# Does the "random Krylov-Bogolyubov theorem" hold in a non-skew-product setting?

Informal description.

Suppose I have a dynamical system $$f$$ defined on the product of a compact space $$X$$ representing the state space of an "experimentally visible" variable and a compact space $$Y$$ representing the state space of an "experimentally invisible" variable. Although $$Y$$ is "invisible", I assume that it is equipped with an equivalence class of probability measures that - like the Riemannian volume measure on a compact smooth manifold - defines "physically accessible" sets of initial conditions in $$Y$$. I assume that $$f$$ has no skew-product structure, meaning that there is allowed to be bi-directional feedback between the visible variable and the invisible variable. A heuristic interpretation of my question below is: If in my experiments I consistently observe the same ergodic statistics for the visible variable, is it necessarily a "physical possibility" that this ergodic statistics comes from an underlying stationary dynamics for the process as a whole?

Precise formulation.

Let $$X$$ and $$Y$$ be compact metric spaces, let $$\pi_X \colon X \times Y \to X$$ be the first coordinate projection, let $$\mu$$ be a Borel probability measure on $$X$$ of full support, let $$\lambda$$ be a Borel probability measure on $$Y$$ of full support, and let $$f \colon X \times Y \to X \times Y$$ be a homeomorphism. Suppose that for every continuous function $$g \colon X \to \mathbb{R}$$, $$\mu \otimes \lambda \left( (x,y) \in X \times Y \, : \, \frac{1}{N} \sum_{n=0}^{N-1} g(\pi_X(f^n(x,y))) \to \int_X g \, d\mu \ \text{ as } \, N \to \infty \right) = 1.$$ Does it follow that there exists an $$f$$-invariant probability measure on $$X \times Y$$ whose $$X$$-marginal is $$\mu$$?

If so, does there necessarily exist a Borel probability measure $$\nu$$ on $$X \times Y$$ with $$\lambda$$-absolutely continuous $$Y$$-projection such that $$\frac{1}{N} \sum_{n=0}^{N-1} f^n_\ast\nu$$ has a subsequence converging weakly to a measure whose $$X$$-marginal is $$\mu$$?

Remark. In the case that $$f$$ has the skew-product structure $$f(x,y)=(\theta(x),\varphi_x(y))$$ [in which case the measure $$\lambda$$ is irrelevant for the first question], an affirmative answer to the first question is given by Corollary 6.13 of Hans Crauel, Random Probability Measures on Polish Spaces - although quoting that result is probably overkill, and I think an affirmative answer to both questions is provided (with $$\nu=\mu \otimes \lambda$$) by just slightly adapting the regular proof of the Krylov-Bogolyubov theorem applied to $$f$$. In terms of application, typically this skew-product setup concerns the scenario that $$X$$ is the state space of an unknown "noise" and $$Y$$ is the "visible" state space - which is opposite to what I'm thinking about in my question.

Okay, I've seen that the proof of an affirmative answer to both questions in the general case (with $$\nu=\mu \otimes \lambda$$) is not very hard:

Since $$X \times Y$$ is compact, we can let $$(k_n)$$ be a strictly increasing sequence of positive integers such that $$\frac{1}{k_n} \sum_{i=0}^{k_n-1} f^i_\ast(\mu \otimes \lambda)$$ is weakly convergent as $$n \to \infty$$. Let $$\mathbb{P}$$ denote the limit. As in the usual proof of the Krylov-Bogolyubov theorem, $$\mathbb{P}$$ is $$f$$-invariant. It remains to show that $$\pi_{X\ast}\mathbb{P}=\mu$$. For any continuous function $$g \colon X \to \mathbb{R}$$, we have \begin{align*} \int_X g \, d(\pi_{X\ast}\mathbb{P}) &= \int_{X \times Y} g \circ \pi_X \, d\mathbb{P} \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g \circ \pi_X \, d(f^i_\ast(\mu \otimes \lambda)) \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \lim_{n \to \infty} \int_{X \times Y} \frac{1}{k_n} \sum_{i=0}^{k_n-1} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \int_{X \times Y} \int_X g \, d\mu \, (\mu \otimes \lambda)(d(x,y)) \\ &\hspace{20mm} \text{by Dom. Conv. Thm. and assumption on \mu and \lambda} \\ &= \int_X g \, d\mu. \end{align*}