I’m looking for subset A of R such that A is a Borel set but A is neither Fσ nor Gδ.
Answer
There are many examples. Here’s one:
Observe first that the rational numbers Q are an Fσ. This is because they are a countable union of points. The irrational numbers R∖Q=⋂q∈QR∖{q} are thus a Gδ. Since both Q and R∖Q are dense and disjoint it follows from the Baire category theorem that Q cannot be a Gδ. [Edit: See also this thread here containing several proofs that Q can’t be a Gδ in R. These proofs explicitly avoid Baire].
The same reasoning shows that F=Q≥0 is an Fσ in [0,∞), but isn’t a Gδ, and that G=R≤0∖Q≤0 is a Gδ in (−∞,0], but isn’t an Fσ. Their union F∪G is then an example of a Borel subset of R which is neither an Fσ nor a Gδ because if it were an Fσ then the same would hold for G=(F∪G)∩(−∞,0), for example. I leave it as an exercise to show that F∪G is both an Fσδ and a Gδσ.
That’s probably the easiest example. A few more (both more interesting but also more involved ones) can be found in this MO thread.
For a much more indepth discussion of such ideas, I recommend looking into one of the following books:

A.S. Kechris, Classical Descriptive Set Theory, Springer GTM 156.

S.M. Srivastava, A course on Borel sets, Springer GTM 180.

J.C. Oxtoby, Measure and Category, Springer GTM 2.
Specifically, look up the sections on the Borel hierarchy.
Attribution
Source : Link , Question Author : Babak Miraftab , Answer Author : Community