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).
I think the first version is more general. It clearly has weaker assumptions, as you noted. As for the conclusions, they both yield measures that correspond to the functional in exactly the same way, and are unique, which are described as following:
- Regular and Borel.
- Locally finite and positive.
In the first case, I think we can safely assume that "positive" is implicit (because it almost always is, if not stated otherwise, and because otherwise the resulting functional would not be positive!). Similarly, if the measure was not locally finite, then I believe that the resulting „functional” would not be bounded (or even finite), so we can also assume that implicitly, so the first version does not provide anything more with the assumptions of the second one.
As for the second case, I believe that there is also implicit assumptions of regularity and Borelness, because otherwise the measure would be unlikely to be unique, because by the first version we already have a Borel measure, but a Borel measure can always (except for trivial cases, e.g. when every set is Borel) be extended to a larger $\sigma$-field, defying uniqueness (remember that the integral of a continuous functions depends only on the Borel part of the measure), and any finite Borel measure on a metric space is already regular, and this should extend to $\sigma$-finite in the obvious way. (Note that locally finite measure on a $\sigma$-compact space is $\sigma$-finite.)
Summing it all up, unless I made some large mistakes:
- The first theorem is more general, because it applies to more cases.
- With the assumptions of the second theorem, they are equally strong.
Best Answer
It is true in general. By the reference provided in the comment above by copper.hat we can write $Tu=\int u d\nu$ for some finitely additive measure $\nu$.
Now let $d\lambda =hd\mu$. Then $\lambda$ is a finite positve measure and $|\int u d\nu|=|Tu| \leq \int |u|d\lambda$. Taking $u$ to be characteristic function of a set $A$ we get $|\nu (A)| \leq \lambda (A)$. This forces $\nu$ to to countably additive and we also get $\nu << \lambda$. Now a simple application of Radon Nikodym Theorem finishes the proof.