Let $\Omega$ be a measurable space equipped with a $\sigma$-ideal $\mathcal{N}$ (though of as the "null sets"). Define the compact Hausdorff space
$$ \tilde{\Omega} := \mathrm{Spec}(L^\infty(\Omega)), $$
where $\mathrm{Spec}$ denotes the Gelfand spectrum. In [this question](https://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu), Terry Tao mentions the following (universal?) property of $\tilde{\Omega}$: "any [modulo-null-sets-equivalence class of?] Baire-measurable map[s] from $\Omega$ to a compact Hausdorff space $K$ can be uniquely identified with a continuous map from $\tilde{\Omega}$ to K." I haven't been able to find a precise statement or proof in the given references (or elsewhere, where would I have to look precisely?). If $K$ is a compact subset of the real numbers, then this is a consequence of Gelfand duality. How would one proceed in the general case?


**EDIT:** Let's ask a (maybe) simpler question. Assume that $\Omega$ is itself a compact Hausdorff space equipped with a Radon probability measure. If necessary, suppose also that $\Omega$ and $K$ are metrizable. Then, via Gelfand duality, we obtain a continuous map
$f: \tilde{\Omega} \to \Omega $. $\tilde{\Omega}$ also inherits the probability measure from $\Omega$ and $f$ is measure preserving. Now the question becomes: Is $f$ a mod $0$ isomorphism of probability spaces? (If this were the case, then a version of the cited segment would be true.)



Representing measurable map to compact space as a continuous map
I think I have found a precise formulation and proof of the cited statement.


Let's suppose for simplicity that $(\Omega, \Sigma, \mathbb{P})$ is a probability space. Define the set of $K$-valued random variables up to almost sure equality (with $K$ some compact space),
$$ \mathrm{RV}(\Omega, K) := \{X: \Omega \to K \mid X \text{ Baire measurable} \}/(=\_{a.e.}).$$


With $\tilde{\Omega}$ as in the question, the desired statement is:


**Theorem.** Let $K$ be a compact metrizable space. Then the ("Gelfand-duality-induced") map
$$ \Psi\_K: \mathrm{RV}(\Omega, K) \to C(\tilde{\Omega}, K), \;\;\; X \mapsto (\tilde{\Omega}\ni \omega \mapsto \Gamma^{-1}((C\_b(K) \ni f \mapsto \omega(f(X))), $$
is a bijection, where
$$ \Gamma^{-1}: \mathrm{Spec}(C\_b(K)) \to K $$
is the inverse Gelfand transform.


**Proof.** As noted already in the question, the claim follows from Gelfand duality when $K$ is a compact subset of $\mathbb{C}$. Moreover, for any compact space $L$, we have that
$$ \mathrm{RV}(\Omega, L^\mathbb{N}) \to \mathrm{RV}(\Omega, L)^\mathbb{N}, \;\;\; X \mapsto (\pi\_n(X))\_{n\in \mathbb{N}}, $$
is bijective. (For injectivity, we use that we consider a *countable* product $L^\mathbb{N}$, while for surjectivity, we use the Stone-Weierstrass theorem and compactness of $L$.) This yields the claim for the case of $K = L^\mathbb{N}$ with $L\subseteq \mathbb{C}$ (cf. the approach suggested by Christian Remling).
Finally, for the general case note that since $K$ is compact and metrizable, it admits an embedding,
$$ \iota: K \to [0,1]^\mathbb{N}, $$
into the Tychonov cube (Tychonov's embedding theorem), as well as a surjection (epimorphism in the category $\mathsf{CompHaus}$ of compact Hausdorff spaces),
$$ \pi: \{0,1\}^\mathbb{N}\to K,$$
from Cantor space (Hausdorff-Alexandrov theorem). This gives us the following commutative diagram (in $\mathsf{Set}$):
$\require{AMScd}$
\begin{CD}
 \mathrm{RV}(\Omega, \{0,1\}^\mathbb{N}) @>>> \mathrm{RV}(\Omega, K) @>>> \mathrm{RV}(\Omega, [0,1]^\mathbb{N}) \\
 @V{\Psi\_{\{0,1\}^\mathbb{N}}}VV @V{\Psi\_K}VV @V{\Psi\_{[0,1]^\mathbb{N}}}VV \\
 C(\tilde{\Omega}, \{0,1\}^\mathbb{N}) @>{C(\tilde{\Omega}, \pi)}>> C(\tilde{\Omega}, K) @>{C(\tilde{\Omega},\,\iota)}>> C(\tilde{\Omega},\, [0,1]^\mathbb{N})
\end{CD}
Since $\tilde{\Omega}$ is extremely disconnected, it is projective in $\mathsf{CompHaus}$ (Gleason's theorem), so $C(\tilde{\Omega}, \pi)$ is surjective. With this, the square on the left gives us surjectivity of $\Psi\_K$, while the square on the right yields injectivity.


