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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

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up vote 4 down vote accepted

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 $\mathcal{B}\subseteq\sigma(\tau')$.

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