[proofplan] We first verify that $\mathcal{I}$ is a sub-$\sigma$-algebra of $\mathcal{B}$ and observe that conditions $(i)$ and $(iii)$ are literal rephrasings of each other in the language of $\mathcal{I}$. We then translate condition $(ii)$ into the assertion that every $\mu$-a.e. $T$-invariant $L^2$ representative is $\mu$-a.e. constant. The substantive implication $(i)\Rightarrow(ii)$ uses rational superlevel sets: a fixed real-valued $L^2$ representative is $\mu$-a.e. constant because each of its rational superlevel sets is almost invariant, hence has measure $0$ or $1$ by ergodicity. The converse $(ii)\Rightarrow(i)$ applies the fixed-function hypothesis to the indicator of an almost invariant set, which forces the set to have measure $0$ or $1$. Combining the implications gives the three-way equivalence. [/proofplan]

[step:Identify the $\sigma$-algebra of almost invariant sets and obtain $(i)\iff(iii)$]We verify that $\mathcal{I}$ is a sub-$\sigma$-algebra of $\mathcal{B}$. First, $X\in\mathcal{I}$ because $T^{-1}X=X$, so $\mu(T^{-1}X\triangle X)=\mu(\varnothing)=0$. Next, let $A\in\mathcal{I}$. Preimages commute with complements, $T^{-1}(X\setminus A)=X\setminus T^{-1}A$, and the symmetric difference is invariant under simultaneously complementing both arguments, $(X\setminus B)\triangle(X\setminus A)=B\triangle A$. Hence pointwise \begin{align*} T^{-1}(X\setminus A)\triangle(X\setminus A)=(X\setminus T^{-1}A)\triangle(X\setminus A)=T^{-1}A\triangle A, \end{align*} so $\mu(T^{-1}(X\setminus A)\triangle(X\setminus A))=\mu(T^{-1}A\triangle A)=0$ and $X\setminus A\in\mathcal{I}$. Finally, let $(A_m)_{m=1}^{\infty}$ be a sequence in $\mathcal{I}$. Pointwise, \begin{align*} T^{-1}\left(\bigcup_{m=1}^{\infty}A_m\right)\triangle\bigcup_{m=1}^{\infty}A_m \subseteq\bigcup_{m=1}^{\infty}(T^{-1}A_m\triangle A_m), \end{align*} because if a point $x$ lies in the left-hand symmetric difference then either $Tx\in\bigcup_m A_m$ and $x\notin\bigcup_m A_m$, or vice versa: in the first case there is some index $m$ with $Tx\in A_m$, while $x\notin A_m$ since $x\notin\bigcup_m A_m$, so $x\in T^{-1}A_m\triangle A_m$; the second case is symmetric. By [Countable Subadditivity](/theorems/1108) applied to the right-hand union, \begin{align*} \mu\left(T^{-1}\bigcup_{m=1}^{\infty}A_m\triangle\bigcup_{m=1}^{\infty}A_m\right) \le\sum_{m=1}^{\infty}\mu(T^{-1}A_m\triangle A_m)=0. \end{align*} Thus $\bigcup_{m=1}^{\infty}A_m\in\mathcal{I}$, and $\mathcal{I}$ is a sub-$\sigma$-algebra of $\mathcal{B}$. By definition, $A\in\mathcal{B}$ is almost invariant in the sense of $(i)$ exactly when $A\in\mathcal{I}$. Hence the assertion in $(i)$ that every almost invariant set $A\in\mathcal{B}$ has $\mu(A)\in\{0,1\}$ is verbatim the assertion in $(iii)$ that every $A\in\mathcal{I}$ has $\mu(A)\in\{0,1\}$. Therefore $(i)\iff(iii)$.[/step]

