This is from Peter Walters' An Introduction to Ergodic Theory. He quotes the "Spectral Theorem for Unitary Operators" which I understand:
If $U$ is a unitary operator on a complex Hilbert space $H$, then for each $f \in H$ there is a unique finite Borel measure $\mu_f$ on the circle $K=\{|z|=1\}\subseteq \mathbb{C}$ such that $$ \langle U^n f, f\rangle = \int_K \lambda^n \mu_f(d\lambda),\qquad \forall n \in \mathbb{Z}. $$
But he also remarks that if $T$ is measure preserving with measure preserving inverse then $U_Tf:= f \circ T$ is a unitary operator on $L^2$. Further (and this is the part I don't understand) if $1$ is the only eigenvalue of $U_T$ with constants the only eigenvectors and $\langle f,1\rangle = 0$, then $\mu_f$ has no atoms. I don't see how to show this. Clearly I can take $n=0$ and the hypotheses tell me $\mu_f(K) = 0$ but I don't see much else that can be said from this. Can I construct a non-constant eigenvector if I knew $\mu_f\{x\}>0$ for some $x$?