$\sigma$-algebra in Riesz representation theorem Let $X$ be a locally compact Hausdorff space and $I$ - a positive linear functional on $C_c(X)$. Then according to the Riesz representation theorem
there exist a $\sigma$-algebra $\mathfrak{M}$ in $X$ and measure $\mu$ on $\mathfrak{M}$ such that $I(f)=\int_X f d \mu$ for $f \in C_c(X)$.
$\mathfrak{M}$ and $\mu$ have the following properties:
(a) $\mathfrak{M}$ contains all Borel sets,
(b) $\mu(V)=\sup \{I(f): f \in C_c(X), 0\leq f \leq 1, \operatorname{supp} f \subset V\}$ for each open $V$, 
(c) $\mu(K) < \infty$  for compact $K$,
(d) $\mu(E)=\inf \{\mu(V): E \subset V, \ V \mbox{ open}\}$ for each $E \in \mathfrak{M}$, 
(e) $\mu(E)=\sup \{\mu(K): K \subset E, \ K \mbox{ compact} \}$ for each open $E$ and for each $E\in \mathfrak{M}$ such that $\mu(E)< \infty$,
(f) $\mu$ is a complete measure on $\mathfrak{M}$. 
Let a $\sigma$-algebra $\mathfrak{N}$ in $X$ and and a measure  $\nu$ on $\mathfrak{N}$ satisfy (a)-(f).
Is it true that $\mathfrak{M}=\mathfrak{N}$?. I think it is easy when $X$ is $\sigma$-compact. In general I don't know.
Thanks.
 A: In general, it is not the case that $\mathfrak{M}$ has to equal $\mathfrak{N}$. However, it is true that $\mathfrak{M}=\mathfrak{N}$ when $X$ is $\sigma$-compact or, more generally, when $\mu$ is $\sigma$-finite.
As a counterexample, consider the case where $X=S\times\mathbb{R}$ for an uncountable discrete space $S$ and with the usual topology on $\mathbb{R}$. We can define $I\colon C_c(X)\to\mathbb{R}$ by
$$
I(f)=\sum_{s\in S}\int_\mathbb{R}f(s,x)\,dx.
$$
Define $\pi_s\colon\mathbb{R}\to X$ by $\pi_s(x)=(s,x)$, and define the following sigma algebras.


*

*$\mathfrak{M}_1$ is the $E\subseteq X$ such that $\pi_s^{-1}(E)\subseteq\mathbb{R}$ is a Lebesgue measurable subset of $\mathbb{R}$ for all $s\in S$ and is Borel measurable for all but countably many $s$. 

*$\mathfrak{M}_2$ is the set of $E\subseteq X$ such that $\pi_s^{-1}(E)$ is Lebesgue measurable for all $s\in S$.


Using $\lambda$ for the Lebesgue measure on $\mathbb{R}$, we can define the measure $\mu$ on $(X,\mathfrak{M}_2)$ by
$$
\mu(E)=\begin{cases}
\sum_{s\in S}\lambda(\pi^{-1}_s(E)),&\textrm{ if }\{s\in S\colon\pi_s^{-1}(E)\not=\emptyset\}\textrm{ is countable} ,\\
\infty,&\textrm{otherwise}.
\end{cases}
$$
It can be seen that taking $\mathfrak{M}=\mathfrak{M}_2$ satisfies all of the required properties, as does taking $\mathfrak{N}=\mathfrak{M}_1$ and $\nu=\mu\vert_\mathfrak{N}$. Note that the sets $E\in\mathfrak{M}_2$ of zero measure must have $\pi^{-1}_s(E)=\emptyset$ for all but countably many $s$, so $(X,\mathfrak{N},\nu)$ is complete. However, for any Lebesgue measurable but non-Borel set $E\subset\mathbb{R}$, then $S\times E$ is in $\mathfrak{M}$ but is not in $\mathfrak{N}$, so these $\sigma$-algebras are different.

Now for the general case of a locally compact set $X$. Given a positive linear functional $I\colon C_c(X)\to\mathbb{R}$, you can define the following $\sigma$-algebras and measures,


*

*$\mathfrak{M}_0$ is the Borel $\sigma$-algebra on $X$, and $\mu_0$ is the unique measure on $(X,\mathfrak{M}_0)$ satisfying the required properties (other than completeness, (f)).

*$(X,\mathfrak{M}_1,\mu_1)$ is the completion of $(X,\mathfrak{M}_0,\mu_0)$.

*$\mathfrak{M}_2=\{E\subseteq X\colon E\cap K\in\mathfrak{M}_1{\rm\ for\ all\ compact\ }K\subseteq X\}$ and $\mu_2$ is the unique extension of $\mu_0$ to $\mathfrak{M}_2$.


These measures are indeed uniquely defined, and we have the following properties (I'm using $\mu^c$ to denote the continuous part of a measure $\mu$. That is, $\mu$ after its atoms have been subtracted out).


*

*For a $\sigma$-algebra $\mathfrak{M}$ on $X$, there exists a measure $\mu$ satisfying the required properties (other than, possibly completeness (f)) if and only if $\mathfrak{M}_0\subseteq\mathfrak{M}\subseteq\mathfrak{M}_2$.