[guided]We begin by giving $\mathcal{I}$ the structure we will need: it contains $X$ and is closed under complements and countable unions, hence is a sub-$\sigma$-algebra of $\mathcal{B}$. The countable-union closure is what lets us push the $\mu$-null hypothesis through the symmetric difference via countable subadditivity. First, $X\in\mathcal{I}$: $T^{-1}X=X$ because the preimage of the whole space under any map is the whole space, so $T^{-1}X\triangle X=\varnothing$ and $\mu(\varnothing)=0$. Next, take $A\in\mathcal{I}$. The key identity for closure under complements is \begin{align*} T^{-1}(X\setminus A)\triangle(X\setminus A)=T^{-1}A\triangle A. \end{align*} This follows from two basic facts: preimages commute with complements, $T^{-1}(X\setminus A)=X\setminus T^{-1}A$, and symmetric differences are invariant under simultaneously complementing both arguments, $(X\setminus B)\triangle(X\setminus A)=B\triangle A$. Applying $\mu$ gives $\mu(T^{-1}(X\setminus A)\triangle(X\setminus A))=0$, so $X\setminus A\in\mathcal{I}$. For closure under countable unions, let $(A_m)_{m=1}^{\infty}$ be a sequence in $\mathcal{I}$. We claim the pointwise inclusion \begin{align*} T^{-1}\left(\bigcup_{m=1}^{\infty}A_m\right)\triangle\bigcup_{m=1}^{\infty}A_m \subseteq\bigcup_{m=1}^{\infty}(T^{-1}A_m\triangle A_m). \end{align*} Indeed, if $x$ lies on the left, then either $Tx\in\bigcup_m A_m$ and $x\notin\bigcup_m A_m$, or vice versa. In the first case there is some $m$ with $Tx\in A_m$, while $x\notin A_m$ since $x\notin\bigcup_m A_m$, so $x\in T^{-1}A_m\triangle A_m$. The second case is symmetric. By [Countable Subadditivity](/theorems/1108), and using that each $\mu(T^{-1}A_m\triangle A_m)=0$, \begin{align*} \mu\left(T^{-1}\bigcup_{m=1}^{\infty}A_m\triangle\bigcup_{m=1}^{\infty}A_m\right) \le\sum_{m=1}^{\infty}\mu(T^{-1}A_m\triangle A_m)=0. \end{align*} Hence $\bigcup_{m=1}^{\infty}A_m\in\mathcal{I}$, completing the verification that $\mathcal{I}$ is a sub-$\sigma$-algebra of $\mathcal{B}$. Finally, comparing definitions: in $(i)$, a set $A\in\mathcal{B}$ is almost invariant exactly when $\mu(T^{-1}A\triangle A)=0$, which by definition of $\mathcal{I}$ is the condition $A\in\mathcal{I}$. So $(i)$ is the statement 'every $A\in\mathcal{I}$ has $\mu(A)\in\{0,1\}$', which is $(iii)$. Thus $(i)\iff(iii)$.[/guided]

[step:Translate the Koopman fixed space into $\mu$-a.e. invariance]Let $[f]$ denote the $\mu$-a.e. equivalence class of a measurable representative $f:X\to\mathbb{K}$ with $\int_X|f|^2\,d\mu<\infty$. We first verify that the Koopman operator $U_T$ is well defined. Since $T$ is measure preserving, the pushforward measure satisfies $\mu\circ T^{-1}=\mu$. By [Change of Variables (general)](/theorems/22) through the pushforward identity, \begin{align*} \int_X|f(Tx)|^2\,d\mu(x)=\int_X|f(y)|^2\,d(\mu\circ T^{-1})(y)=\int_X|f(y)|^2\,d\mu(y)<\infty, \end{align*} so $f\circ T\in L^2(X,\mathcal{B},\mu)$. If $f=g$ $\mu$-a.e., then \begin{align*} \{x\in X:f(Tx)\ne g(Tx)\}=T^{-1}\{y\in X:f(y)\ne g(y)\}, \end{align*} which has $\mu$-measure $0$ because $T$ is measure preserving and the inner set is $\mu$-null. Hence $[f\circ T]$ depends only on $[f]$, and $U_T$ is well defined. For any representative $f$ of $[f]$, the definition of equality of $L^2$ classes gives \begin{align*} U_T[f]=[f] \quad\Longleftrightarrow\quad [f\circ T]=[f] \quad\Longleftrightarrow\quad f\circ T=f\ \mu\text{-a.e.