Inequivalent metrics can give rise to the same class of Borel sets I was going through "Convergence of Probability Measures" by Patrick Billingsley. In Section 1: I encountered the following problem:
Show that inequivalent metrics can give rise to the same class of Borel sets.
My idea is that the 2 metrics generate different topologies but the Sigma algebra generated by them is the same. However I don't know how to go about proceeding to prove this. I guess I need a convincing example.
My Background:
I read Topology from "Topology and modern analysis" by G.F Simmons, the Rudin texts and Billingsley "Probability and Measure". But this still boggles me.
My Searches: I searched for "non equivalent metrics" and "inequivalent metrics" getting 85 and 3 results respectively. But neither helpful nor relevant.
I would appreciate any useful hints, tips and even complete answers (preferably the first two). 
 A: An example is given by the real line, and the real line with the origin replaced by an isolated point.
A: Theorem. Given a separable completely metrisable (Polish) space $( X , \mathcal{T} )$, and any Borel $B \subseteq X$ you can define a new Polish topology $\mathcal{T}_B$ on $X$ which is finer than, and has the same Borel subsets as, the original topology, and in which $B$ is a clopen set.  
Metrics witnessing that $\mathcal{T}$ and $\mathcal{T}_B$ are Polish are clearly inequivalent (as long as $B$ is a non-clopen subsets of $X$).
The following outline comes from Kechris, Classical Descriptive Set Theory.

Claim 1. If $( X , \mathcal{T} )$ is Polish and $F \subseteq X$ is closed, then there is a Polish topology $\mathcal{T}_F$ on $X$ extending the original topology with the same Borel sets in which $F$ is clopen  
proof sketch. Consider the topology $\mathcal{T}_F$ on $X$ generated by the family $$\{ U \cap F : U \subseteq X \text{ is open} \} \cup \{ U \setminus F : U \subseteq X \text{ is closed} \}.$$  This is easily seen to be the topological sum of the subspace topologies on $F$ and $X \setminus F$, which are themselves Polish, and so the sum is as well.  It is easy to see that $\mathcal{T}$ and $\mathcal{T}_F$ have the same Borel sets. $\;$ $\Box$
Claim 2. Suppose $( X , \mathcal{T} )$ is a Polish space and $\langle \mathcal{T}_n \rangle$ is a sequence of Polish topologies on $X$ such that each $\mathcal{T}_n$ is finer than, and has the same Borel sets as, the original topology $\mathcal{T}$.  Then the topology $\mathcal{T}_\infty$ on $X$ generated by $\bigcup_n \mathcal{T}_n$ is Polish and has the same Borel sets as $\mathcal{T}$.  
proof sketch. The diagonal map $f : X \to X^{\mathbb{N}}$ ($f(x) = \langle x \rangle_{n \in \mathbb{N}}$) is clearly injective and has range a closed subset of the product space $\prod_{n \in \mathbb{N}} ( X , \mathcal{T}_n )$.  It is easily seen that the topology on $X$ induced by $f$ is $\mathcal{T}_\infty$, and is Polish since $f[X]$ is a Polish subspace of $\prod_{n \in \mathbb{N}} ( X , \mathcal{T}_n )$. As $\mathcal{T}_\infty$ is generated by a countable family of Borel subsets of $( X , \mathcal{T} )$ it follows that the Borel subsets of $( X , \mathcal{T}_\infty )$ coincide with the Borel subsets of $( X , \mathcal{T} )$. $\;$ $\Box$
From these two is follows that the family $\mathcal{S}$ of all subsets $B \subseteq X$ for which there is a Polish topology on $X$ extending the original topology but with the same Borel sets, and in which $B$ is clopen forms a $\sigma$-algebra containing all closed (open) subsets of $X$.  Thus the Theorem holds. $\;$ $\Box$


For an explicit example, given any closed $F \subseteq \mathbb{R}$ define a new metric $d$ on $\mathbb{R}$ by $$d ( x , y ) = \begin{cases}
| x - y |, &\text{if }x,y \in F \\
| x - y |, &\text{if }x,y \in \mathbb{R} \setminus F\\
1, &\text{if }| F \cap \{ x , y \} | = 1.
\end{cases}$$
A: "Show that inequivalent metrics can give rise to the same class of Borel sets."
any in-finite metric $d$ can be replaced by the metric $\frac{d}{d+1}$, which is finite and both these metrics generate the same topology and hence the same (Borel) sigma algebra
However if you are referring to problem $2$ in Billingley, which reads out "distinct topologies that give rise to the same Borel sets", then this problem is entirely different
what he wants to teach is the fact that Sigma algebra is $>>$ topology generated by a base set for topology.
Hence it might be true that two topologies dont match but their sigma algebra does match, take example of lower limit topology vs the usual topology on $\mathbb{R}$.
