Consider $f \equiv 0$ on $I = [0,1]$. Define the equivalence class of functions, $X$, that are equal to $f$ almost everywhere (this is indeed an equivalence class). In this equivalence class, we have for any $E \subset [0,1]$ such that $m(E) = 0$,
$$\chi_E(x) = \begin{cases}1 & x \in E \\ 0 & x \not \in E \end{cases}$$
and clearly $\chi_E \sim f \implies \chi_E \in X$. Moreover, given $E_1, E_2 \subset I$ both of measure zero, we know that $\chi_{E_1 \cup E_2} \in X$. What if we repeat this process uncountably many times for some well-ordered set $\{E_{a}\}$, verifying at each step that $\chi_{G_{b}} \in X$ (where $G_{b}$ is the aggregate set up through $E_{b}$, so that $G_{b} = E_{a_1} \cup E_{a_2} \dots \cup E_{b}$ obeying the well-ordering of the uncountable set), but so that for
$$G = \cup_a E_a \implies m(G) > 0$$
and we stop when $G_b$ reaches $G$ by this process of including more sets $1$ by $1$ in this uncountable induction. Then the equivalence relation no longer holds.
I have a feeling this isn't so much a measure theory question, but rather a question of countability and uncountability and infinite induction. I think my question boils down to: can algorithms like the above be repeated uncountably many times and still have meaning?
I would appreciate it if someone can fill in the technical details I'm missing about uncountable induction, and how this prevents a contradiction.
So $X$ is the set of functions on $I$ that are $0$ almost everywhere and your "process" takes the characteristic functions of two sets $E_1$, $E_2$ of measure $0$ and returns the characteristic function of their union $E_1 \cup E_2$. It's meaningful to think about iterating this process transfinitely, but you will never reach a fixed point: if $E$ has measure $0$ then you can always find a larger set $E' \supset E$ that also has measure $0$. To get a fixed point, you would need to apply Zorn's lemma, but that is not possible since the union of an arbitrary chain of measure $0$ subsets can have positive measure.