Let $X$ be a compact metric space equipped with a Borel probability measure $\mu$. If $T$ is a continuous ergodic transformation, prove there is a full measure set s.t Birkhoff averages of any continuous function converge on that set.
I am only looking for hints.
My work: First of all, $C(X)$ is separable, so there is a dense sequence $f_k$. It is clear that there is a set of full measure s.t for each $k$ the Birkhoff averages converge to $\int f_k$. At this point I tried to do some limit argument to no avail. I get a double limit and I just cannot seem to justify why you can exchange the two limits. Thus I think I am looking at this problem the wrong way. Especially since I do not see why continuity of $T$ is needed here.
I would much appreciate advice/hints rather than flat out solutions.
Your first intuition is right. As $C(X)$ is separable you can find a dense sequence $f_k$. For every $f_k$ there is a full measure set $X_k$ for which the Birkhoff sums converge to the integrable.
Then a countable intersection of full measure set is a full measure set. So on the set $Y = \cap X_k$ every $f_k$ has its Birkhoff sums converging to its average.
No take a $g \in C(X)$. There is a subsequence $\{g_k\} \subset \{f_k\}$ such that for every $\epsilon$, there is a $K$ such that for every $k>K$, $|f-g_k|_\infty \leq \epsilon$.
So for every $x \in Y$, $|\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g|=|\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-g_k(T^i(x))+\frac{1}{N}\sum_{i=1}^{N}g_k(T^i(x))-\int g_k+\int (g_k-g)|$
But we have that $$ |\int (g_k-g)| \leq \epsilon Vol(X) $$
$$ |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-g_k(T^i(x))|\leq \epsilon $$ And $$ \lim \frac{1}{N}\sum_{i=1}^{N}g_k(T^i(x))-\int g_k = 0 $$
So applying the triangular inequality to detach this three pieces we get that $$ \limsup |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| \leq \epsilon (1+ Vol(X)) $$ as this is true for every $\epsilon$ we have that $$ \limsup |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| =0 $$ And then $$ \lim |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| =0 $$