Proving that the basin of an invariant probability measure has full measure 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
 A: 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.)
