Understanding Riesz-Markov representation theorem

general-topologymeasure-theory

Recently, I have come across Riesz representation theorem in this lecture note.

(Gaans) Let $X$ be a compact Hausdorff space and $E:=\mathcal C_b (X)$ the real vector space of all real-valued continuous bounded maps on $X$. Let $E'$ be the dual of $E$. Let $\varphi \in E'$ be positive and $\|\varphi\|_E = 1$. Then there exists a unique Borel probability measure $\mu$ on $X$ such that
$$
\varphi(f) = \int_X f \mathrm d \mu \quad \forall f \in E.
$$

(Rudin) Let

  • $X$ be a locally compact Hausdorff space,

  • $E:= \mathcal C_c (X)$ the complex vector space of all complex-valued continuous maps on $X$ with compact support, and

  • $\Lambda:E \to \mathbb C$ (not necessarily continuous) positive linear. Here $\Lambda$ is positive means $f \in E \text{ s.t. } f(X) \subset \mathbb R_{\ge 0}$ implies $\Lambda (f) \in \mathbb R_{\ge 0}$.

Then there exists a $\sigma$-algebra $\mathfrak{M}$ on $X$ which contains all Borel sets of $X$, and there exists a unique non-negative measure $\mu$ on $\mathfrak{M}$ such that

  1. $$\Lambda(f) = \int_X f \mathrm d \mu \quad \forall f \in E.$$

  2. $\mu(K) < \infty$ for every compact set $K \subseteq X$.

  3. $\mu(E) = \inf \{\mu(V) \mid E \subset V, V \text{ open}\}$ for all $E \in \mathfrak{M}$.

  4. $\mu(E) = \sup \{\mu(K) \mid K \subset E, K \text{ compact}\}$ for all open set $E \subseteq X$ and for all $E \in \mathfrak{M}$ with $\mu(E) <\infty$.

  5. If $E \in \mathfrak{M}$ such that $A \subset E$, and $\mu(E)=0$, then $A \in \mathfrak{M}$.

My understanding:

    1. means $\mu$ is finite on compact sets.
    1. means $\mu$ is outer regular on every $E \in \mathfrak{M}$.
    1. means $\mu$ is tight on every open sets and on every $E \in \mathfrak{M}$ with $\mu(E) <\infty$.
    1. means $\mu$ is complete.

Rudin's version is powerful in the sense that it applies to even unbounded linear maps.

We can recover Gaans's version as follows. The real vector space of all real-valued continuous bounded maps on $X$ can be seen as a complex vector subspace of all complex-valued continuous bounded maps on $X$. By Hahn-Banach theorem, we can extend $\varphi$ to the whole complex vector space of all complex-valued continuous bounded maps on $X$. Then we apply Rudin's version to get $\mu$ and then restrict $\mu$ to the Borel $\sigma$-algebra of $X$.

Could you confirm if my above understanding is correct?

Best Answer

Your argument is almost correct. However, that the extension to a complex-valued linear function can be represented by a unique measure satisfying these criteria doesn't directly imply that there exists a unique such measure representing the original functional. However, this is not a serious issue; Rudin's proof works for real linear functionals in his proof. More seriously, the existence of a unique tight Borel measure representing the functional is a weaker assumption than the existence of a unique Borel measure representing the functional. In general, it can be strictly weaker.

For general compact Hausdorff spaces, the Baire $\sigma$-algebra, the smallest $\sigma$-algebra making all continuous real-valued functions measurable, can be strictly smaller than the Borel $\sigma$-algebra. Clearly, one could prove a representation theorem in terms of measures on this coarser $\sigma$-algebra. Now, every Baire probability measure on compact Hausdorff space admits a unique extension to a tight Borel probability measure. However, there can exist other extensions to a Borel probability measure that are not tight; not every Borel probability measure on a compact Hausdorff space must be tight. In particular, the uniqueness claim in Gaan's version need not hold.

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