Do a Hausdorff space and its associated completely regular space have the same Borel subsets?

Question

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).)

Answer 1

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). This implies that the initial topology on $X$ induced by $C(X)$ is trivial, and so is its Borel sigma-algebra.

Answer 2

The Borel $\sigma$-algebras may be unequal.

Let $\omega_1$ be the least uncountable ordinal. Let $L_0=[0,\omega_1)\times I/\sim$ be the "long line", where $[0,\omega_1)$ is the space of ordinals less than $\omega_1$ with the order topology, $I=[0,1]$ is the unit interval, and the identifications are: $(\alpha,1)\sim(\alpha+1,0)$. Let $L'=L_0\cup\{*\}$ be the one point compactification of $L_0$.

Such $L'$ is a connected, linearly ordered compact space. Note that the Stone-Cech compactification of $L_0$ is again $L'$. The subspace $\omega_1=[0,\omega_1)\times\{0\}\subseteq L'$ is not Borel. Also for every closed subset $A\subseteq L_0$ that is not bounded above, every continuous function $f:A\to\mathbb{R}$ is eventually constant.

Let $L$ be a refinement of $L'$: a subset $U$ is open in $L$ if it is of the form $U=V\cup (W\setminus\omega_1)$ where $V,W$ are open in $L'$. Here $\omega_1$ is a closed hence a Borel subset of $L$. But if $f:L\to\mathbb{R}$ is continuous then it has to be eventually constant then it is continuous as a function on $L'$. Thus $C_b(L')=C_b(L)$, but $\mathcal{B}(L')\subsetneq\mathcal{B}(L)$.