Number of equivalence classes of functions of real variable with the a.e relation.

Question

What is the cardinal of the set $\mathcal{F}(\mathbb{R};X)/ \sim$ where $\sim$ is the relation $f\sim g \iff \mu(\{x\in \mathbb{R};f(x)\ne g(x)\})=0$ and $|X|=|\mathbb{R}|$? I guess that is $\mathfrak{c}$ but I can't prove it.

Answer 1

The answer is $2^c$....[1].. Let $D$ be the set of uncountable closed subsets of $R$. We have $|R|=c$ and $|d|=c$ for $d\in D$. We can show there exists a family $E$ of pairwise disjoint subsets of $R$ with $|E|=c$, such that $e \cap d \ne \phi$ for all $e \in E ,d \in D$. So if $\phi \ne S \subsetneqq E$ then $\cup S$ is not measurable...[2].. For $ \phi \ne S \subsetneqq E$ , let $f_S :R \to \{0,1 \}$ be the characteristic function of $\cup S$. Now for distinct $S,T$ with $\phi \ne S \subsetneqq E$ and $\phi \ne T \subsetneqq E$, we have $\phi \ne A \subsetneqq E$ where $A= (S-T)\cup (T-S)$, and $\{ x : f_S \ne f_T \} = \cup A$ which is non-measurable, so $\chi _S$ and $\chi _T$ are inequivalent...[3].. I will supply the details about the existence of $E$ if asked. It only depends on |$C|=c$, and $d \in C \implies |d|=c$. I am very very slow at typing in LaTeX and not very good with it yet. The reason that $\cup S$ is not measurable when $\phi \ne S \subsetneqq E$ is that the inner measure of $\cup S$ and the inner measure of $R- \cup S$ are both $0$ because neither of them has an uncountable closed subset.

Answer 2

I believe I have seen this result before: There are $c$ many pairwise disjoint subsets of $[0,1]$ whose Lebesgue outer measures are all $1.$ Assuming this, the cardinality sought in this setting is $2^c.$

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Added later: https://mathoverflow.net/questions/119646/is-there-a-maximum-to-the-amount-of-disjoint-non-measurable-subsets-of-the-unit