The following is from the sixth lecture notes (pp.11-12) of MAGIC010 Ergodic Theory .
The claim:
Let $X$ be a compact metric space, $T:X\to X$ a continuos mapping and define $C(X, \mathbb{R}) = \{f:X\to \mathbb{R}\mid \text{$f$ is continuous}\}$. Then the following are equivalent: 1.) $T$ is uniquely ergodic. 2.) for all $f \in C(X, \mathbb{R})$ there exists a constant $c(f) \in \mathbb{R}$ such that $\lim_{n\to\infty}\sum_{j=0}^{n-1}(f\circ T^j)(x) = c(f)$ uniformly for $x \in X$.
The given proof that $2.) \implies 1.)$ is as follows:
$ \int fd\mu = \lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\int f\circ T^jd\mu = \int \lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}f\circ T^jd\mu = \int c(f)d\mu = c(f)$.
Where to my knowledge the author uses the Dominated Convergence Theorem. But 1.) the DCT requires the bounding function to be non-negative and I don't see any reason why $c(f)$ couldn't be negative. Is there an implicit assumption that $f$ is non-negative or in what other way can the use of DCT be justified? If it is so that DCT is not strictly necessary for the proof that $2.) \implies 1.)$, the given proof that $1.) \implies 2.)$ relies entirely upon DCT.
$|\frac{1}{n}\sum_{j=0}^{n-1}f\circ T^j|\leq M$ if $|f| \leq M$ on $X$. Since $\mu$ is a finite measure the constant $M$ is integrable.