[proofplan] The proof has two directions. If $T$ is ergodic and $f$ is an invariant measurable function, then any nontrivial Borel level set of $f$ would give a measurable set invariant modulo null sets, contradicting ergodicity. Conversely, if a nontrivial invariant set exists, its indicator function is a nonconstant invariant measurable function. The only point requiring care is that all invariance is interpreted modulo $\mathbb P$-null sets. [/proofplan]

[step:Extract a nontrivial Borel level set from a nonconstant measurable function]Let $f:E\to\mathbb C$ be $\mathcal E/\mathcal B(\mathbb C)$-measurable and suppose that $f$ is not equal to a constant $\mathbb P$-a.e. Define the real-valued measurable maps $u:E\to\mathbb R$ by $u(x)=\operatorname{Re}(f(x))$ and $v:E\to\mathbb R$ by $v(x)=\operatorname{Im}(f(x))$. At least one of $u$ and $v$ is not equal to a constant $\mathbb P$-a.e.; otherwise $f=u+iv$ would be equal to a complex constant $\mathbb P$-a.e. Let $g:E\to\mathbb R$ denote one of these two functions that is not a.e. constant. We claim that there exists $t\in\mathbb R$ such that \begin{align*} 0<\mathbb P(\{x\in E:g(x)<t\})<1. \end{align*} Indeed, if no such $t$ existed, then each set $\{g<t\}$ would have probability $0$ or $1$. Define \begin{align*} S:=\{q\in\mathbb Q:\mathbb P(\{x\in E:g(x)<q\})=0\}. \end{align*} Set $a:=\sup S$. If $S$ is empty, then $\mathbb P(\{g<q\})=1$ for every $q\in\mathbb Q$. Hence the countable intersection $\bigcap_{q\in\mathbb Q}\{x\in E:g(x)<q\}$ has probability $1$, but this intersection is empty because $g$ is real-valued. If $S=\mathbb Q$, then $\mathbb P(\{g<q\})=0$ for every $q\in\mathbb Q$. Hence the countable intersection $\bigcap_{q\in\mathbb Q}\{x\in E:g(x)\ge q\}$ has probability $1$, but this intersection is empty because $g$ is real-valued. Hence $a\in\mathbb R$. Let $q\in\mathbb Q$ satisfy $q<a$. Since $a=\sup S$, there exists $r\in S$ with $q<r\le a$. By monotonicity, $\{g<q\}\subset\{g<r\}$, so $\mathbb P(\{g<q\})=0$. Therefore \begin{align*} \mathbb P(\{g<a\})=0, \end{align*} because $\{g<a\}$ is the countable union of the sets $\{g<q\}$ over rational $q<a$. For every rational number $q>a$, the definition of $a$ gives $\mathbb P(\{g<q\})=1$, and hence \begin{align*} \mathbb P(\{g\ge q\})=0. \end{align*} Taking the countable union over rational $q>a$ gives \begin{align*} \mathbb P(\{g>a\})=0. \end{align*} Thus $\mathbb P(\{g=a\})=1$, contradicting the choice of $g$ as not a.e. constant. This proves the claim. Choose $t\in\mathbb R$ with $0<\mathbb P(\{g<t\})<1$. If $g=u$, set $B:=\{z\in\mathbb C:\operatorname{Re}(z)<t\}$. If $g=v$, set $B:=\{z\in\mathbb C:\operatorname{Im}(z)<t\}$. In either case $B$ is a Borel subset of $\mathbb C$, and the measurable set $A:=f^{-1}(B)$ satisfies $0<\mathbb P(A)<1$.[/step]

