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Let $X$ be a smooth oriented manifold of positive dimension. Let $\Lambda$ be a positive linear functional on $C_c(X)$.

By Riesz Representation theorem there exists a unique measure $\mu$ on a $\sigma$-algebra $\mathfrak B$, which sarisfies $I(f)=\int_X f d \mu$ for all $f \in C_c(X)$.

In addition, $\mathfrak B,\mu$ satisfy:

(a) $\mathfrak{B}$ 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 $E \in \mathfrak{B}$,

(e) $\mu(E)=\sup \{\mu(K): K \subset E, \ K \mbox{ compact} \}$ for each open $E$ and for each $E\in \mathfrak{B}$ such that $\mu(E)< \infty$,

(f) $\mu$ is a complete measure on $\mathfrak{B}$.

Is it true that $\mathfrak B(X)$ is not complete w.r.t $\mu$?

Note: It turns out that $\mathfrak B$ is unique. Thus, an equivalent question is whether $\mathfrak B= \mathfrak B(X)$. (Since $\mathfrak B(X)$ satisfies all the conditions except maybe $(f)$).

I guess that if $\Lambda$ is such that the induced measure $\mu$ is absolutely continuous w.r.t some Riemannian measure* (restricted to $\mathfrak B(X)$), then $\mathfrak B(X)$ won't be complete.

*By a Riemannian measure, I mean any measure $\tilde \mu$ which we get from Riesz theorem, by taking $\Lambda(f)=\int_X f Vol_{\mathfrak g}$, where $Vol_{\mathfrak g}$ is the Riemannian volume form of some Riemannian metric $\mathfrak g$ on $X$.

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Asaf Shachar
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1 Answers1

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Consider a measure space $(X, \mathcal{B},\mu)$ and define the following topology on $X$

A set $F\in X$ is closed iff $F=X$ or $\mu(F)=0$.

This generate a topology on $X$ because $(X, \mathcal{B},\mu)$ is complete. In the particular case where $X=\mathbb{R}$, $\mathcal{B}$=Lebesgue's $\sigma-$ algebra and $\mu$=Lebesgue measure, we have a topology on $\mathbb{R}$ such that every closed set have measure zero by definition. The Borel $\sigma$-algebra is contained in the Lebesgue $sigma$-algebra and is complete, because if $\mu(F)=0$ and $\tilde{F}\subset F$ so $\tilde{F}$ is Lebesgue measurable and therefore $\mu(\tilde{F})=0$. This implies that $\tilde{F}$ is closed and therefore a borel set in this new topology.

Euler88 ...
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  • Thanks. However, I am not sure this example works, since the Lebesgue measure is obtained by the Riesz theorem after we start with the standard topology on $\mathbb{R}$ and take $\Lambda(f)=\int_{\mathbb{R}} f$ be the Riemann integral of $f$ for $f \in C_c(X)$. Note that $C_c(X)$, by definition, depends on the topology on $X$ which is given to us. (Anyway, I am actually interested in more regular cases, so I have edited the question a bit) – Asaf Shachar Aug 23 '16 at 16:25