*The measure $\mu$ is determined uniquely by $\mu=\mu_2\vert_\mathfrak{M}$.

*The measure space $(X,\mathfrak{M},\mu)$ just defined is complete if and only if $\mathfrak{M}_1\subseteq\mathfrak{M}\subseteq\mathfrak{M}_2$.

*$\mu^0_c$ is $\sigma$-finite  ⇔  $\mu^0_c(X\setminus U)=0$ for some (open) $\sigma$-compact $U\subseteq X$  ⇒  $\mathfrak{M}_1=\mathfrak{M}_2$.

*$\mathfrak{M}_1=\mathfrak{M}_2$  ⇒ there is an open $\sigma$-compact $U\subseteq X$ such that every compact subset of $X\setminus U$ has zero $\mu^0_c$ measure.


We can prove these statements.
First, the measure $\mu_0$ on $(X,\mathfrak{M}_0)$ satisfying the required properties exists and is unique by the Riesz representation theorem, so I don't need to prove that. Next, $\mathfrak{M}_1$ and $\mu_1$ are uniquely defined because all measure spaces have a unique completion. That $\mathfrak{M}_2$ is a $\sigma$-algebra is just a matter that of checking that it is closed under countable unions and complements.
Furthermore, $\mu_2$ is uniquely defined by the condition
$$
\mu_2(E)=\inf\left\{\mu_0(S)\colon S\supseteq E{\rm\ is\ open}\right\}.
$$
That $\mu_2$ is countably additive follows from the fact that it agrees with $\mu_1$ on $\mathfrak{M}_1$ and that every $E\in\mathfrak{M}_2$ with $\mu_2(E) < \infty$ is also in $\mathfrak{M}_1$. To see this, consider that if $\mu_2(E) < \infty$ then we have an open set $S$ containing $E$ with $\mu_0(S) < \infty$. Then, by condition (e), there are compact sets $K_n\subseteq S$ with $\mu_0(K_n)\to\mu_0(S)$, in which case $S_0=S\setminus\bigcup_n K_n$ is $\mu_0$-null so, by completeness, $S_0\cap E\in\mathfrak{M}_1$. Also, $K_n\cap E\in\mathfrak{M}_1$ by definition. So, $E$ is the union of sets $S_0\cap E$ and $K_n\cap E$ and must itself be in $\mathfrak{M}_1$.
Next, suppose that $\mu$ and $\mathfrak{M}$ satisfy the required properties (other than, possibly, completeness (f)). Then, we have $\mathfrak{M}_0\subseteq\mathfrak{M}$ by (a) and agrees with $\mu$ on all Borel sets by uniqueness of the Riesz representation. Suppose that $E\in\mathfrak{M}$ is contained in a compact set $K$. Then, by (c,d,e), $\mu(E) < \infty$ and there exist open sets $S_n$ containing $E$ with $\mu(S_n)\to\mu(E)$. Then the intersection $S=\bigcap_n S_n$ is Borel, contains $E$ and $\mu(S)=\mu(E)$. Applying the same argument to $K\setminus E$ implies the existence of a Borel set $S^\prime$ containing $K\setminus E$ such that $\mu(S^\prime)=\mu(K\setminus E)$. It follows that $F=S\setminus(K\setminus S^\prime)$ has zero $\mu$-measure and is Borel, so, by completeness, $F^\prime=F\setminus E=S\setminus E$ is in $\mathfrak{M}_1$ and, therefore, $E=S\setminus F^\prime$ is also in $\mathfrak{M}_1$. Now for arbitrary $E\in\mathfrak{M}$ and compact $K$, applying the above argument to $E\cap K\subseteq K$ shows that $E\cap K\in\mathfrak{M}_1$. By definition, this means that $E\in\mathfrak{M}_2$ and $\mathfrak{M}\subseteq{M}_2$ as required. Then, $\mu=\mu_2\vert_\mathfrak{M}$ by (d) and the fact that it agrees with $\mu_0$ on open sets.
Conversely, suppose that $\mathfrak{M}_0\subseteq\mathfrak{M}\subseteq\mathfrak{M}_2$ and $\mu=\mu_2\vert_{\mathfrak{M}}$. Properties (a,b,c) follow from the fact that $\mu_0=\mu\vert_{\mathfrak{M}_0}$ satisfies these. Properties (d,e) follow from the fact that it is the restriction of $\mu_2$ which satisfies these properties.
Now, suppose that $(X,\mathfrak{M},\mu)$ is complete. By the definition of $\mu_1$ and $\mathfrak{M}_1$, any $E\in\mathfrak{M}_1$ can be written as $E=A\cup B$ for Borel $B$ and $A$ contained in a Borel set of zero measure. So, by completeness, $E\in\mathfrak{M}$ and we have shown that $\mathfrak{M}_1\subseteq\mathfrak{M}$.
Conversely, suppose that $\mathfrak{M}$ contains $\mathfrak{M}_1$ and consider $E\in\mathfrak{M}$ with $\mu(E)=0$. Then, $\mu_2(E)=0$. In particular, this is finite, so $E\in\mathfrak{M}_1$ as argued above. By completeness, any $F\subseteq E$ is in $\mathfrak{M}_1$ and, hence, is in $\mathfrak{M}$. So, $(X,\mathfrak{M},\mu)$ is complete.
