Why is the equivalence relation in $L^p$ space of equal almost everywhere valid? I know that on $L^p(\mathbb{R}^n)$, two functions $f$, $g$ are regarded as the same, denoted $f\sim g$, if $f=g$ almost everywhere.
Transitivity of equivalence relation is satisfied, i.e. if $f\sim g$ and $g\sim h$, then $f\sim h$. And "countable transitivity" also holds, i.e., if $f \sim r_1$, $r_1 \sim r_2 \sim r_3 \sim \cdots \sim g$, then $f\sim g$. This is because a countable union of measure zero sets is still of measure zero. Now my question is why "uncountable transitivity" doesn't hold? Functions $\chi_{[0,s]}$ where $s$ ranges over $[0,1]$, cannot be all in the same equivalence class. But it seems plausible to connect any two such functions $\chi_{[0,s]}, \chi_{[0,t]}$by uncountably many equivalence relations. I think it might be because the equivalence relation cannot be used infinitely many times. But I'm still confused and I guess it is a little related to set theory. Could anyone say more about this?
 A: I think you need to be more careful when talking about "countable transitivity".
Let's suppose we have a sequence of functions $f_n$, all belonging to the same equivalence class. If I understand correctly, you want to claim that the limiting function $g = \lim_{n} f_n$ is also in the same equivalence class. This is no longer an assertion about the equivalence relation. It depends on how the notion of convergence interacts with the equivalence relation.
Suppose you are working in the space of sequences of natural numbers, i.e. functions from $\mathbb{N} \to \mathbb{N}$. Define an equivalence relationship on such sequences that says that $a(k) \sim b(k)$ if $\exists M$ such that $\forall m \geq M$, $a(m) = b(m)$. It's easy to find sequences of such sequences, $a_{n}(k)$, all belonging to the same equivalence class, where the pointwise limit $\lim_{n} a_{n}(k)$ converges and is no longer in the same equivalence class, e.g. by recursively adjusting the values of the sequence at higher and higher inputs. In this example, the convergence does not play well with the equivalence relation.
Returning to $L^p$ space, you do have "countable transitivity" because the condition underlying the equivalence relation, disagreeing on sets of measure zero, plays well with countable sequences. However, if you were to define an uncountable sequence of functions, each differing from the last at only a single point, and if you could make sense of the limit as a function, you would have a process that could transform any function into any other function. However, this kind of limit does not play nicely with the equivalence relation, because uncountable unions of measure zero sets may have positive measure.
In summary: the best you can get from transitivity is "finite transitivity", i.e. a finite sequence of objects, each in the same equivalence class as the last object, never leaves the class. Any kind of infinite transitivity depends on how the limits you are taking interact with the specific equivalence relation in question.
A: The transitivity of the equivalence relation is defined as $a \sim b$, $b\sim c$ implies $a \sim c$. This, by induction, only implies "finite transitivity", i.e. $a_1 \sim a_2 \sim \cdots \sim a_n$ implies $a_1 \sim a_n$. The "infinite transitivity" has nothing to do with the equivalence relation.

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*By "countable transitivity" I mean, $f_1$ differs from $f_2$ on a measure zero set $E_1$, $f_2$ differs from $f_3$ on a measure zero set $E_2$, and so on, and let $f_0$ differs from $f_1$ on the union $E_1 \cup E_2 \cup \cdots$, then we still have $f_0 \sim f_1$. This actually follows from the fact that a union of countably many measure zero sets is still of measure zero, and thus $f_0$ only differs from $f_1$ on a measure zero set. Then by how we define two functions are equivalent, we have $f_1 \sim f_0$. There is nothing to do with the transitivity property of the equivalence relation.


*"uncountable transitivity" does not hold at all.
