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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?

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2 Answers 2

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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 $\cup_nA_n\in S$.

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  • $\begingroup$ Thanks for the answer! I guess in my approach one needs to observe that $$\mu\left(\bigcup_{n \in \mathbb{N}} \left(B_n \cap \bigcup_{k \in \mathbb{N}} A_k \right)\right) \geq \mu\left(\bigcup_{n \in \mathbb{N}} B_n \cap A_n \right) = \sum_{n \in \mathbb{N}} \mu (B_n \cap A_n)$$ $\endgroup$
    – Harry
    Jul 30, 2022 at 22:00
  • $\begingroup$ @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 $\endgroup$
    – Snoop
    Jul 30, 2022 at 22:06
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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 $\cup_{k=1}^{\infty} A_k \in S$.

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