Preimage of generated $\sigma$-algebra

Yes, we can use transfinite induction to prove this (formalizing the word "repeatedly"). That would be the bottom-up approach. There is also a top-down approach, using the characterization of $\sigma(C)$ as the smallest $\sigma$-algebra containing $C$.

A key fact here is the the preimage operation commutes with all the set algebra operations: if $f \colon X \to Y$ then

• $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$
• $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$
• $f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B)$. In particular $f^{-1}(Y - A) = X - f^{-1}(A)$.

and so forth. So $f^{-1}\colon P(Y) \to P(X)$ is a lattice homomorphism on the powerset lattices, to use that terminology. More importantly for us, the preimage operation even commutes with infinite unions and intersections.

To show that $f^{−1}(\sigma(C)) \subseteq \sigma(f^{−1}(C))$, you could follow this strategy:

• Show that $f^{-1}(C) \in \sigma(f^{-1}(C))$
• Show that if $f^{-1}(A_i) \in \sigma(f^{-1}(C))$ for $i \in \omega$ then $f^{-1}(\bigcup A_i) \in \sigma(f^{-1}(C))$
• Show that if $f^{-1}(A) \in \sigma(f^{-1}(C))$ then $f^{-1}(Y - A) \in \sigma(f^{-1}(C))$

The point here is that, if we let $D$ be the collection of sets $A$ such that $f^{-1}(A) \in \sigma(f^{-1}(C))$, then $D$ is itself a $\sigma$ algebra containing $C$, which means $\sigma(C) \subseteq D$. But by the definition of $D$ this means $f^{-1}(\sigma(C)) \subseteq \sigma(f^{-1}(C))$.

None of the three bullets will require taking forward images under $f$. For example, for the third one, let $A$ be as stated. This means $f^{-1}(A)$ is in $\sigma(f^{-1}(C))$, which means that $X - f^{-1}(A)$ is also in $\sigma(f^{-1}(C))$. But $f^{-1}(Y - A)$ is exactly $X - f^{-1}(A)$, so we see that $f^{-1}(Y - A)$ is indeed in $\sigma(f^{-1}(C))$.

The underlying point here is that the entire proof is algebraic and that a more general theorem is true: you can replace $f^{-1}$ with any other homomorphism of the powerset lattices that preserves countable unions.

Edit: The original answer that was here is wrong; Carl Mummert's answer contains the correct way to do what I was trying to do. So instead here is the solution by transfinite induction.

Define a sequence of functions $\sigma_i$ where $i$ is an ordinal as follows. For any subset $C$ of the base set $E$ we take $\sigma_0(C) = C$. If $i$ is a successor ordinal, take $\sigma_i(C)$ to be the set of all countable unions or complements of elements of $\sigma_{i-1}(C)$; otherwise $i$ is a limit ordinal and we take $\sigma_i(C) = \bigcup_{j < i} \sigma_j(C)$. For example, $\sigma_{\omega}(C)$ is the set of all sets that can be obtained from $C$ by performing countable union or complement countably many times.

We would like to say that $\sigma(C)$ is the union of the $\sigma_i(C)$ over all ordinals $i$, but the ordinals don't form a set so we can't actually do this. What I believe is true is that we only need to take ordinals of cardinality at most the cardinality of the powerset of the underlying set.

If that's true, then the proof is as follows. It's enough to show that $f^{-1}(\sigma_i(C)) = \sigma_i(f^{-1}(C))$ for all $i$. This is obvious for $i = 0$. If it's true for all ordinals $j < i$, then it's true for $i$ by the obvious argument. And now we are done by the principle of transfinite induction.

Let me remark that, on the one hand, this is more complicated than Carl Mummert's proof because it requires knowledge of ordinals. On the other hand, getting into the nitty-gritty like this really shows how complicated $\sigma$-algebras can be.