An analytic set that is not a Borel set...see this post* from long ago.
Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.
*From: [email protected] (Gerald Edgar) Newsgroups: sci.math Subject: Re: Real Measurable, non-Borel. Date: 7 Oct 1993 08:13:10 -0400 Organization: The Ohio State University, Dept. of Math. Message-ID: [email protected] References: [email protected]
In $\lt$[email protected]> [email protected] (eli lapell) wrote: $\gt$What is a set of real numbers which is measurable but not Borel? Or just not Borel, period ??
An explicit example of a set of real numbers that is measurable (indeed, analytic) but not Borel [due to Lusin, Fundamenta Math. 10 (1927) p. 77]:
the set of all real numbers x with continued fraction expansion
x = a[0] + 1/(a[1] + 1/(...))
such that, for some positive integers r[1] < r[2] < ..., we have a[r[i]] divides
a[r[i+1]] for all i.
Other examples of analytic sets that are not Borel can be given in (complete separable) metric spaces other than the line:
In the space K[0,1] of nonempty compact subsets of [0,1] with the Hausdorff metric: The subset consists of the uncountable compact subsets. [Hurewicz, 1930]
${}$
In the space C[0,1] of real-valued continuous functions on [0,1] with the
unform metric: The subset consists of the differentiable functions.
[Mazurkiewicz, 1936]