Let $\omega^\omega$ be Baire space. If $A,B\subseteq\omega^\omega$ we say that $A$ is Wadge reducible to $B$ (written $A\leq_w B$) if there is a continuous function $f:\omega^\omega\rightarrow\omega^\omega$ with $x\in A$ if and only if $f(x)\in B$. (In other words $A$ is a continuous preimage of $B$). By identifying sets with $A\leq_w B$ and $B\leq_w A$ we induce a partial ordering on the set of corresponding equivalence classes, or Wadge degrees. Under the axiom of determinacy AD, it turns out by the so-called Wadge lemma that this hierarchy is almost linearly ordered, meaning we get a linear order if we identify a degree $a$ and the degree consisting of complements of members of $a$. Clearly the Borel sets will form an initial segment of this hierarchy. What I am curious about is if this is still the case if we drop the AD assumption.

Even without AD, determinacy holds for the Borel sets and so the Borel Wadge degrees will still be almost linear ordered - indeed almost well-ordered. For non-Borel degrees using AC it is possible to get a lot of bad behavior; for example many incomparable Wadge degrees. But all the ways I can figure to do such things is to enumerate all continuous functions and build simultaneously the incomparable sets by diagonalizing against the continuous functions. An argument like this is (I think) rather unlikely to produce a Borel set.

So specifically my question is: is it possible for there to be a non-Borel set which is Wadge-incomparable to some Borel set? This is the same as asking if it is possible for there to be a Borel set $B$ and a non-Borel set $A$ with $B\not\leq_w A$.