} \end{align*} Thus condition $(ii)$ is precisely the assertion that every measurable representative $f:X\to\mathbb{K}$ of an element of $L^2(X,\mathcal{B},\mu)$ satisfying $f\circ T=f$ $\mu$-a.e. is $\mu$-a.e. constant.[/step]

[guided]The goal of this step is to convert the operator-theoretic condition $U_T[f]=[f]$ into the pointwise condition $f\circ T=f$ $\mu$-a.e., which is what the next step will use to apply ergodicity. First we have to make sense of $U_T$ as a map of equivalence classes. Two things must be checked: (a) $f\circ T\in L^2$ whenever $f\in L^2$; (b) $[f\circ T]$ depends only on $[f]$. For (a), the measure-preserving hypothesis $\mu(T^{-1}A)=\mu(A)$ for all $A\in\mathcal{B}$ is equivalent to the assertion that the pushforward measure $\mu\circ T^{-1}$ equals $\mu$. [Change of Variables (general)](/theorems/22) through the pushforward identity then gives \begin{align*} \int_X|f(Tx)|^2\,d\mu(x)=\int_X|f(y)|^2\,d(\mu\circ T^{-1})(y)=\int_X|f(y)|^2\,d\mu(y)<\infty, \end{align*} verifying that $f\circ T\in L^2(X,\mathcal{B},\mu)$. For (b), suppose $f=g$ $\mu$-a.e., i.e. $\mu\{y\in X:f(y)\ne g(y)\}=0$. The set where $f\circ T$ and $g\circ T$ disagree is exactly the preimage of the set where $f$ and $g$ disagree: \begin{align*} \{x\in X:f(Tx)\ne g(Tx)\}=T^{-1}\{y\in X:f(y)\ne g(y)\}. \end{align*} Measure preservation gives this preimage $\mu$-measure $0$, so $f\circ T=g\circ T$ $\mu$-a.e., hence $[f\circ T]=[g\circ T]$. With $U_T$ well defined, the equivalence \begin{align*} U_T[f]=[f] \quad\Longleftrightarrow\quad f\circ T=f\ \mu\text{-a.e.} \end{align*} is just the definition of equality of equivalence classes in $L^2$. So condition $(ii)$ becomes the pointwise statement that every $L^2$ representative with $f\circ T=f$ $\mu$-a.e. is $\mu$-a.e. constant.[/guided]

[step:Use rational superlevel sets to force a fixed real function to be constant, proving $(i)\Rightarrow(ii)$]Assume $T$ is ergodic, and let $[f]\in L^2(X,\mathcal{B},\mu)$ satisfy $U_T[f]=[f]$. We show $[f]$ is $\mu$-a.e. constant. **Reduction to a finite-valued real representative.** Consider first the real case $\mathbb{K}=\mathbb{R}$. Pick any measurable representative $f_0:X\to[-\infty,+\infty]$ of $[f]$ with $\int_X|f_0|^2\,d\mu<\infty$. Since $|f_0|^2$ is finite $\mu$-a.e. (an extended-real function with finite Lebesgue integral takes the value $\pm\infty$ only on a $\mu$-null set), the set \begin{align*} N_\infty=\{x\in X:|f_0(x)|=\infty\} \end{align*} satisfies $\mu(N_\infty)=0$. Define a new representative \begin{align*} f:X&\to\mathbb{R}\\ x&\mapsto\begin{cases}f_0(x),&x\notin N_\infty,\\ 0,&x\in N_\infty.