Example of non-unique extension of measure for the pre-integral in the Stone-Daniell Theorem Prerequisites
Definition 1 Let $\mathcal{L}$ be a non-empty collection of real-valued functions on a set $X$. Then $\mathcal{L}$ is a real vector space iff for all $f,g\in\mathcal{L}$ and $c\in\mathbb{R},cf+g\in\mathcal{L}$. Let $f\lor g:=\mathrm{max}(f,g),f\land g:=\mathrm{min}(f,g)$. A vector space $\mathcal{L}$ of functions is called a vector lattice iff for all $f$ and $g$ in $\mathcal{L}$, $f\lor g\in\mathcal{L}$. Then also $f\land g\equiv-(-f\lor-g)\in\mathcal{L}$. The vector lattice $\mathcal{L}$ will be called a Stone vector lattice iff for all $f\in\mathcal{L}, f\land1\in\mathcal{L}$.
Definition 2 Given a set $X$ and a vector lattice $\mathcal{L}$ of real functions on $X$, a pre-integral is a function $I$ from $\mathcal{L}$ into $\mathbb{R}$ such that:
(a) $I$ is linear: $I(cf+g)=cI(f)+I(g)$ for all $c\in\mathbb{R}$ and $f,g\in\mathcal{L}$.
(b) $I$ is nonegative, in the sense that whenever $f\in\mathcal{L}$ and $f\geq0$ (everywhere on $X$), then $I(f)\geq0$.
(c) $I(f_n)\downarrow0$ whenever $f_n\in\mathcal{L}$ and $f_n(x)\downarrow0$ for all $x$.
Definition 3 For the rest of this post, assume given a set $X$, a vector lattice $\mathcal{L}$ of real functions on $X$, and a pre-integral $I$ on $\mathcal{L}$. For any two functions $f$ and $g$ in $\mathcal{L}$ with $f\leq g$ (that is $f(x)\leq g(x)$ for all $x$), let
\begin{equation}
[f,g):=\{\langle x,t\rangle\in X\times\mathbb{R}:f(x)\leq t<g(x)\}.
\end{equation}
Let $\mathcal{S}$ be the collection of all $[f,g)$ for $f\leq g$ in $\mathcal{L}$. Define $\nu$ on $\mathcal{S}$ by $\nu([f,g)):=I(g-f)$. So if $g\geq0$, then $I(g)=\nu([0,g))$, and for any $f\in\mathcal{L},I(f)=\nu([0,f^+))-\nu([0,f^-))$.
Theorem (A. C. Zaanen) $\nu$ extends to a countably additive measure on the $\sigma$-algebra $\mathcal{T}$ generated by $\mathcal{S}$.
Theorem (Stone-Daniell) Let $I$ be a pre-integral on a Stone vector lattice $\mathcal{L}$. Then there is a measure $\mu$ on $X$ such that $I(f)=\int fd\mu$ for all $f\in\mathcal{L}$. The measure $\mu$ is uniquely determined on the smallest $\sigma$-ring $\mathcal{B}$ for which all functions in $\mathcal{L}$ are measurable.
Proof Let $\mathcal{M}$ be the collection of all sets $f^{-1}((1,\infty))$ for $f\in\mathcal{L}$. Then $\mathcal{M}$ contains, for any $f\in\mathcal{L}$ and $r>0$, the sets $f^{-1}((r,\infty))=(f/r)^{-1}((1,\infty))$ and $f^{-1}((-\infty,-r))=(-f)^{-1}((r,\infty))$. Since the intervals $(-\infty,-r)$ and $(r,\infty)$ for $r>0$ generate the $\sigma$-ring of Borel subsets of $\mathbb{R}$ not containing 0, $\mathcal{M}$ generates the $\sigma$-ring $\mathcal{B}$ defined in the statement of the theorem.
...
Problem
Give an example where the measure $\mu$ in the Stone-Daniell theorem has more than one extension to the smallest $\sigma$-algebra $\mathcal{A}$ for which all functions in $\mathcal{L}$ are measurable, and where $\mu$ is bounded on the smallest $\sigma$-ring $\mathcal{R}$ for which the functions are measurable.
