A random variable, measurable w.r.t. the class of invariant events, is invariant

Question

I would like to deeply understand the following proof; however, many things are not really obvious to me. I would appreciate if somebody could explain the solution in detail. Thank you.

Problem (Durrett, 2010, Chapter 7, Exercise 7.1.1)

Let $\mathcal{I}$ be the class of invariant events, that is, for any $A \in \mathcal{I}$, $P(A \Delta \varphi^{-1}(A)) = 0$ (also known as almost invariant). Prove that if $X$ is measurable with respect to $\mathcal{I}$, $X$ is invariant, that is, $X \circ \varphi = X$ a.s.

Solution

The set of invariant random variables contains the indicator functions $1_A$ with $A \in \mathcal{I}$ (why?) and is closed under pointwise limits (why?), so all $X \in \mathcal{I}$ are invariant (why?).

Regards, Ivan

Answer 1

1. The set $\{\omega,\chi_A\circ \varphi(\omega)\neq \chi_A(\omega)\}$ is exactly $A\Delta\varphi^{-1}(A)$, which have measure $0$. Hence there is equality almost everywhere.

2. Let $X_n\to X$ pointwise, and $N_n$ of measure $0$ such that $X_n\circ\varphi(\omega)=X_n(\omega)$ for $\omega\in\Omega\setminus N_n$. Let $N:=\bigcup_nN_n$: it has measure $0$. Then for all $\omega\in\Omega\setminus N$, we have $X\circ \varphi(\omega)=X(\omega)$.

3. Approximate $X$ pointwise by linear combinations of characteristic functions of invariant sets.