\end{cases} \end{align*} Then $f$ is $(\mathcal{B},\mathcal{B}(\mathbb{R}))$-measurable, finite-valued, and equals $f_0$ outside the $\mu$-null set $N_\infty$, so $[f]=[f_0]$. The condition $f_0\circ T=f_0$ $\mu$-a.e. transfers to $f\circ T=f$ $\mu$-a.e. because modification on the $\mu$-null set $N_\infty$ changes both sides of the equation only on a $\mu$-null set, namely on $N_\infty\cup T^{-1}N_\infty$, which has $\mu$-measure $0$ by measure preservation and subadditivity. Define the exceptional set \begin{align*} N=\{x\in X:f(Tx)\ne f(x)\}\in\mathcal{B},\qquad\mu(N)=0. \end{align*} **Rational superlevel sets are almost invariant.** Let $\mathbb{Q}\subseteq\mathbb{R}$ denote the rational numbers. For each $q\in\mathbb{Q}$ define \begin{align*} A_q=\{x\in X:f(x)>q\}\in\mathcal{B}, \end{align*} which is measurable because $f$ is. For $x\in X\setminus N$ we have $f(Tx)=f(x)$, so \begin{align*} x\in T^{-1}A_q &\iff f(Tx)>q\\ &\iff f(x)>q\\ &\iff x\in A_q, \end{align*} whence $T^{-1}A_q\triangle A_q\subseteq N$. Therefore $\mu(T^{-1}A_q\triangle A_q)\le\mu(N)=0$, i.e. $A_q\in\mathcal{I}$. By the ergodicity hypothesis, \begin{align*} \mu(A_q)\in\{0,1\}\quad\text{for every }q\in\mathbb{Q}. \end{align*} **Extracting a real constant.** Define \begin{align*} S=\{q\in\mathbb{Q}:\mu(A_q)=1\}\subseteq\mathbb{Q}. \end{align*} We show $S$ is nonempty. If $S=\varnothing$, then $\mu(A_{-m})=0$ for every $m\in\mathbb{N}$. Because $f$ is finite-valued, $X=\bigcup_{m=1}^{\infty}A_{-m}$, and [Countable Subadditivity](/theorems/1108) yields \begin{align*} 1=\mu(X)\le\sum_{m=1}^{\infty}\mu(A_{-m})=0, \end{align*} a contradiction. Hence $S\ne\varnothing$. Next we show $S\ne\mathbb{Q}$. If $S=\mathbb{Q}$, then $\mu(A_m)=1$ for every $m\in\mathbb{N}$, so by countable subadditivity \begin{align*} \mu\left(X\setminus\bigcap_{m=1}^{\infty}A_m\right) =\mu\left(\bigcup_{m=1}^{\infty}(X\setminus A_m)\right) \le\sum_{m=1}^{\infty}\mu(X\setminus A_m)=0, \end{align*} so $\mu\bigl(\bigcap_{m=1}^{\infty}A_m\bigr)=1$. But finite-valuedness of $f$ forces $\bigcap_{m=1}^{\infty}A_m=\varnothing$, contradicting $\mu(X)=1$. The set $S$ is downward closed in $\mathbb{Q}$: if $q,r\in\mathbb{Q}$ with $q<r$ and $r\in S$, then $A_r\subseteq A_q$, so \begin{align*} 1=\mu(A_r)\le\mu(A_q)\le 1, \end{align*} and $q\in S$. Since $S$ is nonempty, downward closed, and not all of $\mathbb{Q}$ (in particular, bounded above by some rational $r\notin S$), the real supremum \begin{align*} c_*=\sup_{\mathbb{R}}S\in\mathbb{R} \end{align*} is finite. Every rational $q<c_*$ lies in $S$ (otherwise $c_*$ would not be the supremum, by downward closure), and every rational $q>c_*$ lies outside $S$. Hence $\mu(A_q)=1$ for $q\in\mathbb{Q}\cap(-\infty,c_*)$ and $\mu(A_q)=0$ for $q\in\mathbb{Q}\cap(c_*,\infty)$. Define the full-measure pinning set \begin{align*} B_-&=\bigcap_{\substack{q\in\mathbb{Q}\\ q<c_*}}A_q,\\ B_+&=\bigcap_{\substack{q\in\mathbb{Q}\\ q>c_*}}(X\setminus A_q),\\ B&=B_-\cap