My efforts:
We will make use of the following lemma.
Lemma Let $\mathcal{R}$ be a $\sigma$-ring of subsets of a set $X$. Let $\mathcal{S}$ be the $\sigma$-algebra generated by $\mathcal{R}$.
(1) $\mathcal{S}$ consists of all sets in $\mathcal{R}$ and all complements of sets in $\mathcal{R}$.
(2) Let $\mu$ be countably additive from $\mathcal{R}$ into $[0,\infty]$. For any set $C\subset X$ let $\mu_*(C):=\mathrm{sup}\{\mu(B):B\subset C,B\in\mathcal{R}\}$ (inner measure). Then $\mu_*$ restrict to $\mathcal{S}$ is a measure, which equals $\mu$ on $\mathcal{R}$.
(3) The extension of $\mu$ to a measure on $\mathcal{S}$ is unique if and only if either $\mathcal{S}=\mathcal{R}$ or $\mu_*(X\setminus A)=+\infty$ for all $A\in\mathcal{R}$.
By the lemma, the example must satisfy $\mathcal{A}\neq\mathcal{B}$ and $\mu_*(X\setminus B)<+\infty$ for some $B\in\mathcal{B}$. We can firstly find such a $\mathcal{B}$ and $\mu$, and then construct a Stone lattice and pre-integral.
First I tried $X=\mathbb{R}$, $\mathcal{B}$ consisting of all Borel sets in $\mathbb{R}\setminus(-1,1)$, and $\mu$ the Lebesgue measure. $\mathcal{B}$ is a $\sigma$-ring. $\mathcal{A}$ consists of all Borel sets in $\mathbb{R}$, so $\mathcal{A}\neq\mathcal{B}$. Pick $B=\mathbb{R}\setminus(-1,1)\setminus(2,3)$. Then $\mu_*(X\setminus B) =\mu_*((-1,1)\cup(2,3)) =\mu((2,3))=1<+\infty$. So this $\mathcal{B}$ and $\mu$ work. Next we need to construct the Stone lattice and pre-integral. We need to find a Stone vector lattice $\mathcal{L}$ of $f$ defined on $\mathbb{R}$ such that $\mathcal{B}$ is the smallest $\sigma$-ring for which all functions in $\mathcal{L}$ are measurable. This seems impossible if we look at the proof of the Stone-Daniell Theorem. We can let one $f^{-1}((1,\infty))$ lie outside $(-1,1)$. But another one must be inside with appropriate $r$ by $(f/r)^{-1}((1,\infty))$.
Any easy example of a $\sigma$-ring which is not a $\sigma$-algebra would be the collection of all countable subsets of an uncountable set. So I tried   $X=\mathbb{R}$, $\mathcal{B}$ being the collection of all countable subsets of $\mathbb{R}$, and $\mu$ the Lebesgue measure. But $\mu_*(X\setminus B)=+\infty$ for all $B\in\mathcal{B}$.
 A: Let $\mathcal{L}=\{f: \Bbb R \rightarrow \Bbb R : f \text{ is continuous and } \textrm{supp} f \subseteq [0,1]\}$.
(In other words, $f \in \mathcal{L}$ if and only if $f$ is defined on $X=\Bbb R$, $f$ is continuous and $f$ is zero outside $[0,1]$).
It is easy to see that $\mathcal{L}$ is a Stone vector lattice.  Let us define the function $I$ from $\mathcal{L}$ into $\Bbb R$ by $I(f) = \int f \, d\lambda$, where $\lambda$ is the Lebesgue measure. It is easy to see that $I$ is a pre-integral.
Now, let $\mathcal{B}$ be the smallest $\sigma$-ring for which all functions in $\mathcal{L}$ are measurable. It is not difficult to prove that $\mathcal{B}= \{ B \subseteq \Bbb R: B \text{ is a Borel subset of } [0,1]\}$. There is an unique measure $\mu$ defined on $\mathcal{B}$, such that $I(f) = \int f \, d\mu$. Namely, $\mu$ is the Lebesgue measure on the Borel subsets of $[0,1]$.
