $n\geq 1$. Then (1.8.2) implies that, for any $n\geq 1$, $$A=A_{2n}\cup\left[\bigcup^\infty_{k=2n}G_k\right]=A_{2n}\cup\left[\bigcup^\infty_{k=n}G_{2k}\right]\cup\left[\bigcup^\infty_{k=n}G_{2k+1}\right]\mbox{.}$$ Hence, $$\tag{1.8.3}\mu^*(A)\leq\mu^*(A_{2n})+\sum^\infty_{k=n}\mu^*(G_{2k})+\sum^\infty_{k=n}\mu^*(G_{2k+1)}\mbox{.}$$ From the definition of the sets $A_n$ we have: if $x\in G_{2k}$, $y\in G_{2k+2}$, then $$\rho(x,B^c)>\frac{1}{2k+1},\quad\rho(y,B^c)<\frac{1}{2k+2}\mbox{.}$$ Consequently, $$\rho(G_{2k},G_{2k+2})\geq\frac{1}{2k+1}-\frac{1}{2k+2}>0\mbox{.}$$ Using the relation $A_{2n}\supset\bigcup\limits^{n-1}_{k=1}G_{2k}$ and the property (vi) of metric outer measures, we get $$\mu^*(A)\geq\mu^*(A_{2n})\geq\mu^*\left(\bigcup^{n-1}_{k=1}G_{2k}\right)=\sum^{n-1}_{k=1}\mu^*(G_{2k})\mbox{.}$$ This shows that the series $$\tag{$\dagger$}\sum^\infty_{k=1}\mu^*(G_{2k})$$ is convergent. Similarly one shows that the series $$\sum^\infty_{k=1}\mu^*(G_{2k+1})$$ is convergent. Taking $n\rightarrow\infty$ in (1.8.3), we then obtain $$\mu^*(A)\leq\lim_{n\rightarrow\infty}\mu^*(A_{2n})\mbox{.}$$ This proves (1.8.1).

I want to ask the convergence of ($\dagger$). If $\mu^*(A)<\infty$, then the series definitely converges. But there is no assumption about that.


1 Answer 1


If $\mu^*(A)=\infty>\lim\mu^*(A_{2n})$, then $\mu^*(A)\leq\lim\mu^*(A_{2n})$, a contradiction. Hence if $\mu^*(A)=\infty$, the lemma must hold; and we can suppose $\mu^*(A)<\infty$.


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