TITLE: Non-Borel sets without axiom of choice
QUESTION [67 upvotes]: This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel  sets would be a Hamel basis, which needs axiom of choice, and examples of non Lebesgue measurable set would be Vitali sets, which seems to be unprovable without axiom of choice. Then I saw the answer of François G. Dorais. His construction of an uncountable $\mathbb{Q}$-independent set in $\mathbb R$ does not require axiom of choice. Which leaves a faint hope for the following:

Is it possible to construct without using the axiom of choice examples of non Borel sets?

There is a classic example of an analytic but non-Borel set due to Lusin, described by Gerald Edgar here, but it is not clear to me whether it needs axiom of choice since it seems to be putting restrictions on higher and higher terms in the continued fraction expansion. Since I do not know logic and set theory, I hoped of asking the experts.

REPLY [20 votes]: There are some very nice examples of non-Borel sets. Two that I particularly like are the differentiable functions (as a subset of the space of continuous functions on [0,1], say) and the set of all infinite graphs that contain an infinite clique (as a subset of the set of all graphs with vertex set $\mathbb{N}$ with the product topology). In general, a continuous image of a Borel set need not be Borel, and many natural non-Borel sets arise this way.