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 $T' \subseteq T$.

Is it true that $\mathcal B(X,T') = \mathcal B(X,T)$?

(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).)

generatedby the subsets of the form $U \setminus Z$? Because it is not clear to me why the arbitrary union of such subsets would again be of this form. Also, what is the connection with my question? (I ask about the equality of two Borel $\sigma$-algebras, not two topologies.) $\endgroup$