Is it possible to prove the **non-completeness** of the Borel-Lebesgue measure on $\mathbb{R}$ (restricted to the Borel $\sigma$-algebra) **without** the full axiom of choice, but still **with Countable Choice** ? 
It seems to be the case when I read https://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice, but I was unable to really prove it.