Inverse images and $\sigma$-algebras

Solution 1:

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.