Robin Chapman [introduced me](https://mathoverflow.net/questions/32126/function-with-range-equal-to-whole-reals-on-every-open-set/32127#32127) to [Conway's Base 13 Function](http://en.wikipedia.org/wiki/Conway_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?



Is Conway's base-13 function measurable?
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.


