TITLE: On isomorphic measurable spaces
QUESTION [5 upvotes]: Let $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ be measurable spaces. Suppose furthermore that $X$ is a topological space, with topology $\tau$, and that $\mathcal{A}$ is the $\sigma$-algebra generated by $\tau$.
Now suppose that $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ are isomorphic, i.e. there exists a bijection $f: X \rightarrow Y$ such that both $f$ and $f^{-1}$ are measurable. We can define a topology $\tau'$ on $Y$ by
$$\tau' := \{ f(V): V \in \tau \}.$$
My question is the following : is $\mathcal{B}$ necessarily generated by the topology $\tau'$?
It is easy to see that $\mathcal{B}$ contains the Borel $\sigma$-algebra generated by $\tau'$, but can it be strictly larger?
Thank you,
Malik

REPLY [5 votes]: No, it can't be larger. The family $\mathcal{A}'=\{f^{-1}(B):B\in\sigma(\tau')\}$ is a $\sigma$-algebra on $X$ that contains $\tau$ and hence also $\sigma(\tau)=\mathcal{A}$. Hence $\{f(A):A\in\mathcal{A}'\}\supseteq\mathcal{B}$. But by construction, $\{f(A):A\in\mathcal{A}'\}=\sigma(\tau')$ and therefore $\mathcal{B}\subseteq\sigma(\tau')$.