B_+. \end{align*} Each intersection is countable because $\mathbb{Q}$ is countable, so $B_-,B_+,B\in\mathcal{B}$. By countable subadditivity, \begin{align*} \mu(X\setminus B_-)&\le\sum_{\substack{q\in\mathbb{Q}\\ q<c_*}}\mu(X\setminus A_q)=0,\\ \mu(X\setminus B_+)&\le\sum_{\substack{q\in\mathbb{Q}\\ q>c_*}}\mu(A_q)=0, \end{align*} so $\mu(B)=1$. Let $x\in B$. If $f(x)<c_*$, then by [Density of Rationals](/theorems/740) there is $q\in\mathbb{Q}$ with $f(x)<q<c_*$; since $q<c_*$ we have $x\in B_-\subseteq A_q$, forcing $f(x)>q$, contradiction. Hence $f(x)\ge c_*$. If $f(x)>c_*$, then density gives $q\in\mathbb{Q}$ with $c_*<q<f(x)$; since $q>c_*$ we have $x\in B_+\subseteq X\setminus A_q$, forcing $f(x)\le q$, contradiction. Hence $f(x)\le c_*$. Therefore $f(x)=c_*$ for every $x\in B$, and $\mu(B)=1$ gives $f=c_*$ $\mu$-a.e. **The complex case.** Suppose now $\mathbb{K}=\mathbb{C}$. Choose a measurable representative $f:X\to\mathbb{C}$ of $[f]$ with $\int_X|f|^2\,d\mu<\infty$ and $f\circ T=f$ $\mu$-a.e. Then $\operatorname{Re}f,\operatorname{Im}f:X\to\mathbb{R}$ are measurable, satisfy $|\operatorname{Re}f|,|\operatorname{Im}f|\le|f|$, hence both lie in $L^2(X,\mathcal{B},\mu)$ (with real scalars). The identity $f\circ T=f$ $\mu$-a.e. gives, by taking real and imaginary parts, \begin{align*} (\operatorname{Re}f)\circ T&=\operatorname{Re}f\ \mu\text{-a.e.},\\ (\operatorname{Im}f)\circ T&=\operatorname{Im}f\ \mu\text{-a.e.} \end{align*} Applying the real case to each yields real constants $a_*,b_*\in\mathbb{R}$ with $\operatorname{Re}f=a_*$ and $\operatorname{Im}f=b_*$ $\mu$-a.e., so $f=a_*+ib_*$ $\mu$-a.e.[/step]

[guided]Assume $T$ is ergodic and that $[f]\in L^2(X,\mathcal{B},\mu)$ satisfies $U_T[f]=[f]$, so by the previous step we may work with a representative $f$ satisfying $f\circ T=f$ $\mu$-a.e. We prove $f$ is $\mu$-a.e. constant. We begin with the real case $\mathbb{K}=\mathbb{R}$. Why are we allowed to assume $f$ is finite-valued? Because $\int_X|f|^2\,d\mu<\infty$ forces $|f|<\infty$ on a set of full measure: if $|f|=\infty$ on a measurable set $E$ with $\mu(E)>0$, the integral over $E$ would be $\infty$, contradicting integrability. Redefining $f$ to be $0$ on the $\mu$-null set $N_\infty=\{|f|=\infty\}$ produces a measurable representative of the same equivalence class that is everywhere finite. Modifying $f$ on a $\mu$-null set does not affect the $\mu$-a.e. invariance $f\circ T=f$: both sides change only on the $\mu$-null set $N_\infty\cup T^{-1}N_\infty$ (the second piece is $\mu$-null by measure preservation), so the set where they disagree changes only by a $\mu$-null set. Without loss of generality, then, $f:X\to\mathbb{R}$ is finite-valued, measurable, and $f\circ T=f$ outside a measurable $\mu$-null exceptional set \begin{align*} N=\{x\in X:f(Tx)\ne f(x)\},\qquad\mu(N)=0. \end{align*} The key idea is to translate the statement '$f$ is $\mu$-a.e. constant' into a statement about sets, where ergodicity