[guided]The goal of this step is to turn the failure of being a.e. constant into an actual measurable set of probability strictly between $0$ and $1$. Since $f$ is complex-valued, we first reduce to a real-valued function. Define \begin{align*} u:E&\to\mathbb R \end{align*} by $u(x)=\operatorname{Re}(f(x))$ and define \begin{align*} v:E&\to\mathbb R \end{align*} by $v(x)=\operatorname{Im}(f(x))$. Both maps are $\mathcal E/\mathcal B(\mathbb R)$-measurable because the real and imaginary part maps $\mathbb C\to\mathbb R$ are $\mathcal B(\mathbb C)/\mathcal B(\mathbb R)$-measurable and $f$ is $\mathcal E/\mathcal B(\mathbb C)$-measurable. If both $u$ and $v$ were equal to constants a.e., say $u=c_1$ a.e. and $v=c_2$ a.e., then $f=c_1+ic_2$ a.e. This is excluded. Hence one of $u$ and $v$ is not equal to a constant a.e.; call that real-valued measurable function $g:E\to\mathbb R$. We now show that some strict sublevel set of $g$ has intermediate probability. Suppose the contrary: for every $t\in\mathbb R$, the set $\{x\in E:g(x)<t\}$ has probability either $0$ or $1$. Define \begin{align*} a:=\sup\{q\in\mathbb Q:\mathbb P(\{x\in E:g(x)<q\})=0\}. \end{align*} The use of rational numbers is important because it lets us take countable unions, so countable additivity applies. Let \begin{align*} S:=\{q\in\mathbb Q:\mathbb P(\{x\in E:g(x)<q\})=0\}. \end{align*} The set $S$ is neither empty nor all of $\mathbb Q$. If $S$ were empty, then $\mathbb P(\{g<q\})=1$ for every rational $q$. Taking the countable intersection over $q\in\mathbb Q$, the set \begin{align*} \bigcap_{q\in\mathbb Q}\{x\in E:g(x)<q\} \end{align*} would have probability $1$. But this intersection is empty: for each $x\in E$, the real number $g(x)$ is larger than some rational number. If $S=\mathbb Q$, then $\mathbb P(\{g<q\})=0$ for every rational $q$, so $\mathbb P(\{g\ge q\})=1$ for every rational $q$. Taking the countable intersection over $q\in\mathbb Q$, the set \begin{align*} \bigcap_{q\in\mathbb Q}\{x\in E:g(x)\ge q\} \end{align*} would have probability $1$. But this intersection is empty: for each $x\in E$, the real number $g(x)$ is smaller than some rational number. Thus $a=\sup S$ is a real number. Let $q\in\mathbb Q$ satisfy $q<a$. The supremum property gives some $r\in S$ with $q<r\le a$; otherwise $q$ would be an upper bound for $S$, contradicting $q<a=\sup S$. Since $r\in S$, we have $\mathbb P(\{g<r\})=0$, and monotonicity of measure applied to $\{g<q\}\subset\{g<r\}$ gives \begin{align*} \mathbb P(\{g<q\})=0. \end{align*} Since \begin{align*} \{g<a\}=\bigcup_{\substack{q\in\mathbb Q, q<a}}\{g<q\}, \end{align*} [countable subadditivity](/theorems/1108) gives \begin{align*} \mathbb P(\{g<a\})=0. \end{align*} For every rational $q>a$, the definition of $a$ gives \begin{align*} \mathbb P(\{g<q\})=1, \end{align*} and therefore \begin{align*} \mathbb P(\{g\ge q\})=0. \end{align*} Since \begin{align*} \{g>a\}=\bigcup_{\substack{q\in\mathbb Q, q>a}}\{g\ge q\}, \end{align*} we obtain \begin{align*} \mathbb P(\{g>a\})=0. \end{align*} Consequently $\mathbb P(\{g=a\})=1$, so $g$ is equal to the constant $a$ a.e., contradicting the choice of $g$. Therefore there is $t\in\mathbb R$ with \begin{align*} 0<\mathbb P(\{g<t\})<1. \end{align*} If $g=\operatorname{Re}(f)$, choose the Borel half-plane \begin{align*} B:=\{z\in\mathbb C:\operatorname{Re}(z)<t\}. \end{align*} If $g=\operatorname{Im}(f)$, choose the Borel half-plane \begin{align*} B:=\{z\in\mathbb C:\operatorname{Im}(z)<t\}. \end{align*} Then $A:=f^{-1}(B)$ is measurable, and by construction \begin{align*} 0<\mathbb P(A)<1. \end{align*} This is the nontrivial measurable set that will contradict ergodicity once we prove it is invariant modulo null sets.[/guided]

