I'm trying to solve problem 4.1.5 from the book Foundations on ergodic theory from M. Viana and K. Oliveira. The problem is stated as follows:
Let $M$ be a metric space. We call the basin of an invariant probability measure $\mu$ the set $B(\mu)$ of all points $x\in M$ such that $$ \lim\limits_{n\to\infty} \frac{1}{n} \sum_{j=0}^{n-1} \varphi (f^j(x))=\int \varphi d\mu$$ for every bounded continuous function $\varphi:M\rightarrow \mathbb{R}$. Check that the basin is an invariant set. Moreover, if $\mu$ is ergodic then $B(\mu)$ has full $\mu$-measure.
I managed to prove that $B(\mu)$ is indeed invariant and the book has also a proposition that states that if $\mu$ is ergodic, then for every $\varphi\in L^1(\mu)$, the average time $$\tilde{\varphi}(x)=\lim\limits_{n\to\infty} \frac{1}{n} \sum_{j=0}^{n-1} \varphi(f^{j}(x))$$ agrees with $\int \varphi d\mu$ for $\mu$ almost every point $x\in M$.
My problem is that with this setting I know the condition in the definition on $B(\mu)$ holds for almost every $x\in M$, but the points $x$ which fail to match the condition might depend on the choice of the function $\varphi$. I would really appreciate if you could help me get rid of that dependence or show me another approach to prove that $B(\mu)$ has full measure.
I'm a PHD student and I'm allowed to use any standard fact from measure theory, functional analysis or probability theory(as seen in books like Folland, Royden, Durret, Brezis).
Thanks in advance
I don't think this is true without more assumptions.
This counterexample might be overkill, but here goes. Let $\kappa$ be a measurable cardinal, which means there is a countably additive probability measure $\mu$ defined for every subset of $\kappa$, taking only the values 0 and 1, and such that $\mu(\{x\})=0$ for every singleton. We can replace $\kappa$ with $\kappa \times \{0,1\}$ since they have the same cardinality. Let $M = \kappa \times \{0,1\}$ equipped with the discrete metric, so $\mu$ is a Borel probability measure on $M$.
Let $f : M \to M$ be defined by $f(x,0) =(x,1)$ and $f(x,1)=(x,0)$, so it just swaps the 0s and 1s. The measure $\mu$ is trivially ergodic since every set has measure 0 or 1. Fix an arbitrary $y \in \kappa$ and let $\varphi$ be the function with $\varphi(y,1)=1$, $\varphi(y,0)=0$, and $\varphi =0$ elsewhere. Then $\varphi$ is bounded and continuous (since $M$ has the discrete metric). It is easy to see that $\frac{1}{n} \sum_{j=0}^{n-1} \varphi (f^j(p))= 1/2$ for $p=(y,0)$ and $p=(y,1)$, but since $\mu(\{(y,1)\})=0$ we have $\int \varphi\,d\,mu = 0$. So $(y,0)$ and $(y,1)$ are not in the basin. Since $y$ was arbitrary, the basin is empty.
(I don't want to get into the set-theoretic assumptions about large cardinals. I'll just say that this example requires working in an axiom system that is a little stronger than ZFC, but is widely believed to still be consistent.)