An uncountable chain of equivalence relations

Question

First, an example: We know that, for two real valued, Lebesgue-integrable functions, the relation "equals almost everywhere" is an equivalence relation. In particular, if $f_0$ is Lebesgue-integrable, then if one defines $f_1$ to be different from $f_0$ at exactly one point, then $f_0 \sim f_1$. But suppose we have an uncountable chain of such equivalence relations, could we somehow relate $f_0$ to some $f$ such that $f_0$ and $f$ differ on an uncountable set, and perhaps of non-zero measure?

Caveats that I'm thinking: one cannot make such an uncountable chain of equivalence relations. But surely with something equivalent to the axiom of choice, we can do this? My point is not really about the above example, but really more about making a lot of equivalences at once, and how much "a lot" can be.

Edit: Why I mentioned the axiom of choice (admittedly something I am really shaky about; I would welcome anyone to "school" me on the topic): An equivalence relation on a set (non-empty) set $X$ is a subset $R_2$ of the Cartesian product $X\times X$, where $a \sim b$ iff $(a,b) \in X \times X$. One can extend this to $X \times X \times X$, where $(a, b, c) \in R_3 \subseteq X^3$ iff $a \sim b \sim c$ (with $R_3$ the subset of equivalent stuff, and the extension done by the transitivity property). So then we have an infinite product $X^I$ where $I$ is uncountable, and for this to be guaranteed to be non-empty I need the axiom of choice. Of course, that doesn't say anything about subset $R_I$ of equivalent things... [Edit 3: Refer to Mike Haskel's comment for why this is flawed]

Edit 2: Going back to the example given, I could also consider using the well-ordering axiom -- I take a set of measure not zero, start enumerating all its elements by the axiom, and come up with my (dubious) chain by fudging elements belonging to the set.

Answer 1

The technique you're suggesting with "long chains of equivalent functions" can't work. We already know that, though, from how you posed the question: if it worked the way you want, we could use the technique to show that any two functions are equal almost everywhere, and that's not true. So let's explore your suggestion a bit and see exactly where it breaks down. For a little bit, we can forget about functions and measure theory: let's just say we're working with some set $X$ with an equivalence relation $\sim$.

The idea you're trying to use seems to be that of a "path" between points in $X$: we want to say that, e.g., if we know $x_0 \sim x_1 \sim x_2 \sim \ldots \sim x_n$, we can conclude $x_0 \sim x_n$. The question is whether it makes sense to consider infinite paths, and what such paths might look like. The answer is that there is a reasonable definition for what an infinite path would be. However, it doesn't allow us to do what you want.

(You mention the well-ordering axiom in one of your edits, so I'll assume you're familiar with ordinals and transfinite induction.)

Definition: A path is a sequence $(x_i)_{i < \alpha}$ of points in $X$, indexed by some nonzero ordinal $\alpha$, such that, for all $i > 0$, there is some $j < i$ such that $x_j \sim x_i$.

That is exactly what we need to prove by induction that every $x_i$ in the path is equivalent to $x_0$:

• If $i = 0$, then obviously $x_i \sim x_0$.
• If $i > 0$, then assume we know that, for all $j < i$, $x_j \sim x_0$. By the definition of path, we know that there is some $j < i$ such that $x_j \sim x_i$. From the induction hypothesis and the transitivity of $\sim$ we can conclude that $x_0 \sim x_i$.

If we weaken the definition to just say we're indexed by some ordered set, we won't be able to conclude that everything in the path is equivalent.

Now, let's move back to measure theory and look at what you're trying to do. You're trying to build a long path where, whenever in its construction we need to have an equivalence, we get that equivalence by having the functions differ at no more than one point. You're then trying to use the length of the path (i.e., the size of the index ordinal) to make up for the fact that each individual step is so small. As it turns out, ordinals don't work that way: even when they're very very large "going up," an essential fact about ordinals is that they're still finite "going down."

Theorem. A linear (i.e., total) order is furthermore a well order if and only if it contains no infinite descending chains $a_0 > a_1 > a_2 > \ldots$.

So, say we have some long path of functions $(f_i)_{i < \alpha}$ where, for all $i < \alpha$, $f_i \sim f_j$ for some $j < i$ by way of differing on only a single point. If we look at any $i_0 > 0$, then, there is some $i_1 < i_0$ where $f_{i_0}$ and $f_{i_1}$ differ only on a single point. If $i_1 > 0$ still, then do the same thing to get an $i_2$, etc. The point is that, no matter what, we will have $i_n = 0$ in only finitely many steps. So, in fact $f_{i_0}$ (and therefore any function in our sequence, since $i_0$ was arbitrary) differs from $f_0$ on only finitely many points!