Construction of a Borel measurable function mapping Borel set to non-Borel set I hope to construct an example of a function $f: \mathbb{R}^1\to\mathbb{R}^1$, which is Borel measurable ($f^{-1}(B) \in \mathcal{B}^1, \forall B \in \mathcal{B}^1$), but does not preserve Borel measurability, i.e., an image of a Borel set is non-Borel. I tried as follows:
First pick up a non-Borel set $N \subset [0, 1]$, then I define $f(x) = \sup\{y: y \le x, y\in N\}$. As a piecewise constant non-decreasing function, it is guaranteed that $f(x)$ is Borel measurable. And $f([0,1]) = N$ maps $[0,1]$ to a non-Borel set. 
Could anyone tell me whether it is a valid example or not? My example seems much simpler than the existing example for this problem, which involves Polish space etc that I cannot understand.

Your conclusion that $f([0, 1]) = N$ is false [or at least non-trivial]. Note that the supremum of a set doesn't always lie in that set. You would need a property like closedness to conclude this - but all closed sets are Borel.