applies. The natural set-valued data attached to $f$ is its family of superlevel sets. Let $\mathbb{Q}\subseteq\mathbb{R}$ denote the rational numbers and for each $q\in\mathbb{Q}$ define \begin{align*} A_q=\{x\in X:f(x)>q\}. \end{align*} Measurability of $f$ gives $A_q\in\mathcal{B}$. Why use only rationals? Because we will take a countable intersection of full-measure events, and countability is exactly what makes such an intersection still full-measure. For $x\in X\setminus N$, $f(Tx)=f(x)$, so \begin{align*} x\in T^{-1}A_q &\iff f(Tx)>q\\ &\iff f(x)>q\\ &\iff x\in A_q. \end{align*} Thus $T^{-1}A_q\triangle A_q\subseteq N$, so $\mu(T^{-1}A_q\triangle A_q)\le\mu(N)=0$ and $A_q$ is almost invariant. Ergodicity gives \begin{align*} \mu(A_q)\in\{0,1\}\quad\text{for every }q\in\mathbb{Q}. \end{align*} Now define the set of rational thresholds whose superlevel sets have full measure: \begin{align*} S=\{q\in\mathbb{Q}:\mu(A_q)=1\}. \end{align*} We claim $S$ is nonempty and not all of $\mathbb{Q}$. If $S=\varnothing$, then $\mu(A_{-m})=0$ for every $m\in\mathbb{N}$. But finiteness of $f$ means every $x\in X$ satisfies $f(x)>-m$ for some $m\in\mathbb{N}$, so $X=\bigcup_{m=1}^{\infty}A_{-m}$. By [Countable Subadditivity](/theorems/1108), \begin{align*} 1=\mu(X)\le\sum_{m=1}^{\infty}\mu(A_{-m})=0, \end{align*} impossible. Hence $S\ne\varnothing$. If $S=\mathbb{Q}$, then $\mu(A_m)=1$ for every $m\in\mathbb{N}$, and countable subadditivity gives \begin{align*} \mu\left(X\setminus\bigcap_{m=1}^{\infty}A_m\right) =\mu\left(\bigcup_{m=1}^{\infty}(X\setminus A_m)\right) \le\sum_{m=1}^{\infty}\mu(X\setminus A_m)=0, \end{align*} making $\bigcap_{m=1}^{\infty}A_m$ full-measure. But '$f(x)>m$ for every $m\in\mathbb{N}$' is impossible for a finite real number, so this intersection is empty, contradicting $\mu(X)=1$. Hence $S\ne\mathbb{Q}$. Furthermore $S$ is downward closed: if $q,r\in\mathbb{Q}$ with $q<r$ and $r\in S$, then $A_r\subseteq A_q$ and \begin{align*} 1=\mu(A_r)\le\mu(A_q)\le 1, \end{align*} forcing $q\in S$. Since $S$ is nonempty, downward closed, and bounded above (it omits some rational $r\notin S$, and by downward closure $r$ is an upper bound), the real supremum \begin{align*} c_*=\sup_{\mathbb{R}}S \end{align*} is finite. The supremum together with downward closure gives: every rational $q<c_*$ is in $S$ (so $\mu(A_q)=1$), and every rational $q>c_*$ is outside $S$ (so $\mu(A_q)=0$). We extract the candidate value of $f$ by forming the countable intersections \begin{align*} B_-&=\bigcap_{\substack{q\in\mathbb{Q}\\ q<c_*}}A_q,\\ B_+&=\bigcap_{\substack{q\in\mathbb{Q}\\ q>c_*}}(X\setminus A_q),\\ B&=B_-\cap B_+. \end{align*} Countability of $\mathbb{Q}$ is essential: a union of countably many $\mu$-null sets is $\mu$-null. Concretely, \begin{align*} \mu(X\setminus B_-)&\le\sum_{\substack{q\in\mathbb{Q}\\ q<c_*}}\mu(X\setminus A_q)=0,\\ \mu(X\setminus B_+)&\le\sum_{\substack{q\in\mathbb{Q}\\ q>c_*}}\mu(A_q)=0, \end{align*} so $\mu(B)=1$. Finally, fix $x\in B$. If $f(x)<c_*$, [Density of Rationals](/theorems/740) gives $q\in\mathbb{Q}$ with $f(x)<q<c_*$; since $q<c_*$ the point $x\in B_-$ must lie in $A_q$, i.e. $f(x)>q$, contradiction. If $f(x)>c_*$, density gives $q\in\mathbb{Q}$ with $c_*<q<f(x)$; since $q>c_*$ the point $x\in B_+$ must lie in $X\setminus A_q$, i.e. $f(x)\le q$, contradiction. Therefore $f(x)=c_*$ for all $x\in B$, and $\mu(B)=1$ gives $f=c_*$ $\mu$-a.e. The complex case reduces to the real one: if $f:X\to\mathbb{C}$ is square-integrable with $f\circ T=f$ $\mu$-a.e., then $\operatorname{Re}f$ and $\operatorname{Im}f$ are real-valued and square-integrable (each is bounded in modulus by $|f|$), and each is fixed $\mu$-a.e. on the same null set on which $f$ is fixed. Applying the real case to each gives real constants $a_*,b_*$ with $\operatorname{Re}f=a_*$ and $\operatorname{Im}f=b_*$ $\mu$-a.e., so $f=a_*+ib_*$ $\mu$-a.e.[/guided]

[step:Use fixed indicators to force almost invariant sets to have measure $0$ or $1$, proving $(ii)\Rightarrow(i)$]Assume every $[f]\in L^2(X,\mathcal{B},\mu)$ with $U_T[f]=[f]$ is $\mu$-a.e. constant. Let $A\in\mathcal{B}$ satisfy $\mu(T^{-1}A\triangle A)=0$. Define the indicator map \begin{align*} \mathbb{1}_A:X&\to\{0,1\}\subseteq\mathbb{K}\\ x&\mapsto\begin{cases}1,&x\in A,\\ 0,&x\notin A.\end{cases} \end{align*} This is measurable because $\mathbb{1}_A^{-1}(\{1\})=A\in\mathcal{B}$ and $\mathbb{1}_A^{-1}(\{0\})=X\setminus A\in\mathcal{B}$, and \begin{align*} \int_X|\mathbb{1}_A(x)|^2\,d\mu(x)=\mu(A)\le\mu(X)=1<\infty, \end{align*} so $\mathbb{1}_A$ represents an element of $L^2(X,\mathcal{B},\mu)$. Pointwise, $\mathbb{1}_A(Tx)=1\iff Tx\in A\iff x\in T^{-1}A$, so $\mathbb{1}_A\circ T=\mathbb{1}_{T^{-1}A}$. Two indicators differ exactly on the symmetric difference of their underlying sets: \begin{align*} \{x\in X:\mathbb{1}_{T^{-1}A}(x)\ne\mathbb{1}_A(x)\}=T^{-1}A\triangle A. \end{align*} The right-hand side has $\mu$-measure $0$ by almost invariance, so $\mathbb{1}_A\circ T=\mathbb{1}_A$ $\mu$-a.e., i.e. $U_T[\mathbb{1}_A]=[\mathbb{1}_A]$. By the standing hypothesis, there is $c\in\mathbb{K}$ with $\mathbb{1}_A=c$ $\mu$-a.e. Let \begin{align*} E=\{x\in X:\mathbb{1}_A(x)=c\}\in\mathcal{B},\qquad\mu(E)=1. \end{align*} Since $\mathbb{1}_A(X)\subseteq\{0,1\}$, the equality $\mathbb{1}_A=c$ on the nonempty set $E$ forces $c\in\{0,1\}$. If $c=0$, then for every $x\in E$ we have $\mathbb{1}_A(x)=0$, so $x\notin A$. Thus $A\cap E=\varnothing$, i.e. $A\subseteq X\setminus E$, and \begin{align*} \mu(A)\le\mu(X\setminus E)=1-\mu(E)=0, \end{align*} so $\mu(A)=0$. If $c=1$, then $E\subseteq A$, hence $X\setminus A\subseteq X\setminus E$ and \begin{align*} \mu(X\setminus A)\le\mu(X\setminus E)=0, \end{align*} so $\mu(A)=\mu(X)-\mu(X\setminus A)=1$. In either case $\mu(A)\in\{0,1\}$. Since $A$ was an arbitrary almost invariant measurable set, condition $(i)$ holds.[/step]

