Halmos "Measure Theory" section 6 problem 3 This question reads: if the smallest normal class containing a class E is denoted by N(E), then , for every semiring P, N(P) = S(P).
Has anyone had any joy with this? I've been working under the assumption that the proof of this assertion goes very much like that of Theorem B above (if R is a ring, then M(R) = S(R)) So I'm hunting for a class K(F) to take a role similar to that in the theorem. But all my choices fail, either to be closed to (even finite) unions of discrete sets, or to satisfy "N ⊂ K(E) ∀ E in N implies that N is a semiring".
Am I taking completely the wrong approach? Any hints appreciated.
 A: Denote $\mathcal{R}$ the ring generated by the semiring $\mathcal{P}.$ Then the class $\mathcal{R}$ is bigger than the class $\mathcal{P}.$ Consequently, $\mathcal{N}(\mathcal{P})\subset \mathcal{N}(\mathcal{R}).$  On the other hand, each element of $\mathcal{R}$ is a disjoint union of finite number of elements from $\mathcal{P}$ (Halmos, $\S5$, exercise 3). As $\mathcal{N}(\mathcal{P})$ is closed under the formation of countable disjoint unions and contains $\mathcal{P},$ it contains $\mathcal{R}.$ Hence, $\mathcal{R}\subset \mathcal{N}(\mathcal{P})$ and $\mathcal{N}(\mathcal{R})\subset \mathcal{N}(\mathcal{P}).$ Finally, $\mathcal{N}(\mathcal{P})= \mathcal{N}(\mathcal{R}).$ So, it is enough to consider the case when $\mathcal{P}=\mathcal{R}$ is a ring. 
Next step is to show that $\mathcal{N}(\mathcal{R})$ is closed under finite intersections. At first we'll check that for every $E\in \mathcal{R}$ and every $F\in \mathcal{N}(\mathcal{R})$ $E\bigcap F\in \mathcal{N}(\mathcal{R}).$ To do this denote $\mathcal{K}_1(E)=\{F\in \mathcal{N}(\mathcal{R}): E\bigcap F\in \mathcal{N}(\mathcal{R})\}$ and check that it is a normal class. 1st property: assume that $F_n\in \mathcal{K}_1(E),$ $F_n\supset F_{n+1}$  then $E\bigcap F_n\in \mathcal{N}(\mathcal{R}),$ $E\bigcap F_n\supset E\bigcap F_{n+1}.$  As $\mathcal{N}(\mathcal{R})$ is a normal class, one has $\bigcap^{\infty}_{n=1} (E\bigcap F_n)=E\bigcap(\bigcap^{\infty}_{n=1} F_n)\in \mathcal{N}(\mathcal{R}).$ So, $\bigcap^{\infty}_{n=1} F_n \in \mathcal{K}_1(E)$ and $\mathcal{K}_1(E)$ is closed under the formation of countable decreasing intersections. 2nd property ($\mathcal{K}_1(E)$ is closed under the formation of countable disjoint unions) can be checked completely in the same way. Also note that $\mathcal{K}_1(E)$ is a normal class for any choice of the set $E$. In our case $E\in \mathcal{R},$ so $\mathcal{R}\subset \mathcal{K}_1(E)$ ($E\in \mathcal{R}$ and $\mathcal{R}$ is a ring). Hence, $\mathcal{N}(\mathcal{R})\subset \mathcal{K}_1(E)$ and actually $\mathcal{N}(\mathcal{R})=\mathcal{K}_1(E).$ This means that for each $E\in \mathcal{R}$ and every $F\in \mathcal{N}(\mathcal{R})$ $E\bigcap F\in \mathcal{N}(\mathcal{R}).$ Now consider $E\in \mathcal{N}(\mathcal{R})$ and define $\mathcal{K}_1(E)$ in the same way. We already proved that for each $F\in \mathcal{R}$ $E\bigcap F\in \mathcal{N}(\mathcal{R}).$ In other words, we have that $\mathcal{R}\subset \mathcal{K}_1(E).$ But we already proved that $\mathcal{K}_1(E)$ is a normal class, so $\mathcal{N}(\mathcal{R})=\mathcal{K}_1(E).$ It means that for all $E,F\in \mathcal{N}(\mathcal{R})$ $E\bigcap F\in \mathcal{N}(\mathcal{R}).$ So, $\mathcal{N}(\mathcal{R})$ is closed under finite intersections. 
Now we'll show that $\mathcal{N}(\mathcal{R})$ is closed under countable intersections (not necessarily decreasing!). Indeed, assume that $F_n\in \mathcal{N}(\mathcal{R}), \ n\geq 1.$ Then $E_n=F_1\bigcap \ldots \bigcap F_n\in\mathcal{N}(\mathcal{R})$ (finite intersections). Evidently, $E_n\supset E_{n+1}.$ $\mathcal{N}(\mathcal{R})$ is closed under decreasing countable intersections, so $\bigcap^{\infty}_{n=1} E_n=\bigcap^{\infty}_{n=1} F_n\in \mathcal{N}(\mathcal{R}).$ So, $\mathcal{N}(\mathcal{R})$ is closed under the formation of countable disjoint unions and countable intersections. 
The point is to show that $\mathcal{N}(\mathcal{R})$ is closed under differences. At first consider $E\in \mathcal{R}$ and denote $\mathcal{K}_2(E)=\{F\in \mathcal{N}(\mathcal{R}): E\setminus F\in \mathcal{N}(\mathcal{R})\}.$ $\mathcal{K}_2(E)$ is closed under disjoint unions: $E\setminus \bigcup^{\infty}_{n=1} F_n=\bigcap^{\infty}_{n=1}
(E\setminus F_n)$ (as $\mathcal{N}(\mathcal{R})$ is closed under arbitrary countable intersections!).  $\mathcal{K}_2(E)$ is closed under countable intersections: $E\setminus \bigcap^{\infty}_{n=1} F_n=\bigcup^{\infty}_{n=1}
(E\setminus F_n)=\bigcup^{\infty}_{n=1} (F_1\bigcap \ldots \bigcap F_{n-1} \bigcap E\setminus F_n)$. Hence,     $\mathcal{K}_2(E)$ is a normal class and for all $E\in \mathcal{R},$ $F\in \mathcal{N}(\mathcal{R})$ $E\setminus F\in \mathcal{N}(\mathcal{R}).$ Then it is an easy matter to prove that $\mathcal{N}(\mathcal{R})$ is closed under differences and is actually a $\sigma-$ring.
A: Problem 6.2 was to show that a σ-ring is a normal class; and a normal ring is a σ-ring. This is very easy; I assume it below.
$()$ is a σ-ring; 6.2 shows it is normal, so $()\supset ()$. Suppose $A,B∈$ ; then since $$ is a semiring $A-B$ is a finite union of disjoint sets of $$, i.e., $A-B∈()$. Then $A\cup B=B\cup(A-B)$ is a disjoint union of sets of $()$ so $A\cup B∈()$. Hence $()\subset ()$ and therefore $(())\subset ()$ as it is the minimal normal class containing $()$. But 6.2 shows that $(())$ is a σ-ring containing $$ so $()\subset (())\subset ()$. Hence $()=()$ as required.
