A question on invariant measures

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.

I presume that the state space $X$ is compact. Then your set is precisely the set of functions $f\in C(X)$, for which the Cesaro averages $$ C_n f(x) = \frac1{n+1} \Bigl( f(x) + f(Tx) + \dots + f(T^n x) \Bigr) $$ uniformly converge to a constant.

This is, obviously, a generalization of a characterization of uniquely ergodic systems as those for which all Cesaro averages uniformly converge to constants, see Uniform convergence of Birkhoff averages and unique ergodicity, and the proof is more or less the same.

The idea is to use the ``Krylov-Bogolyubov'' measures $$ \kappa_{x,n} = \frac1{n+1} \Bigl(\delta_x + \delta_{Tx} + \dots + \delta_{T^n x} \Bigr) $$

Indeed, if $C_n f$ does not converge uniformly to a constant function, then there are sequences of points $x_k,y_k\in X$ and of integers $m_k,n_k$ such that $\kappa_{x_k,m_k}$ (resp., $\kappa_{y_k,n_k}$) weakly converges to a $T$-invariant measure $\mu$ (resp., $\nu$) with $\langle f,\mu \rangle \neq \langle f,\nu \rangle$. Conversely, if $\langle f,\mu \rangle \neq \langle f,\nu \rangle$ then by Birkhoff's ergodic theorem $C_n f$ can not uniformly converge to a constant.

As in RW's answer, I assume $X$ is compact, and $C(X)$ means continuous real-valued functions (the extension to complex continuous functions is routine). But now assume something much stronger: $X$ is zero-dimensional.

Then, modulo coboundaries (functions of the form $f - T\circ f$), and assuming something like recurrence (or some condition known to guarantee an obvious pre-ordering is a valid partial ordering, see below), these are just the functions that are infinitesimals plus a constant function; simply note that $f- \int f\,d\mu$ (for these special $f$s) vanishes at all the measures (or more accurately, vanishes at all the positive linear functionals on $C(X)$ that are induced by the invariant measures). This describes precisely what are known as the infinitesimal subgroup of the (pre-) ordered group $C(X)/(\text{image}(I-T))$. [See towards the end of

M Michael Boyle and David E Handelman Orbit equivalence, flow equivalence and ordered cohomology Israel Journal of Mathematics December 1996, Volume 95, Issue 1, pp 169-210.]

Most of this deals with integer-valued continuous functions---which of course are interesting only for highly disconnected spaces; but at one point we consider real-valued functions. Somewhat later, it was shown by Nick Ormes that the ordered group (here a vector space) is generically laced with infinitesimals.]

Without something like recurrence, the pre-ordering is not likely to be a proper partial ordering, and in that case the answer still consists of the infinitesimals, but with a slightly different interpretation.