Show that the given family is a $\sigma$-algebra

Let $$(\Omega, \mathcal{F}, \mu)$$ be a probability space and let $$f : \Omega \rightarrow \Omega$$ be $$\mathcal{F}$$-measurable and such that $$\mu(f^{-1}(A)) = \mu(A)$$ for all $$A \in \mathcal{F}$$. I am interested in showing that the set $$S = \{A \in \mathcal{F} : \mu(f^{-1}(A) \triangle A) = 0 \}$$ is a $$\sigma$$-algebra. Here $$\triangle$$ denotes the symmetric difference, i.e. $$A \triangle B = (A \cup B) \setminus (A \cap B)$$.

Since $$f^{-1}(\Omega) = \Omega$$, it is easy to see that $$\Omega \in S$$. Moreover, using the fact that $$f^{-1}(A^c) = (f^{-1}(A))^c$$ one can also show that for any $$A \in S$$ we have $$A^c \in S$$. Now let $$A_n \in \mathcal{F}$$, $$n \in \mathbb{N}$$, be pairwise disjoint. Then also $$B_n := f^{-1}(A_n)$$, $$n \in \mathbb{N}$$, are pairwise disjoint as well as $$B_n \cup A_n$$, $$n \in \mathbb{N}$$ and $$B_n \cap A_n$$, $$n \in \mathbb{N}$$. We can now consider \begin{align} \mu\left(f^{-1}\left(\bigcup_{n \in \mathbb{N}} A_n\right) \triangle \bigcup_{n \in \mathbb{N}} A_n\right) &= \mu\left(\bigcup_{n \in \mathbb{N}} f^{-1}(A_n) \triangle \bigcup_{n \in \mathbb{N}} A_n\right) = \mu\left(\bigcup_{n \in \mathbb{N}} B_n \triangle \bigcup_{n \in \mathbb{N}} A_n\right) \\ &= \mu\left(\bigcup_{n \in \mathbb{N}} B_n \cup \bigcup_{n \in \mathbb{N}} A_n\right) - \mu\left(\bigcup_{n \in \mathbb{N}} B_n \cap \bigcup_{n \in \mathbb{N}} A_n\right) \\ &= \mu\left(\bigcup_{n \in \mathbb{N}} B_n \cup A_n \right) - \mu\left(\bigcup_{n \in \mathbb{N}} \left( B_n \cap \bigcup_{k \in \mathbb{N}} A_k \right)\right) \\ &= \sum_{n \in \mathbb{N}} \mu( B_n \cup A_n ) - \sum_{n \in \mathbb{N}} \sum_{k \in \mathbb{N}} \mu( B_n \cap A_k). \end{align} Is it possible to show that the double sum is equal to $$\sum_{n \in \mathbb{N}} \mu( B_n \cap A_n )?$$ Is the fact that $$\mu(f^{-1}(A)) = \mu(A)$$ for all $$A \in \mathcal{F}$$ any helpful?

Observe that $$A\Delta B=(A/B)\cup(B/A)$$. This simplifies the argument, as the two sets are disjoint. Let $$A \in S$$, then \begin{aligned}\mu(f^{-1}(A^c)\Delta A^c)&=\mu(f^{-1}(A^c)\cap A)+\mu(A^c\cap f^{-1}(A))=\\ &=\mu(A/f^{-1}(A))+\mu(f^{-1}(A)/A)=\\ &=\mu(f^{-1}(A)\Delta A)=0\end{aligned} so that $$A^c \in \Sigma$$. Now let $$(A_n)_{n \in \mathbb{N}}\subseteq S$$. We have \begin{aligned}\mu(f^{-1}(\cup_nA_n)\Delta \cup_nA_n)&=\mu(f^{-1}(\cup_nA_n)\cap(\cap_nA_n^c))+\mu(\cup_nA_n\cap f^{-1}(\cap_nA_n^c))=\\ &=\mu(\cup_n(f^{-1}(A_n)\cap(\cap_nA_n^c)))+\mu(\cup_n(A_n\cap (\cap_nf^{-1}(A_n^c))))\leq\\ &\leq \sum_{n\in \mathbb{N}}\mu(f^{-1}(A_n)\cap A_n^c))+\sum_{n \in \mathbb{N}}\mu(A_n\cap f^{-1}(A_n^c))=\\ &=\sum_{n \in \mathbb{N}}(\mu(f^{-1}(A_n)\cap A_n^c)+\mu(A_n\cap f^{-1}(A_n^c)))=\\ &=\sum_{n \in \mathbb{N}}\mu(f^{-1}(A_n)\Delta A_n)=0\end{aligned} thus $$\cup_nA_n\in S$$.