[guided]Assume every $L^2$ equivalence class fixed by $U_T$ is $\mu$-a.e. constant. To prove ergodicity we take an arbitrary almost invariant $A\in\mathcal{B}$, so $\mu(T^{-1}A\triangle A)=0$, and show $\mu(A)\in\{0,1\}$. The strategy is to convert the set $A$ into a function via its indicator, apply the hypothesis to the indicator, and read back the conclusion. Define \begin{align*} \mathbb{1}_A:X&\to\{0,1\}\subseteq\mathbb{K}\\ x&\mapsto\begin{cases}1,&x\in A,\\ 0,&x\notin A.\end{cases} \end{align*} We check the $L^2$ hypothesis. Measurability: $\mathbb{1}_A^{-1}(\{1\})=A\in\mathcal{B}$ and $\mathbb{1}_A^{-1}(\{0\})=X\setminus A\in\mathcal{B}$. Square-integrability uses that $\mu$ is a probability measure: \begin{align*} \int_X|\mathbb{1}_A(x)|^2\,d\mu(x)=\int_A 1\,d\mu(x)=\mu(A)\le\mu(X)=1<\infty. \end{align*} So $\mathbb{1}_A$ represents an element of $L^2(X,\mathcal{B},\mu)$. Next we check the fixed-class condition. Pointwise, $\mathbb{1}_A(Tx)=1$ exactly when $Tx\in A$, i.e. when $x\in T^{-1}A$. Therefore $\mathbb{1}_A\circ T=\mathbb{1}_{T^{-1}A}$. Two indicators of measurable sets differ exactly on the symmetric difference of those sets: \begin{align*} \{x\in X:\mathbb{1}_{T^{-1}A}(x)\ne\mathbb{1}_A(x)\}=T^{-1}A\triangle A, \end{align*} which has $\mu$-measure $0$ by almost invariance. Hence $\mathbb{1}_A\circ T=\mathbb{1}_A$ $\mu$-a.e., so $U_T[\mathbb{1}_A]=[\mathbb{1}_A]$. By the standing hypothesis, there exists $c\in\mathbb{K}$ with $\mathbb{1}_A=c$ $\mu$-a.e. Let $E=\{x\in X:\mathbb{1}_A(x)=c\}$, so $\mu(E)=1$; in particular $E\ne\varnothing$. The value $c$ must lie in the range $\mathbb{1}_A(E)\subseteq\{0,1\}$, so $c\in\{0,1\}$. The two cases give the two possible measures of $A$. If $c=0$: every $x\in E$ has $\mathbb{1}_A(x)=0$, i.e. $x\notin A$. So $E\cap A=\varnothing$, $A\subseteq X\setminus E$, and $\mu(A)\le\mu(X\setminus E)=1-\mu(E)=0$. If $c=1$: every $x\in E$ has $x\in A$, so $E\subseteq A$, $X\setminus A\subseteq X\setminus E$, and $\mu(X\setminus A)\le 0$, giving $\mu(A)=1$. Either way $\mu(A)\in\{0,1\}$. Since $A$ was an arbitrary almost invariant set, this is exactly condition $(i)$.[/guided]

[step:Combine the implications to obtain the three-way equivalence]The $\sigma$-algebra identification step established $(i)\iff(iii)$. The rational superlevel-set step established $(i)\Rightarrow(ii)$. The indicator step established $(ii)\Rightarrow(i)$. Together, the last two yield $(i)\iff(ii)$, and combining with $(i)\iff(iii)$ gives the three-way equivalence $(i)\iff(ii)\iff(iii)$.[/step]

[guided]The three implications established above close the equivalence: \begin{align*} (i)&\iff(iii)&&\text{($\sigma$-algebra identification step),}\\ (i)&\Rightarrow(ii)&&\text{(rational superlevel-set step),}\\ (ii)&\Rightarrow(i)&&\text{(indicator step).} \end{align*} The second and third together give $(i)\iff(ii)$. Combined with the first, this yields the three-way equivalence $(i)\iff(ii)\iff(iii)$, which is the conclusion of the theorem.[/guided]