91Ó°ÊÓ

A set function \(\phi\) is called additive if \(\phi(A \cup B) = \phi(A) + \phi(B)\) for any two disjoint sets A and B in the ring. We want to show that the function \(\phi\) is an additive set function on the ring of elementary subsets of \((0,1]\). Consider any two disjoint elementary sets \(A\) and \(B\) in \(\mathscr{R}\). Since they are elementary sets, they can be expressed as finite unions of disjoint intervals, i.e., \(A = \bigcup_i A_i\) and \(B = \bigcup_j B_j\), where \(A_i\) and \(B_j\) are disjoint intervals. Now, \(\phi(A)\) and \(\phi(B)\) are the sum of the lengths of their constituent intervals. Since \(A\) and \(B\) are disjoint, their union \(A \cup B\) consists of the same intervals, thus the same sum of lengths. Thus, we have $$ \phi(A \cup B) = \phi(A) + \phi(B), $$ which proves that \(\phi\) is an additive set function.

A set function \(\phi\) is called regular if for any positive number \(\varepsilon\), and any set \(E\) in the ring, there exists an elementary set \(A\) such that \(E \subseteq A\) and \(\phi(A) - \phi(E) < \varepsilon\). To show that \(\phi\) is not a regular set function, we need to find an example where this condition is not satisfied. Let's consider the set \(E = (0,1]\). Note that \(E \in \mathscr{R}\). Now, let \(A = [a,1]\), where \(a \in (0,1)\). It is clear that \(E \subseteq A\). However, for any elementary set \(A = [a,1]\), we have $$ \phi(A) - \phi(E) = \phi((a,1]) - \phi((0,1]) = (1 - a) - (1 + 1) = - a - 1 \not< 0. $$ Since \(\phi(A) - \phi(E) < 0\), the set function \(\phi\) is not regular.

A set function \(\phi\) is called countably additive if for any countable collection of pairwise disjoint sets \(\{E_n\}\) in the ring, we have \(\phi(\bigcup_n E_n) = \sum_n \phi(E_n)\). Let's assume that \(\phi\) can be extended to a countably additive set function on a \(\sigma\)-ring. Consider the sets \(E_n = (0, \frac{1}{n}]\), where \(n = 1,2,3,\dots\). These sets are in the \(\sigma\)-ring and are pairwise disjoint. We can write $$ (0,1] = \bigcup_{n=1}^\infty E_n. $$ By countable additivity, we should have $$ \phi((0,1]) = \sum_{n=1}^\infty \phi(E_n). $$ Now, observe that \(\phi(E_n) = 1 + \frac{1}{n}\) for each \(n\). Therefore, $$ \phi((0,1]) = \sum_{n=1}^\infty \left(1 + \frac{1}{n}\right). $$ However, the sum on the right-hand side is not convergent, since the series \(\sum_{n=1}^\infty \frac{1}{n}\) is a harmonic series, which is known to diverge. This contradicts the assumption that \(\phi\) can be extended to a countably additive set function on a \(\sigma\)-ring. Thus, the function \(\phi\) cannot be extended to a countably additive set function on any \(\sigma\)-ring.