Are there more Lebesgue measurable or more non Lebesgue measurable functions?
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The sets are the same cardinality. Given a single unmeasurable function $g$, and any measurable function $f$, $f\rightarrow f+g$ is a 1-1 function from the set of measurable functions to the set of unmeasurable functions. On the other hand, let $C$ be the Cantor set (or any measure-zero subset of $\mathbb R$ with the same cardinality as $\mathbb R$.) Then let $\phi:\mathbb R \rightarrow C$ be a 1-1 correspondence, and, for any unmeasurable function $f:\mathbb R \rightarrow \mathbb R$, define $f^\phi:\mathbb R \rightarrow \mathbb R$ as $f^\phi(x)=0$ if $x\not\in C$ and $f^\phi(x)=f(\phi^{-1}(x))$ if $x\in C$. Then $f^\phi$ is measurable (since it's support is a subset of $C$) and $f^\phi = g^\phi$ if and only if $f=g$. Therefore, we have a 1-1 map from the set of unmeasurable functions to the set of measurable functions. So we have that the two sets are the same cardinality. If we define an equivalence relation $f \cong g$ if $\{x:f(x)\neq g(x)\}$ is measure zero, are the two sets, modulo this equivalence, still the same? Also, is there a Baire category-like sense in which the non-measurable functions are "larger?" |
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The total number of functions from $\mathbb{R}$ to itself is (assuming the Axiom of Choice for the cardinal arithmetic): $$\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_02^{\aleph_0}} = 2^{2^{\aleph_0}} = |\mathcal{P}(\mathbb{R})|.$$ I.e., it is equal to the cardinality of the power set of $\mathbb{R}$. As Thomas Andrew notes, if $N$ is any subset of $\mathbb{R}$ that is of measure $0$, then any function that is supported in $N$ is Lebesgue measurable; in particular, all characteristic functions of subsets of $N$ are Lebesgue measurable. There are $$|\mathcal{P}(N)| = 2^{|N|}$$ such functions. Letting $C$ be the Cantor set (or any measurable uncountable set of measure $0$), we obtain a lower bound for the set of Lebesgue measurable functions of $$\Bigl|\{\chi_A\mid A\subseteq C\}\Bigr| = \Bigl|\mathcal{P}(C)\Bigr| = 2^{|C|} = 2^{2^{\aleph_0}}.$$ Thus, the set of Lebesgue measurable functions has cardinality $2^{2^{\aleph_0}}$. In particular, there can be no more non-measurable functions than measurable functions. Conversely, still assuming the Axiom of Choice, let $V$ be a Vitali subset of $\mathbb{R}$ contained in $[0,1]$, and let $A=V+2$, so that $A$ is a nonmeasurable subset of $\mathbb{R}$, $A\subseteq [2,3]$. If $D$ is any subset of the Cantor set, then $A\cup D$ is nonmeasurable, so $\chi_{A\cup C}$ is a nonmeasurable function. Therefore, there are at least $$|\mathcal{P}(C)| = 2^{|C|} = 2^{2^{\aleph_0}}$$ nonmeasurable functions, and so there are exactly that many. Therefore, assuming the Axiom of Choice, the cardinality of the set of Lebesgue-measurable functions from $\mathbb{R}$ to $\mathbb{R}$, and the cardinality of the set of non-Lebesgue-measurable functions from $\mathbb{R}$ to $\mathbb{R}$, are equal, and equal the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. I believe the use of the Axiom of Choice is important here, since Solovay proved that it is consistent with ZF that all subsets of $\mathbb{R}$ are Lebesgue measurable, in which case every function would be Lebesgue measurable. |
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MORE in some sense... In ZF we can explicitly write down countably many Lebesgue measurable functions. (For example, a rational constant.) But we cannot explicitly write down (and prove in ZF) even one non-Lebesgue measurable function. SO in this sense there are more measurable functions! |
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