An analytic set that is not a Borel set...see [this post][1]* from long ago.

Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.


  [1]: https://groups.google.com/forum/#!topic/sci.math/oEh4FzmmGxQ

>*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 $\[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]