$\newcommand{\mc}{\mathcal}$ $\newcommand{\C}{\mathbb C}$
Let $(X, \mc X, \mu)$ be a probability space and $T:X\to X$ be an invertible measure preserving transformation. Let $U_T$ be the associated Koopman operator, and we will write $Tf$ in place of $U_Tf$. We define eigenvalues and eigenfunctions of this measure preserving systems in two cases.
The $L^\infty$ case. We say $\lambda\in \C$ is an eigenvalue if there is a nonzero $f\in L^\infty$ such that $Tf=\lambda f$. Given an eigenvalue $\lambda$, we say $f\in L^\infty$ is an eigenfunction if $Tf=\lambda f$. Let $\mc F_\infty$ be the collection of all the eigenfunctions in $L^\infty$.
The $L^2$ case. We say $\lambda\in \C$ is an eigenvalue if there is a nonzero $f\in L^2$ such that $Tf=\lambda f$. Given an eigenvalue $\lambda$, we say $f\in L^2$ is an eigenfunction if $Tf=\lambda f$. Let $\mc F_2$ be the collection of all the eigenfunctions in $L^2$.
It is clear that $\mc F_\infty\subseteq \mc F_2$. Let $\mc X_\infty$ and $\mc X_2$ be the $\sigma$-algebras generated by $\mc F_\infty$ and $\mc F_2$ respectively. So we have $\mc X_\infty\subseteq \mc X_2$.
Question. Is the containment $\mc X_\infty\subseteq \mc X_2$ actually an equality in general?
If we assume that $T$ is ergodic then any eigenfunction $f\in L^2$ has the property that $|f|$ is constant $\mu$-a.e. So there is no difference between $L^2$-eigenfunctions and $L^\infty$-eigenfunctions in the case that $T$ is ergodic.
For the general case I thought of appealing to ergodic decomposition (by assuming that $(X, \mc X)$ is regular). Let $\mc E$ be the $\sigma$-algebra of all the $T$-invariant sets (up to a set of measure $0$), and $x\mapsto \mu_x^{\mc E}:X\to \mathscr P(X)$ be the ergodic decomposition of $\mu$, where $\mathscr P(X)$ denotes the set of all the probability measures on $X$. Assume that $\mc E$ is countably generated. Each ($\mu$-almost each actually) $\mu_x^{\mc E}$ is an ergodic $T$-invariant measure with the property that $\mu_{Tx}^{\mc E} = \mu_x^{\mc E}$. Further, if we define $[x]_{\mc E}=\bigcap_{x\in E\in \mc E} E$, then $\mu_x^{\mc E}([x]_{\mc E})=1$.
Now let $f\in L^2$ be an arbitrary eigenfunction corresponding to an eigenvalue $\lambda$. Then the map $f1_{[x]_{\mc E}}$ is an eigenfunction in $L^2(X, \mc X, \mu_x^{\mc E})$ and therefore, by the fact that $\mu_x^{\mc E}$ is $T$-ergodic, it is an eigenfunction in $L^\infty(X, \mc X, \mu_x^{\mc E})$. In fact, $f1_{[x]_{\mc E}}$ is in $L^\infty(X, \mc X, \mu)$. So if there were only finitely many ergodic components then we could write $f$ as a sum of finitely many $L^\infty$-eigenfunctions. However, this is not clear when we have infinitely many ergodic components.