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I want to show that if a set of sets $E$ generates a $\sigma$-algebra, and any $S\in E$ has $f^{-1}(S)$ is Borel measurable, then $f^{-1}(S_0)$ is measurable for any $S_0$ in the $\sigma$-algebra.

I tried using the definition of measurable sets but couldn't get it from here. Think it may be related to Borel algebras?

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  • $\begingroup$ It's a bit hard to follow the thread of this Question. Generally the issue of whether preimages are "measurable" is critically tied to the function $f$ that defines those preimages, Indeed you do say something about function $f$ and its preimages, but you sort of bury this important nugget at the end of a long "hypothetical". I suggest editing to separate more clearly the setup (definition of sigma algebra generated by $E$ as a given) and the hypothesized properties of the preimages $f^{-1}(S_0)$ for any $S_0$ in that generated sigma algebra. $\endgroup$ – hardmath Apr 7 '16 at 1:18
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Let $\mathcal M$ be the collection of sets $A$ in the codomain of $f$ such that $f^{-1}(A)$ is Borel. This contains $E$. To show it is a $\sigma$-algebra, it suffices to note that it contains $\emptyset$, $f^{-1} (A^c) = (f^{-1}(A))^c$ (so it's closed under complement), and $f^{-1}\left( \bigcup_j A_j \right) = \bigcup_j f^{-1}(A_j)$ (so it's closed under countable unions). Therefore $\mathcal M$ must contain your $\sigma$-algebra.

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