TITLE: A question on invariant measures
QUESTION [6 upvotes]: Let $(X, \mathcal{B}, T)$  be a topological dynamical system and $M(X, T)$ be the set of all invariant measures.
I do not know is there some nice functional characterization of  the following set
$\{f\in C(X): \text{the set}\{\int fd\mu: \mu \in M(X,T)\}\text{is a singleton} \}$.
Any comments and remarks will be appreciated.

REPLY [7 votes]: Your set consists exactly of weak coboundaries plus constants.
A function $f\in C(X)$ is called a coboundary if $f = h \circ T - h$ for some $h\in C(X)$. A function is called a weak coboundary if it is a uniform limit of coboundaries. 

Proposition A. $f \in C(X)$ is a weak coboundary if and only if $\int f d\mu=0$ for every $\mu \in M(X,T)$.

Proof. The "only if" part is trivial, so let's prove the "if" part. Assume $\int f d\mu=0$ for every $\mu \in M(X,T)$. Denote by $f^{(n)} := f + f\circ T + \dots +f\circ T^{n-1}$ the $n$-th Birkhoff sum of $f$. As explained in RW's answer, the sequence of Cesàro averages $\frac{1}{n}f^{(n)}$ converges uniformly to $0$. So the proposition will be a consequence of the following lemma:

Lemma B. For every $f \in C(X)$ and every $n$, the function $\frac{1}{n}f^{(n)} - f$ is a coboundary. 

It's an entertaining exercise to prove the lemma; but if you're in a hurry, here's the solution.
Proof. Let $h := \frac{1}{n}\sum_{i=0}^{n-1} f^{(i)}$ (where $f^{(0)}:=0$). Then
$$
h \circ T - h = \frac{1}{n}\sum_{i=0}^{n-1} \left(f^{(i)}\circ T-f^{(i)}\right) = \frac{1}{n}\sum_{i=0}^{n-1} \left(f \circ T^i - f\right)=\frac{f^{(n)}}{n} - f.
$$
Remarks: The terminology weak coboundary comes from the 2002 Inventiones paper by Bousch and Jenkinson. See their paper for an explicit example of a weak coboundary that is not a coboundary. I think that Proposition A and Lemma B above are kind of folklore. Proposition A is Lemma 3 from Bousch-Jenkinson; for a more complete statement (and a proof using Hahn-Banach), see Proposition 2.13 in Katok-Robinson's survey. Lemma B is used as obvious in Bousch-Jenkinson; I learned it from Andrés Navas. We have used these ideas in non-commutative settings.