Now let $\mathcal{A}$ be the smallest $\sigma$-algebra  for which all functions in $\mathcal{L}$ are measurable. It is not difficult to prove that $\mathcal{A} = \mathcal{B} \cup \{B \cup (\Bbb R \setminus [0,1]): B \in \mathcal{B}\} $.
Note that, for all $c \in \Bbb R$, the set function defined on $\mathcal{A}$ by
$\mu_c(B) = \mu(B)$ and $\mu_c(B \cup (\Bbb R \setminus [0,1])) = \mu(B)+c$,  for all $B \in \mathcal{B}$, is a measure and it is an extension of $\mu$ to $\mathcal{A}$. Moreover, for all $c \in \Bbb R$,   $I(f) = \int f \, d\mu_c$. So we see that the extension of $\mu$ to $\mathcal{A}$ is not uniquely determined.
A: A $\sigma$-ring is a collection $\mathcal{R}$ of sets, with $\emptyset\in\mathcal{R}$, such that $A\setminus B\in\mathcal{R}$ for any $A\in\mathcal{R}$ and $B\in\mathcal{R}$ , and such that $\bigcup_{j\geq1}A_j\in\mathcal{R}$ whenever $A_j\in\mathcal{R}$ for $j=1,2,\ldots$. So any $\sigma$-algebra is a $\sigma$-ring. Conversely, a $\sigma$-ring $\mathcal{R}$ of subsets of a set $X$ is a $\sigma$-algebra in $X$ if and only if $X\in\mathcal{R}$. For example, the set of all countable subsets of $\mathbb{R}$ is a $\sigma$-ring which is not a $\sigma$-algebra. If $f$ is a real-valued function on a set $X$, and $\mathcal{R}$ is a $\sigma$-ring of subsets of $X$, then $f$ is said to be measurable for $\mathcal{R}$ iff $f^{-1}(B)\in\mathcal{R}$ for any Borel set $B\subset\mathbb{R}$ not containing 0. (If this is true for general Borel sets, then $f^{-1}(\mathbb{R})=X\in\mathcal{R}$ implies $\mathcal{R}$ is a $\sigma$-algebra).
For any $c\in\mathbb{R}$, define the function $f_c$ on $X=(-1,1)$ by
\begin{equation}
f_c = \begin{cases}
c, & x>0 \\
0, & x\leq0.
\end{cases}
\end{equation}
Define $\mathcal{L}:=\{f_c|c\in\mathbb{R}\}$. Obviously $\mathcal{L}$ is a Stone vector lattice.
\begin{equation}
f_c^{-1}((1,\infty)) = \begin{cases}
\emptyset, & c\leq1 \\
(0,1), & c>1.
\end{cases}
\end{equation}
\begin{equation}
f_c^{-1}((-\infty,-1)) = \begin{cases}
\emptyset, & c\geq-1 \\
(0,1), & c<-1.
\end{cases}
\end{equation}
Thus the smallest $\sigma$-ring for which each $f_c$ is measurable is $\mathcal{B}=\{\emptyset,(0,1)\}$.
\begin{equation}
I(f_c)=\int^1_{-1}f_cd\mu=\int^1_0f_cd\mu=c\int^1_0d\mu=c\mu((0,1)).
\end{equation}
Thus $\mu$ is determined by $\mu(\emptyset)=0$ and $\mu((0,1))=I(f_1)$.
The smallest $\sigma$-algebra generated by $\mathcal{B}$ is $\mathcal{A}=\{\emptyset,(0,1),(-1,1),(-1,0]\}$. The extension $\bar{\mu}$ of $\mu$ to \$\mathcal{A}$ is nonunique. The following are two examples.
(1) $\bar{\mu}((-1,0])=1,\bar{\mu}((-1,1))=1+I(f_1)$.
(2) $\bar{\mu}((-1,0])=2,\bar{\mu}((-1,1))=2+I(f_1)$.