• Thanks for the answer! I guess in my approach one needs to observe that $$\mu\left(\bigcup_{n \in \mathbb{N}} \left(B_n \cap \bigcup_{k \in \mathbb{N}} A_k \right)\right) \geq \mu\left(\bigcup_{n \in \mathbb{N}} B_n \cap A_n \right) = \sum_{n \in \mathbb{N}} \mu (B_n \cap A_n)$$ Jul 30, 2022 at 22:00
• @Harry: attention at the definition of $\sigma$-algebra: you may ultimately want to prove the last assertion for $(A_n)_n$ not necessarily pairwise disjoint Jul 30, 2022 at 22:06

Note that a non-empty family $$\mathcal{D}$$ of subsets of $$\Omega$$ that is closed under complements and countable unions of pairwise-disjoint sets, then $$\mathcal{D}$$ is called a Dynkin system. An important obsevation pertaining to this problem is that a Dynkin system may or may not be a $$\sigma$$-algebra. This means that a Dynkin system may not be closed under countable unions of not necessarily disjoint sets.

Now returning to the question of showing $$S$$ is a $$\sigma$$-algebra, a concise solution is available by using indicator function notation. Indeed, note that

$$S = \{ A \in \mathcal{F} : \text{\mathbf{1}_{A} = \mathbf{1}_A \circ f holds \mu-a.s.}\}.$$

This is because $$A \triangle B$$ is precisely the exceptional set on which $$\mathbf{1}_{A} = \mathbf{1}_B$$ fails. Then

1. $$\mathbf{1}_{\Omega} = \mathbf{1}_{\Omega} \circ f$$ everywhere, hence $$\Omega \in S$$.

2. If $$\mathbf{1}_A = \mathbf{1}_A \circ f$$ a.s., then $$\mathbf{1}_{A^c} = \mathbf{1}_{\Omega} - \mathbf{1}_{A} = (\mathbf{1}_{\Omega} \circ f) - (\mathbf{1}_{A} \circ f) = \mathbf{1}_{A^c} \circ f \qquad \text{\mu-a.s.}$$ So, $$A^c \in S$$.

3. Suppose $$(A_n)_{n=1}^{\infty}$$ is a family of events in $$S$$. Then $$\sum_{k=1}^{n} \mathbf{1}_{A_k} = \sum_{k=1}^{n} \mathbf{1}_{A_k} \circ f$$ holds $$\mu$$-a.s., and so, \begin{align*} \mathbf{1}_{\cup_{k=1}^{n} A_k} = \min\left\{1, \sum_{k=1}^{n} \mathbf{1}_{A_k} \right\} = \min\left\{1, \sum_{k=1}^{n} \mathbf{1}_{A_k} \circ f \right\} = \mathbf{1}_{\cup_{k=1}^{n} A_k} \circ f \qquad \text{\mu-a.s.} \end{align*} Letting $$n \to \infty$$ and noting that $$\mathbf{1}_{\cup_{k=1}^{n} A_k}$$ converges to $$\mathbf{1}_{\cup_{k=1}^{\infty} A_k}$$ everywhere, it follows that \begin{align*} \mathbf{1}_{\cup_{k=1}^{\infty} A_k} = \mathbf{1}_{\cup_{k=1}^{\infty} A_k} \circ f \qquad \text{\mu-a.s.} \end{align*} Therefore $$\cup_{k=1}^{\infty} A_k \in S$$.