Is Conway's base-13 function measurable? Robin Chapman introduced me to Conway's Base 13 Function. Now, my real analysis is a tiny bit rusty, so maybe my question has a really simple and quick answer, but here it goes:
Consider the support set of the base-13 function, is the set Lebesgue measurable? And if so, does it have non-zero measure?
 A: The function is easily seen to be Borel, since the graph of the function can be defined using only natural number quantifiers. In particular, a number is in the support if and only if there is a last place in its digits where $+$ or $-$ appears, followed eventually by ., but then after this, +, - and . do not appear again. That is,  


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*$x\in S\iff\exists n_1\exists n_2\gt n_1\forall m\geq n_1$ the $m^{\rm th}$ digit of $x$ in base 13 is neither $+$ nor $-$ nor ., except at $n_1$, where it is either $+$ or $-$ and and $n_2$, where it is .


Any set of reals that is definable using only natural number quantification is Borel, since existential quantification over the naturals corresponds to a countable union and universal quantification corresponds to countable intersection. Such sets lie in the arithmetic hierarchy, which is a very low part of the hyperarithmetic hiearchy, which leads ultimately to the Borel sets. 
The same idea shows that the graph of the function, as a set of pairs, is Borel. 
A: It seems that the definition of the function doesn't use the axiom of choice. This implies that the support set should be Lebesgue measurable.
A: Call the support set $S$.  The answer is yes it is Lebesgue measurable and no, it has zero measure.  It is even Borel measurable, which would take a tiny bit more effort to prove.
Note that $S$ is included in the set $T$ of numbers in which the "digits" '+','-', and '.' appear finitely many times in the "base 13 expansion".  But almost all numbers are normal, hence have all digits appearing infinitely often.  So the Borel set $T$ has measure zero, hence $S$ is Lebesgue measurable with measure zero.
A: The function is certainly Borel measurable, and hence so it its support set.  The function $f_k(x)$ which gives the $k$th tredigit of $x$ is clearly a Borel function, and it should be a straightforward if tedious exercise to write the base-13 function in terms of the $f_k$, using only common Borel functions (arithmetic, max/min, indicators) and limits.  
Of course the fact that no "nonconstructive" arguments are used in its definition is strong evidence that such a procedure should be possible, and if you know enough logic, as others have mentioned, this can even be a proof.  I just wanted to point out that one can prove the measurability from first principles with a little elbow grease.
