# Relation between topological and measurable subspaces

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!

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 $$M'\subseteq \Sigma(\tau')$$.