In 1918 Sierpiński constructed a Lebesgue measurable real-valued function on $[0,1]$ which isn't bounded above by any Borel function (I couldn't find the original reference, but here is a pdf of a reprint in Russian). His construction is quite hands on and I was wondering whether there was a more abstract approach.
Since there are only $2^{\aleph_0}$ Borel functions but $2^{2^{\aleph_0}}$ Lebesgue measurable functions, it would seem reasonable to expect some sort of diagonalization argument to work.
But for this to work I would need a sort of universal Borel function. Specifically, is there a Lebesgue measurable function $F\colon \mathbb{R}^2\to\mathbb{R}$ such that for every Borel function $f\colon \mathbb{R}\to\mathbb{R}$ there is some $y_f$ such that $f(x)=F(x,y_f)$ for all $x$?
Of course, such a function exists if we drop the Lebesgue measurability requirement. But if we can get such a Lebesgue measurable universal function then the diagonalization I suggested above actually goes through to give a Sierpiński-style counterexample.
Such a function exists by a simple cardinality argument. Let $C$ be the standard Cantor set, or any other set which has zero Lebesgue measure and cardinality of the reals. As you mention, the set of all Borel functions on the real line has the same cardinality as the reals, so there exists a bijective map from $C$ to the set of Borel functions, $y \mapsto f_y$. Define $F(x,y) = 0$ if $y \notin C$, and $F(x,y) = f_y(x)$ if $y\in C$. This map satisfies the condition you are looking for.