If $T$ has continuous spectrum and $(f, 1)=0$ then $\mu_{f}$ has no atoms. I'm reading Peter Walters' An Introduction to Ergodic Theory, and I don't understand his remark.
Following is a sort of lemma, called Spectral Theorem for Unitary Operators, to prove the equivalence of weak-mixing and having continuous spectrum:

Theorem 1.25 (Spectral Theorem for Unitary Operators) in above book
Suppose $U$ is a unitary operator on a complex Hilbert space $\mathcal{H}$. Then for each $f \in \mathcal{H}$ there exists a unique finite Borel measure $\mu_{f}$ on $K = \{ z \in \mathbb{C} \ : \ |z| = 1\}$ such that
$$(U^{n}f, f) = \int _{K} \lambda ^{n} d\mu_{f}(\lambda) \ \ \ \ \ \ \ \forall n \in \mathbb{Z}.$$

Then he remark that:

If $T$ is an invertible measure-preserving transformation then $U_{T}$ is unitary, and if $T$ has continuous spectrum and $(f, 1) = 0$ then $\mu_{f}$ has no atoms.

This is what I don't understand. How can we show that under such conditions, $\mu_{f}$ has no atoms?
I found a related reference, and here Theorem 1.4 is related to what I want to know. And he left my question as an easy(?) exercise.
Can somebody help me please?
 A: I wrote following for the similar question here.
It may helps you:

Theorem.
Let $U$ be a (linear) isometry of a separable Hilbert space $\mathcal{H}$, and $H_{e}(U)$ be the closure of the subspace spanned by the eigenvectors of $U$.
Then for $f \in \mathcal{H}$, $$f \perp H_{e}(U) \Rightarrow \mu_{f} \mbox{ has no atoms.}$$

Proof.
Let $f \perp H_{e}(U)$ and $W = e^{-2\pi i x}U$. Then by von Neumann ergodic theorem, let $$g = \lim \limits _{n \rightarrow \infty} {1 \over n} \sum \limits _{k = 0} ^{n-1} W^{k}f.$$ Then $Wg = g$ so that $g$ is an eigenvector of $U$ with eigenvalue $e^{2\pi i x}$ if $g \neq 0$.
Now we have $$\left\langle g, f \right\rangle  = \lim \limits _{n \rightarrow \infty} {1 \over n} \sum \limits _{k = 0} ^{n-1} e^{-2\pi i x k } \left\langle U^{k}f, f \right\rangle = \lim \limits _{n \rightarrow \infty} \int_{0} ^{1} {1 \over n} \sum \limits _{k = 0} ^{n-1} e^{-2\pi i (x-y) k } \mu_{f}(dy) = \mu_{f}(x).$$
Hence if $\mu_{f}(x) > 0$, then $g \in H_{e}(U)$ and $f \not\perp g$, which contradicts to the assumption $f \perp H_{e}(U)$. This completes the proof.


