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Let $(X, \tau)$ be a topological subspace and let $\Sigma(\mathcal{C})$ denote the $\sigma$-algebra generated by $\mathcal{C}$. Suppose $E \in \Sigma(\tau)$, and consider the topological structure: $$ \tau' = \{F \in P(E): \exists U \in \tau: F=U \cap E\} $$ and the measurable structure: $$ M' = \{F \in P(E): \exists U \in \Sigma(\tau): F=U \cap E\} $$ inherited by $E$ from $X$.

I would like to know what is the relation between $M'$ and $\Sigma(\tau')$. It is clear that $\Sigma(\tau') \subseteq M'$ because if $F \in \tau'$ then $F=U \cap E$ for some $U \in \tau \subseteq \Sigma(\tau)$ so that $F \in M'$, but what about the converse inclusion?


As always, any comment or answer is much appreciated and let me know if I can explain myself clearer!

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To prove the reverse inclusion, let $C$ be the set of $A\subseteq X$ such that $A\cap E\in \Sigma(\tau')$. Then $C$ is a $\sigma$-algebra and $\tau\subseteq C$, so $\Sigma(\tau)\subseteq C$. This exactly says that $M'\subseteq \Sigma(\tau')$.

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