Theorem 38.8.6 (Uniqueness of $L^{\infty}$).label Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_{X} \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^{\infty}(\mu; \complex) \to L^{\infty}(\nu; \complex)$ such that:
- (a)
$\Phi$ is a *-isomorphism.
- (b)
$\Phi|_{C(X; \complex)}$ is the identity.
then:
- (1)
$\mu$ and $\nu$ are equivalent.
- (2)
$L^{\infty}(\mu; \complex) = L^{\infty}(\nu; \complex)$.
- (3)
$\Phi$ is the identity map.
Proof, [Theorem 21.4, Zhu93]. As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^{\infty}(\mu; \complex)$.
(1): By Lemma 38.8.5, there exists $f \in L^{1}(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the Riesz Representation Theorem implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent.
(3): By Lemma 38.8.5, $\Phi$ is also weak*-continuous. By Goldstine’s Theorem, $C(X; \complex)$ is weak*-dense in $L^{\infty}(\mu; \complex)$ and $L^{\infty}(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^{\infty}(\mu; \complex)$.$\square$
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