Let \(\mu^*\) be an outer measure on \(X\) and let \(\mathcal M(\mu^*)\) be its Carathéodory \(\sigma\)-algebra. Show that the restriction of \(\mu^*\) to \(\mathcal M(\mu^*)\) is continuous from below, and continuous from above when the first set has finite outer measure.
Give an outer measure for which these continuity statements fail on arbitrary subsets of \(X\).
Solution
By Carathéodory’s theorem, \(\mu\coloneqq\mu^*|_{\mathcal M(\mu^*)}\) is a measure, so it is countably additive on \(\mathcal M(\mu^*)\). If \(E_1\subseteq E_2\subseteq\cdots\) are Carathéodory measurable, set \(F_1=E_1\) and \(F_n=E_n\setminus E_{n-1}\) for \(n\ge2\). Then the \(F_n\) are pairwise disjoint and measurable, and \[ \mu^*\qty{\bigcup_{n\ge1}E_n} =\sum_{n\ge1}\mu^*(F_n) =\lim_{N\to\infty}\sum_{n=1}^N\mu^*(F_n) =\lim_{N\to\infty}\mu^*(E_N). \]
If \(E_1\supseteq E_2\supseteq\cdots\) are measurable and \(\mu^*(E_1)<\infty\), put \(F_n=E_1\setminus E_n\). Then \(F_n\uparrow E_1\setminus\bigcap_nE_n\), so continuity from below and finite additivity give \[ \mu^*(E_1)-\mu^*\qty{\bigcap_nE_n} =\lim_{n\to\infty}\mu^*(F_n) =\mu^*(E_1)-\lim_{n\to\infty}\mu^*(E_n). \] Cancellation is legitimate because \(\mu^*(E_1)<\infty\), yielding continuity from above.
Neither statement is true for an arbitrary outer measure on arbitrary sets. On \(X=\NN\), define \[ \mu^*(A)= \begin{cases} 0,&A=\emptyset,\\ 1,&0<|A|<\infty,\\ 2,&|A|=\infty. \end{cases} \] This is an outer measure: monotonicity is immediate, and countable subadditivity follows because an infinite union either contains an infinite member or has infinitely many nonempty finite members. For \(E_n=\{1,\ldots,n\}\), \[ \mu^*(E_n)=1\quad\text{for all }n, \qquad \mu^*\qty{\bigcup_nE_n}=\mu^*(\NN)=2, \] so continuity from below fails. For \(G_n=\{n,n+1,\ldots\}\), \[ \mu^*(G_n)=2\quad\text{for all }n, \qquad \mu^*\qty{\bigcap_nG_n}=0, \] so continuity from above fails even though \(\mu^*(G_1)=2<\infty\).