Let's see. Assume measurability axiom - every subset of R has Lebesgue measure. As we can use the usual construction of unmeasurable set on L intersect R, our only escape option is that it has zero measure.

So if we assume measurability, L intersect R is a dense zero-measure subset, just like Q. These are the reals we can know individually, but not the reals-as-a-whole that we know...