Are there more Lebesgue measurable or more non Lebesgue measurable functions? Are there more Lebesgue measurable or more non Lebesgue measurable functions?
Does anybody see how to answer this. Please do tell.
 A: 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?"
A: 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. 
A: 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!