[step:Show that an invariant function has invariant Borel preimages modulo null sets] Assume that $f:E\to\mathbb C$ is $\mathcal E/\mathcal B(\mathbb C)$-measurable and satisfies $f\circ T=f$ $\mathbb P$-a.e. Define the null set $N:=\{x\in E:f(T(x))\ne f(x)\}$. Then $\mathbb P(N)=0$. Let $B\subset\mathbb C$ be Borel and set $A:=f^{-1}(B)$. Since $f$ is measurable, $A\in\mathcal E$. For every $x\in E\setminus N$, \begin{align*} x\in T^{-1}(A)\iff T(x)\in A\iff f(T(x))\in B\iff f(x)\in B\iff x\in A. \end{align*} Hence $T^{-1}(A)\triangle A\subset N$, and therefore $\mathbb P(T^{-1}(A)\triangle A)=0$. [/step]

[step:Use ergodicity to force invariant measurable functions to be constant] Assume that $T$ is ergodic. Let $f:E\to\mathbb C$ be $\mathcal E/\mathcal B(\mathbb C)$-measurable and satisfy $f\circ T=f$ $\mathbb P$-a.e. Suppose, for contradiction, that $f$ is not equal to any constant $\mathbb P$-a.e. By the first step, there exists a Borel set $B\subset\mathbb C$ such that the measurable set $A:=f^{-1}(B)$ satisfies $0<\mathbb P(A)<1$. By the second step, $\mathbb P(T^{-1}(A)\triangle A)=0$. Ergodicity then implies $\mathbb P(A)\in\{0,1\}$, contradicting the strict inequalities above. Hence $f$ is equal to a constant $\mathbb P$-a.e. [/step]

[step:Build a nonconstant invariant function from a nontrivial invariant set] Assume that every $\mathcal E/\mathcal B(\mathbb C)$-measurable function $f:E\to\mathbb C$ satisfying $f\circ T=f$ $\mathbb P$-a.e. is equal to a constant $\mathbb P$-a.e. Let $A\in\mathcal E$ satisfy $\mathbb P(T^{-1}(A)\triangle A)=0$. Define the indicator function $\mathbb 1_A:E\to\mathbb C$ by $\mathbb 1_A(x)=1$ if $x\in A$ and $\mathbb 1_A(x)=0$ if $x\notin A$. Since $A\in\mathcal E$, the function $\mathbb 1_A$ is measurable. For $x\notin T^{-1}(A)\triangle A$, membership of $x$ in $A$ is equivalent to membership of $T(x)$ in $A$, and therefore \begin{align*} \mathbb 1_A(T(x))=\mathbb 1_A(x). \end{align*} Since $\mathbb P(T^{-1}(A)\triangle A)=0$, this proves \begin{align*} \mathbb 1_A\circ T=\mathbb 1_A \end{align*} $\mathbb P$-a.e. By the hypothesis on invariant functions, $\mathbb 1_A$ is equal to a constant $\mathbb P$-a.e. If $\mathbb 1_A=c$ $\mathbb P$-a.e. for some $c\in\mathbb C$, then $c\in\{0,1\}$. Indeed, if $c\notin\{0,1\}$, then $\mathbb 1_A(x)\ne c$ for every $x\in E$, contradicting equality a.e. If $c=0$, then $\mathbb P(A)=0$. If $c=1$, then $\mathbb P(A)=1$. Thus every invariant set modulo null sets has probability $0$ or $1$, so $T$ is ergodic. [/step]