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Let $\Sigma$ be a $\sigma$-algebra over $\mathbb R$ and $\mathcal A \subset \mathcal P(\mathbb R)$. Let also $f: \mathbb R \to \mathbb R$ be any function.

If $\mathcal A$ generates $\Sigma$, is it true that $\widetilde{f^{-1}}(\mathcal A)$ generates $\widetilde{f^{-1}}(\Sigma)$? That is, do these symbols commute:

$$\sigma(\widetilde{f^{-1}}(\mathcal A)) = \widetilde{f^{-1}}(\sigma(\mathcal A))\quad?$$

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    What does the tilde represent here?2011-09-10
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    @Arturo Magidin $f^{−1}$ only exists if $f$ is bijective; $\widetilde{f^{−1}}$, inverse image, always exists. But indeed, I have a strong impression that's not such an important distinguishment, since even my teacher doesn't care — and you don't either, as I see below.2011-09-10
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    @Luke: $f^{-1}$ is often used for the induced function on the power sets, but precisely to avoid confusion I *explicitly* said what I meant by it. It's not that "I don't care", it's that from context it is clear what is meant. For example, Halmos (in *Naive Set Theory*) first introduces the notation $\underline{f}$ and $\overline{f}$ for the direct and inverse image functions. But because the inverse image function takes **subsets** as arguments whereas the inverse function (if it exists) takes **elements** as arguments, there is almost never any ambiguity.2011-09-10
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    @Arturo Magidin Exactly, that's why it's easy to distinguish, and the tilde is not crucial.2011-09-11
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    Possible duplicate: http://math.stackexchange.com/questions/7881/preimage-of-generated-sigma-algebra (although I am late :) )2011-11-23

1 Answers 1

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For any set-theoretic function $f\colon A\to B$, the inverse image function $f^{-1}\colon\mathcal{P}(B)\to\mathcal{P}(A)$ given by $f^{-1}(Y) = \{a\in A\mid f(a)\in Y\}$ is extremely well-behaved relative to set operations. In particular, for all $X,Y\subseteq B$ and all families $\{X_i\}\subseteq \mathcal{P}(B)$, $$\begin{align*} f^{-1}(X\cup Y) &= f^{-1}(X)\cup f^{-1}(Y),\\ f^{-1}(X\cap Y) &= f^{-1}(X)\cap f^{-1}(Y),\\ f^{-1}(\cup X_i) &= \cup f^{-1}(X_i),\\ f^{-1}(\cap X_i) &= \cap f^{-1}(X_i),\\ f^{-1}(X-Y) &= f^{-1}(X) - f^{-1}(Y),\\ f^{-1}(X^c) &= (f^{-1}(X))^c,\\ f^{-1}(X\triangle Y) &= f^{-1}(X)\triangle f^{-1}(Y) \end{align*}$$ where $\triangle$ is the symmetric difference. As such, the inverse image of a family that generates a $\sigma$-algebra will generate the inverse image of the $\sigma$-algebra generated: you can justify the details by looking at the "bottoms-up" description of the $\sigma$-algebra generated by a family that appears in Asaf's answer to this question.

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    A bit quicker, and not dependent on having a transfinite hierarchy construction, is to use the "good sets principle" (google it). This will give a proof from the better known top-down formulation of $\sigma$-algebras.2011-09-12
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    Hello, could you add more detail? I tried your approach but got stuck. @DaveL.Renfro2017-11-13
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    @xFioraMstr18 4: See [Example of the good set principle](https://math.stackexchange.com/questions/2064953/example-of-the-good-set-principle) AND [Proving that a class of good sets forms a $\sigma$-field.](https://math.stackexchange.com/questions/2304152/proving-that-a-class-of-good-sets-forms-a-sigma-field) AND [$\sigma$-algebra produced by a subclass of a class.](https://math.stackexchange.com/questions/1651937/sigma-algebra-produced-by-a-subclass-of-a-class) AND [p. 11 here](http://math.utoledo.edu/~dwhite1/d_6800/measurable.pdf) AND [p. 10 here](https://ee.stanford.edu/~gray/arp.pdf).2017-11-13