Proving that $m(\{f^{-1}(S) : S \in A\}) = \{f^{-1}(S) : S \in m(A)\}$ (monotone classes)

Question

Let $X,Y$ be sets, $A \subseteq \mathcal{P}(Y)$ and $f: X \to Y$. Then $$m(\{f^{-1}(S) : S \in A\}) = \{f^{-1}(S) : S \in m(A)\}$$ Where $m(A)$ denotes the monotone class generated by $A$.

For proving the inclusion $\subseteq$ it is enough to show that $$\{f^{-1}(S) : S \in m(A)\}$$ is a monotone class. Hence we take an increaing sequence $(f^{-1}(S_n))_{n \in \mathbb{N}}$ where $S_n \in m(A)$ an look at its union. I mean trivially $$\bigcup_{n \in \mathbb{N}}f^{-1}(S_n) = f^{-1}\left( \bigcup_{n \in \mathbb{N}}S_n\right)$$ So if we could show $\bigcup_{n \in \mathbb{N}}S_n \in m(A)$ we are done. For this $(S_n)_{n \in \mathbb{N}}$ must be increasing, which is not per se the case, isn't it? I asked the one which wrote this exercise and he replied that it is the same as for $\sigma(A)$ ($\sigma$-algebra generated by $A$) no further assumptions on the nature of $f$ have to be made. But I mean, $\sigma$-algebras are a richer structure than monotone classes, hence it could be that we have to assert more in this case. My assumption would be that $f$ has to be surjective. So my question is: Is this assumption necessary or is it right that it is like in the case of $\sigma$-algebras? Does the proof work nevertheless and I do not see it?

Answer 1

The fact that the sets $f^{-1}(S_n)$ are increasing implies that the sets $f( f^{-1}(S_n))=S_n \cap f(X)$ are also increasing. As you already remarked, if $f$ were surjective this would imply that the sequence $(S_n)_{n \in \mathbb{N}}$ is increasing and the result would follow.

I believe the following is a counterexample to your claim when $f$ is not surjective.

Counterexample:

Let $f: (0,1) \to \mathbb{R}$ be the inclusion. Thus, if $S \subset \mathbb{R}$ is a set then $f^{-1}(S)$ is simply $S\cap (0,1)$. Define the following collection of subsets of $\mathbb{R}$: $$ A = \{ (q\, , \,q+1) : q \in \mathbb{Q} \} \cup \{\mathbb{R}\} $$ So $A$ is the collection of intervals of length $1$ with rational endpoints. This is a monotone class since there are no nested sets except the trivial $(q,q+1) \subset \mathbb{R}$. Thus we have $m(A)=A$.

The idea of the construction is that, even though the sets in $A$ are not nested, we can arrange for their intersection with $(0,1)$ (i.e. their preimage under $f$ ) to be nested.

Take an irrational number $a \in (0,1)$. Take a decreasing sequence $\{ q_n \}$ of rational numbers $q_n \in (0,1)$ which converges to $a$. Define $S_n =(q_n, q_n+1) \in A$. Then $f^{-1}(S_n) = (q_n,1)$ is an increasing sequence of sets since $\{q_n\}$ is decreasing. Since $q_n \to a$, we have $\bigcup_{n} (q_n , 1) = (a,1)$.

However, there is no set $S \in A$ such that $f^{-1}(S) = (a,1)$ since $a$ is irrational! Thus $\{ f^{-1}(S) : S\in m(A) \}$ is not a monotone class.