[proofplan]
Ergodicity is the assertion that invariant $L^2$ functions are constants. Weak mixing is the Koopman-von Neumann lemma together with the spectral fact that atoms are eigenvectors. Strong mixing is the Rajchman spectral characterisation. The final absolute-continuity sufficient condition follows from the Riemann-Lebesgue lemma.
[/proofplan]
[step:Identify ergodicity with the fixed space]
Assume first that $T$ is ergodic. Let $f\in H$ satisfy $U_Tf=f$. Write $f=u+iv$ with real-valued measurable representatives $u,v$. Then
\begin{align*}
u\circ T=u,
\qquad
v\circ T=v
\end{align*}
$\mu$-almost everywhere. For each rational $q$, the set $\{u>q\}$ is invariant modulo $\mu$-null sets. By the [Equivalence of Ergodicity Conditions](/theorems/3444), it has measure $0$ or $1$. The usual rational-level-set argument then gives a constant $c_u\in\mathbb{R}$ such that $u=c_u$ almost everywhere. The same argument gives $v=c_v$ almost everywhere. Hence $f=(c_u+ic_v)\mathbf{1}$ in $H$.
Conversely, assume
\begin{align*}
\ker(U_T-I_H)=\operatorname{span}_{\mathbb{C}}\{\mathbf{1}\}.
\end{align*}
If $A\in\mathcal{B}$ satisfies $\mu(T^{-1}A\triangle A)=0$, then $U_T\mathbb{1}_A=\mathbb{1}_A$ in $H$. Hence $\mathbb{1}_A=c\mathbf{1}$ almost everywhere for some $c\in\mathbb{C}$. Since an indicator takes only the values $0$ and $1$, $c\in\{0,1\}$, and therefore $\mu(A)\in\{0,1\}$. By the invariant-set criterion in the [Equivalence of Ergodicity Conditions](/theorems/3444), $T$ is ergodic.
[/step]
[step:Identify weak mixing spectrally]
By the [Koopman-von Neumann Lemma](/theorems/3457), $T$ is weakly mixing if and only if $U_T$ has no non-constant eigenfunctions, equivalently if and only if $U_T|_{H_0}$ has no eigenvectors.
For $f\in H_0$, the spectral theorem gives
\begin{align*}
\langle U_T^n f,f\rangle=\widehat{\sigma_f}(n).
\end{align*}
Wiener's Fourier-coefficient criterion gives
\begin{align*}
\lim_{N\to\infty}\frac1N\sum_{n=0}^{N-1}|\widehat{\sigma_f}(n)|^2
=
\sum_{\lambda\in\mathbb{T}}\sigma_f(\{\lambda\})^2.
\end{align*}
Thus $\sigma_f$ has an atom at $\lambda$ exactly when the spectral projection of $f$ onto the $\lambda$-eigenspace is non-zero. Equivalently, all zero-mean spectral measures are atomless exactly when $U_T|_{H_0}$ has no eigenvectors. This proves the weak-mixing characterisation.
[/step]
[step:Identify strong mixing spectrally]
The [Spectral Characterisation of Strong Mixing](/theorems/3456) says exactly that $T$ is strongly mixing if and only if every zero-mean spectral measure $\sigma_f$ is Rajchman:
\begin{align*}
\widehat{\sigma_f}(n)\to0.
\end{align*}
This proves the necessary and sufficient spectral condition for strong mixing.
[/step]
[step:Derive the absolute-continuity sufficient condition]
Assume that, for every $f\in H_0$, the measure $\sigma_f$ is absolutely continuous with respect to Haar probability measure $m_{\mathbb{T}}$ on $\mathbb{T}$. Then there exists $\rho_f\in L^1(\mathbb{T},m_{\mathbb{T}})$ such that
\begin{align*}
d\sigma_f=\rho_f\,dm_{\mathbb{T}}.
\end{align*}
By the [Riemann Lebesgue Lemma](/theorems/526),
\begin{align*}
\widehat{\sigma_f}(n)
=
\int_{\mathbb{T}}z^n\rho_f(z)\,dm_{\mathbb{T}}(z)
\longrightarrow0.
\end{align*}
So every zero-mean spectral measure is Rajchman. By the preceding step, $T$ is strongly mixing.
[/step]
