Let $(X,T)$ be a Hausdorff topological space. Let $C\_b(X)$ be its algebra of continuous bounded functions. Let $T'$ be the initial topology on $X$ given by $C\_b(X)$. It is known that $T=T'$ if and only if $(X,T)$ is completely regular. If $\mathcal B(X,T)$ and $\mathcal B(X,T')$ are the Borel $\sigma$-algebras generated by the two topologies, we have $\mathcal B(X,T') \subseteq \mathcal B(X,T)$ because $T' \subseteq T$.



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> Is it true that $\mathcal B(X,T') = \mathcal B(X,T)$?
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(Motivation: I am trying to pull a Borel regular measure from $(X,T')$ back to $(X,T)$. I could use Henry's extension theorem or follow Bourbaki and extend an additive set-function of compact subsets, but my secret hope is that in this very convenient setting the two $\sigma$-algebras coincide, so I don't need to resort to heavy artillery (in particular, Henry's extension theorem requires Zorn's lemma).)



Do a Hausdorff space and its associated completely regular space have the same Borel subsets?
There are Hausdorff spaces all of whose real-valued functions are constant. A classical example (of a countable space) was given by Uryshon ([*Über die Mächtigkeit der zusammenhängenden Mengen*, Math.Annalen, 1925](https://eudml.org/doc/159111)). This implies that the initial topology on $X$ induced by $C(X)$ is trivial, and so is its Borel sigma-algebra.


