Let $X=(X_n)_{n \in \mathbb{Z}}$ and $Y=(Y_n)_{n \in \mathbb{Z}}$ be two sequences of i.i.d. random variables on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that $X$ is independent of $Y$. For $k \in \mathbb{Z}$ consider the shift operator $\theta_k$ which maps a sequence $(x_n)_{n \in \mathbb{Z}}$ in $\mathbb{R}^{\mathbb{Z}}$ to $(x_{n+k})_{n \in \mathbb{Z}}$. Let the function $$ f : \mathbb{R}^{\mathbb{Z}} \times \mathbb{R}^{\mathbb{Z}} \rightarrow \mathbb{R}^{\mathbb{Z}} $$ be measurable and such that $$ \theta_k f(X, Y) = f (\theta_k X, \theta_k Y) \quad \forall k \in \mathbb{Z}. \tag{1} $$ Let $\mu_1$, $\mu_2$ denote the distributions of $X$ and $Y$ and let $\mu := \mu_1 \otimes \mu_2$ be the distribution of $(X, Y)$. $(1)$ means that $$ \theta_k f(x, y) = f(\theta_k x, \theta_ky) \quad \mu \text{-a.s.} \ \forall k \in \mathbb{Z}. $$ Using the fact that $\mu$ is stationary on $(\mathbb{R}^\mathbb{Z} \times \mathbb{R}^\mathbb{Z}, \mathcal{B}(\mathbb{R})^{\mathbb{Z}} \otimes \mathcal{B}(\mathbb{R})^{\mathbb{Z}})$ (with respect to $(\theta_k \times \theta_l)_{k, l \in \mathbb{Z}}$), one can show that the image measure $\mu \circ f^{-1}$ is also stationary on $(\mathbb{R}^\mathbb{Z}, \mathcal{B}(\mathbb{R})^{\mathbb{Z}})$ (with respect to $(\theta_k)_{k \in \mathbb{Z}}$). I am further interested in the following:
Question: Is the image measure $\mu \circ f^{-1}$ ergodic? We know that $\mu \circ f^{-1}$ is ergodic iff $$ \lim_{n \rightarrow \infty} \frac{1}{(2n+1)} \sum_{k \in [ -n, n ] \cap \mathbb{Z}} \mu \left( f^{-1} (\theta_k^{-1} ( A ) \cap B) \right) = \mu(f^{-1}(A)) \mu(f^{-1}(B)) \quad \forall A, B \in \mathcal{B}(\mathbb{R})^{\mathbb{Z}}. \tag{2} $$
To show this, we can write \begin{align*} \mu \left( f^{-1} (\theta_k^{-1} ( A ) \cap B) \right) &= \mu \left( f^{-1} (\theta_k^{-1} ( A ) ) \cap f^{-1}(B) \right) \\ &= \mu \left( \{ (x, y) : f(x, y) \in \theta_k^{-1}(A) \} \cap f^{-1}(B) \right) \\ &= \mu (\{ (x, y) : \theta_k f(x, y) \in A \} \cap f^{-1}(B)) \\ &= \mu (\{ (x, y) : f(\theta_k x, \theta_k y) \in A \} \cap f^{-1}(B)) \\ &= \mu \left( \{ (x, y) : (\theta_k x, \theta_k y) \in f^{-1}(A) \} \cap f^{-1}(B) \right) \\ &= \mu \left( (\theta_k \times \theta_k)^{-1} ( f^{-1}(A) ) \cap f^{-1}(B) \right). \tag{3} \end{align*}
If $\mu$ is ergodic (is it?), then we also know that $$ \lim_{n \rightarrow \infty} \frac{1}{(2n+1)^{\color{red}2}} \sum_{(k, l) \in [ -n, n ]^{\color{red}2} \cap \mathbb{Z}^{\color{red}2}} \mu \left( (\theta_k \times \theta_l)^{-1} ( E ) \cap F \right) = \mu(E) \mu(F) \quad \forall E, F \in \mathcal{B}(\mathbb{R})^{\mathbb{Z}} \otimes \mathcal{B}(\mathbb{R})^{\mathbb{Z}}. \tag{4} $$
Now, is it possible to use $(3)$ and $(4)$ to obtain $(2)$?
This is simpler and more general. Suppose you have any ergodic probability measure preserving system $(X, \mathcal F, \mu, T)$ and $F:X \to Y$ maps it to another system $(Y, \mathcal G, \nu, S)$, where (1) $A \in \mathcal G$ implies that $F^{-1}(A) \in \mathcal F$, (2) the measures satisfy $\nu=\mu\circ F^{-1}$, and (3) the transformations satisfy $S \circ F=F \circ T$. Then the second system is ergodic as well. Just use the definition of ergodicity. Indeed, if $A \in \mathcal G$ is invariant, i.e., $S^{-1}(A)=A$, then $B=F^{-1}(A)$ satisfies $T^{-1}(B)=B$, so $\nu(A)=\mu(B) \in \{0,1\}$.