# $\sigma$-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.

-
When $X$ is $\sigma$-compact you can prove that the answer is yes by making use of the properties of perfect measures. The answer is no for the general case though. You would need the addition criterion that any set $E$ is in the sigma algebra whenever $E\cap K$ is in it for all compact $K$. –  George Lowther Sep 4 '11 at 22:34
@George Care to flesh that out into a full answer? –  Byron Schmuland Sep 6 '11 at 22:50
@Byron: Yes, I'll write it up now –  George Lowther Sep 7 '11 at 0:36
@Richard: Actually, the second part of (e) is unnecessary, as it follows from (d) and the first pat of (e). –  George Lowther Sep 9 '11 at 1:11

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).

-
This is fantastic. I will first digest this, then wait a decent interval and award the bounty. P.S. Isn't is very late where you are? Sorry for keeping you up all hours! –  Byron Schmuland Sep 7 '11 at 2:36
Yes, it was late, and its late again so this will have to wait until tomorrow. I will come back and finish this off though... –  George Lowther Sep 8 '11 at 1:12
No problem, I appreciate your help. I'm slowly working through the details... –  Byron Schmuland Sep 8 '11 at 1:14
In your example $S\times \{0\}\notin\mathfrak{N}$ so $\mathfrak{N}$ doesn't contain all Borel sets. This example doesn't quite match the original problem. –  Byron Schmuland Sep 9 '11 at 17:10
1. Yes, that's the definition of support en.wikipedia.org/wiki/Support_of_a_measure. 2. It's only a contradiction if I is not countable. –  George Lowther Sep 25 '11 at 0:03