Exercise 11.15

Question

Let $\mathscr{R}$ be the ring of all elementary subsets of $(0,1]$. If $0 <
    a \le b \le 1$, define
    \begin{equation*}
      \phi([a,b]) = \phi([a,b)) = \phi((a,b]) = \phi((a,b)) = b-a,
    \end{equation*}
    but define
    \begin{equation*}
      \phi((0,b)) = \phi((0,b]) = 1 + b
    \end{equation*}
    if $0 < b \le 1$. Show that this gives an additive set function $\phi$ on
    $\mathscr{R}$, which is not regular and which cannot be extended to a
    countably additive set function on a $\sigma$-ring.

Amit Awasthi · Accepted Answer

\begin{proof}
      To show the function is additive, we only have to note there cannot be two
      disjoint intervals $A,B$ that share $0$ as the lower endpoint.
      \par $\phi$ is not regular since for any set $(0,a]$ where $a < 1$, we
      we have $\phi((0,a]) = 1+a$ but any closed subset $F \subset (0,a]$ does not
      contain $0$, hence $\phi(F) \le a$. $\phi$ cannot be extended to a
      countably additive set function since
      \begin{equation*}
        (0,1] = \bigcup_{n=0}^\infty \left(
        \frac{1}{2^{n+1}},\frac{1}{2^{n}}\right]
      \end{equation*}
      but $\phi((0,1]) = 2$ while $\phi$ on the right set is $1$.
    \end{proof}

Exercise 11.15

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Exercise 11.15

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