Does the following condition imply ergodicity? Let $(X,\mathcal F,μ,T)$ be a dynamical system (i.e. μ is a probability measure and Τ is μ-preserving) and $\mathcal S\subset\mathcal F$ be a family of sets such that for any $A \in \mathcal F$ and $ε>0$ there exists a $B \in \mathcal S$ with $μ(A\triangle B)\lt ε$.
Assume that  $\lim_{N\to \infty}\frac{1}{N}\sum_{n=0}^N μ(Α\cap T^{-n}B)>0$ , for all $A,B\in S$, with positive measure. Can we conclude that the system is Ergodic?
The motivation behind this question is to understand the concept of ergodicity better, by finding "minimal" conditions for it to occur.
I know, for example, that if the above limit is not only positive but equal to $μ(Α)μ(Β)$ then the system is ergodic and vice versa!
So I am thinking that it is natural to either construct a counterexample (which I tried to do, but failed) or to prove that if the limit is positive it necessarily equals $μ(Α)μ(Β)$.
Another fact; Since Von Neumann's mean ergodic theorem implies that
$$\lim_{N\to \infty}\frac{1}{N}\sum_{n=0}^N μ(Α\cap T^{-n}B)=\int_A\mathbf{E}[1_B|\mathcal{I}]dμ$$
where $\mathbf{E}[1_B|\mathcal{I}]$ is the conditional expectation of $1_B$ with respect to the σ-algebra $\mathcal{I}=\{D\in \mathcal F: T^{-1}D=D\}$, then for the implication to be true it would suffice to show that if for all $A,B\in \mathcal S$ with positive measure we have that $\int_A\mathbf{E}[1_B|\mathcal{I}]dμ>0$, then we specifically have the equality $\int_A\mathbf{E}[1_B|\mathcal{I}]dμ=μ(Α)μ(Β)$.
 A: The answer is no. Take $X$ to be a disjoint union $X_1\sqcup X_2$ of invariant subsystems of positive measure with $T\restriction_{X_1}$ and $T\restriction_{X_2}$ ergodic. Define $\mathcal{S}$ as
$\{B\in\mathcal{F}\mid \mu(B\cap X_1),\mu(B\cap X_2)>0\}$.
Moreover, assume that both $X_1$ and $X_2$ have subsets of arbitrarily small positive measures. Then $\mathcal{S}$ clearly satisfies the condition outlined in the question. Now suppose $A,B\in\mathcal{S}$. They can be partitioned as $A_1\sqcup A_2$ and $B_1\sqcup B_2$ where $A_i:=A\cap X_i$ and $B_i:=B\cap X_i$ are with positive measure. Now we have
$$
\mu(A\cap T^{-n}B)=\mu\left((A_1\cap T^{-n}B_1)\sqcup (A_2\cap T^{-n}B_2)\right)
=\mu(A_1\cap T^{-n}B_1)+\mu(A_2\cap T^{-n}B_2).
$$
The ergodic average then splits into the sum of two ergodic averages for $T\restriction_{X_1}$ and $T\restriction_{X_2}$. These systems (equipped with rescaled probability measures $\frac{1}{\mu(X_1)}.\mu\restriction_{X_1}$ and $\frac{1}{\mu(X_2)}.\mu\restriction_{X_2}$) are ergodic. Therefore
$$
\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\mu(A\cap T^{-n}B)=
\frac{\mu(A_1)\mu(B_1)}{\mu(X_1)}+\frac{\mu(A_2)\mu(B_2)}{\mu(X_2)}>0.
$$
