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

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).)

• I am wondering whether the following space work or no? consider the topology $T$ on $\mathbb{R}$ consist of all $U\setminus Z$ where $U$ is a Euclidean open set and $Z$ is of zero Lebesgue measure. The T_Borel set is the space of all Lebesgue measurable sets. So is it true that $T'$ is the standard Euclidean topology? May 6, 2018 at 14:57
• Do you mean to say that $T$ is the topology generated by 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.) May 6, 2018 at 15:04
• No I mean the topology consist of (not generated by) all such sets. It can be shown that it is a topology. But after I post the previous comment, I realized that the space is completely regular so it does not work for your question.Sorry! but I was hopping that T' would be the standard Euclidean topology. If it were the case, it would be a counter example to your question. right? May 6, 2018 at 15:09
• This link contain a proof that the collection is really a topology mathforum.org/kb/message.jspa?messageID=5961793 May 6, 2018 at 15:22

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.
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 $\mathcal{B}(L')\subsetneq\mathcal{B}(L)$.