The statement is let $T: X \to X$ be a continuous transformation on a compact metrizable space $X$. If $\mu$ is a $T-invariant$ borel probability measure that is also purely atomic then it is a (possible infinite) convex combination of measures of the form $\frac{1}{N} \sum_{i=0}^{N-1} \delta_{T^i(x)}$.
Proof: $\mu$ is invariant iff $\int f \circ T d \mu = \int f d \mu$ for every continuous f. Now suppose $\mu=\sum_{i=1}^{\infty} p_i \delta_{x_i}$ with $\sum p_i =1$. We can assume that the $p_i$ are decreasing. From the above integral condition we get $\sum p_i f(T^{i+1}(x))=\sum p_i f(T^{i}(x))$ so that $T(x_1)$ is an atom with measure $p_1$. From this we get that $T^{N_1}(x_1)=x_1$ and each atom $T^i(x)$ has the same measure. The desired result follows from simple induction.
I don't get why you can conclude that $T(x_1)$ is an atom with measure $p_1$ or any of the statements after that. Any help would be appreciated.
I don't get the insistence on arguing through continuous functions. Really what is being said is that
$$ \mu(\{T(x_1)\})=T^*\mu(\{T(x_1)\})=\mu(T^{-1}(T(x_1))=\sum_{i=1}^{\infty} 1_{T(x_i)=T(x_1)} p_i\geq p_1 $$ so $T(x_1)$ is an atom. Furthermore, since the $p_i$ are decreasing, we must have $\mu(T(x_1))=p_1$. By induction, $\mu(T^n(x_1))=p_1$ for all $n$. Furthermore, we get that the restriction of $T$ to the orbit of $x_1$, that is to say $\mathcal{O}(x_1):=\{T^n(x_1)|n\in \mathbb{N}\}$, must be injective (why?).
Hence if $x_1, T(x_1),...,T^n(x_1)$ are distinct, then $$ 1\geq \mu(X)\geq \mu(\{x_1,...,T^n(x_1)\})=(n+1)p_1 $$ Hence, there must be some $n<m\leq \frac{1-p_1}{p_1}$ such that $T^n(x_1)=T^m(x_1)=T^n(T^{n-m}(x_1))$. However, since $T$ is injective on the orbit of $x_1$, this implies that $x_1=T^{m-n}(x_1)$.
Now $\mu(\cdot)=\mu(\cdot |\mathcal{O}(x_1))\mu(\mathcal{O}(x_1))+\mu(\cdot|X\setminus \mathcal{O}(x_1))(1-\mu(\mathcal{O}(x_1)))$ and we just proved that $$ \mu(\cdot |\mathcal{O}(x_1))=\frac{1}{m-n}\sum_{i=0}^{m-n-1} \delta_{T^{-i}(x_1)} $$ Can you see how to continue from here?