Generalize Riesz–Markov–Kakutani representation theorem from $C_c(X)$ to $C_0(X)$

compactnessgeneral-topologymeasure-theory

I'm reading about Riesz–Markov–Kakutani representation theorem from this page.

Let

  • $X$ be a locally compact Hausdorff space.
  • $C_c(X)$ the space of continuous compactly supported complex-valued functions on $X$.
  • $C_0(X)$ the space of continuous functions on $X$ which vanish at infinity.

A non-negative Borel measure $\mu$ on $X$ is called regular if and only if

  • $\mu(K)<\infty$ for every compact $K$;
  • For every Borel set $E$, $\mu(E)=\inf \{\mu(U) \mid E \subseteq U, U \text { open}\}$.
  • The relation $\mu(E)=\sup \{\mu(K) \mid K \subseteq E, K \text { compact}\}$
    holds whenever $E$ is open, or when $E$ is Borel and $\mu(E)<\infty$.

A complex Borel measure $\mu$ on $X$ is called regular if the non-negative measure $|\mu|$ is regular as defined above.

  • Theorem 1: For any positive linear functional $\psi$ on $C_c(X)$, there is a unique regular non-negative Borel measure $\mu$ on $X$ such that
    $$
    \forall f \in C_{c}(X): \quad \psi(f)=\int_{X} f \mathrm d \mu.
    $$

  • Theorem 2: For any continuous linear functional $\psi$ on $C_0(X)$, there is a unique regular complex Borel measure $\mu$ on $\mathrm{X}$ such that
    $$
    \forall f \in C_{0}(X): \quad \psi(f)=\int_{X} f \mathrm d \mu
    $$
    The operator norm of $\psi$ is the total variation of $\mu$, i.e., $\|\psi\|=|\mu|(X)$. Finally, $\psi$ is positive if and only if $\mu$ is non-negative.

Then there is a paragraph

One can deduce Theorem 2 about linear functionals from the Theorem 1 about positive linear functionals by first showing that a bounded linear functional can be written as a finite linear combination of positive ones.

I know that

  • Let $X$ be a locally compact Hausdorff space. Then the space of continuous functions with compact support is dense in that of continuous functions vanishing at infinity w.r.t. $\| \cdot \|_\infty$. ref.

but could not figure out how to infer Theorem 2 from Theorem 1. Could you elaborate on this point?

Best Answer

If $\psi$ is a continuous linear functional on $C_0(X)$ its restriction to $C_c(X)$ is a continuous linear functional. Hence, there exits $\mu$ as in Theorem 1 with $\psi (f)=\int fd\mu$ for all $f \in C_c(X)$. All that remains it to show that the equation holds for all $f \in C_0(X)$. [The norm of $\psi$ is equal to the norm of its restriction to $C_c(X)$ since $C_c(X)$ is dense in $C_0(X)$]. If $f \in C_0(X)$ then there is a sequence $(f_n)$ in $C_c(X))$ with $f_n \to f$ uniformly. Since $\psi (f_n)=\int f_nd\mu$ for all $n$ we can take limit as $n \to \infty$ to finish the proof. [Recall that $|\int [f_n-f]d\mu|\leq \|f_n-f\|_{\infty} d|\mu|$].

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