# Friedman Lemma 8.1.1

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.

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 $$\mu^*(A)<\infty$$.