Only the final statements giving (separately) necessary and sufficient conditions for $\mathfrak{M}_1=\mathfrak{M}_2$ remain. The set $A=\{x\in X\colon \mu_0(\{x\}) > 0\}$ is Borel measurable. In fact, every subset $S\subseteq A$ is Borel. Note that the set of $x\in S$ such that $\mu_0(\{x\})\ge1/n$ can only have a finite intersection with any compact set (by (c)) so must be closed and, therefore, $S$ is a union of countably many closed sets and hence is Borel. So, a set $S\subseteq X$ automatically has $S\cap A\in\mathfrak{M}_0$, and $S\in\mathfrak{M}_1$ if and only if $S\setminus A\in\mathfrak{M}_1$. So, the definition of $\mathfrak{M}_1$ (and $\mathfrak{M}_2$) only depends on the continuous part of $\mu_0$, defined by $\mu_0^c(E)=\mu_0(E\setminus A)$.
Now, suppose that $\mu_0^c(X\setminus U)=0$ for a $\sigma$-compact set $U=\bigcup_{n\ge1}K_n$ (compact $K_n$). Setting $K_0=X\setminus U$ then $\mu_0^c(K_n) < \infty$ for all $n\ge0$ and $X=\bigcup_{n\ge0}K_n$. So $\mu_0^c$ is $\sigma$-finite. Next, suppose that $\mu_0^c$ is $\sigma$-finite, so that $X=\bigcup_{n\ge1}E_n$ for Borel $E_n$ with $\mu^c_0(E_n) < \infty$. By (e), each $E_n$ is the union of countably many compact sets and a set of zero $\mu^c_0$-measure. So, taking the union over $n$, $X$ is also of the form $E\cup\bigcup_{n\ge1}K_n$ for compact $K_n$ and $\mu^c_0(E)=0$. Then $U=\bigcup_nK_n$ is $\sigma$-compact and $\mu^c_0(X\setminus U)=0$. We can assume that $U$ is also open because, in a locally compact space, every compact set (and hence every $\sigma$-compact set) is contained in an open $\sigma$-compact set.
Then, for any $S\in\mathfrak{M}_2$, we have $S\cap K_n\in\mathfrak{M}_1$ by definition and $S\cap E\in\mathfrak{M}_1$ as it has zero $\mu^c_0$-measure. Therefore, being the union of $S\cap E$ and $S\cap K_n$, $S$ is in $\mathfrak{M}_1$. This shows that $\mathfrak{M}_1=\mathfrak{M}_2$.
Let's now prove the final statement. Using Zorn's lemma, choose a maximal collection $\{K_i\colon i\in I\}$ of pairwise disjoint compact sets such that $\mu^c_0(K_i) > 0$. We can find open $\sigma$-compact $U_i$ containing $K_i$ with $\mu^c_0(U_i) < \infty$ (by (d)), and set $U=\bigcup_iU_i$. By maximality of the collection $\{K_i\}$, every compact subset $K$ of $X\setminus U$ has zero $\mu_0^c$-measure. Let's show that $\mathfrak{M}_1=\mathfrak{M}_2$ implies that $I$ is countable, so that $U$ is $\sigma$-compact.
Let $\tilde K_i$ be the support of $\mu_0^c$ restricted to $K_i$ (i.e., the smallest closed subset of $K_i$ with $\mu_0^c(K_i)=\mu_0^c(\tilde K_i)$). This means that $\mu_0^c(S\cap \tilde K_i) > 0$ for any open set $S$ with $S\cap \tilde K_i\not=\emptyset$. So, every open set with finite $\mu_0^c$-measure can only intersect countably many of the $\tilde K_i$. By (d), this also means that any set with zero $\mu_0^c$-measure can only intersect countably many of the $\tilde K_i$. As every $S\in\mathfrak{M}_1$ is the union of a Borel set and a set with zero measure, it must satisfy $S\cap \tilde K_i\in\mathfrak{M}_0$ for all but countably many $i$. If we can find sets $S_i\subseteq \tilde K_i$ with $S_i\in\mathfrak{M}_1\setminus\mathfrak{M}_0$ then $S=\bigcup_i S_i$ will be in $\mathfrak{M}_2$. As $\mathfrak{M}_1=\mathfrak{M}_2$ then this means that $S_i=S\cap \tilde K_i\in\mathfrak{M}_0$ for all but countably many $i$, so $I$ itself is countable. This is the point where we need to assume that $\mu^c_0$ is a continuous measure, so that it was necessary so subtract out the atoms. The existence of $S_i$ follows from the fact that any finite atomless measure on the Borel $\sigma$-algebra of a compact Hausdorff space cannot be complete (I'll state this fact without proof though - left as an exercise - because this answer is getting way too long).
