Nowhere continuous function for every equivalence class

Question

Since our calculus lectures, we know that there are nowhere continuous functions (like the indicator function of the rationals). However, if we change this Dirichlet function on a set of measure zero, then we get a continuous function (the zero function). So my question ist

Does there exist a function $f: \mathbb{R}\rightarrow \mathbb{R}$ such that every function which equals $f$ almost everywhere (with respect to the Lebesgue measure) is nowhere continuous? Is it possible to choose $f$ borel-measurable?

Answer 1

See my answer to this question for the construction of an $F_\sigma$ set $M\subset\mathbb R$ such that $0\lt m(M\cap I)\lt m(I)$ for every finite interval $I,$ where $m$ is the Lebesgue measure.

Let $f$ be the indicator function of $M.$ Clearly $f$ is Borel-measurable. If $g(x)=f(x)$ almost everywhere, then $g^{-1}(0)$ and $g^{-1}(1)$ are everywhere dense, whence $g$ is nowhere continuous.

Answer 2

Any discontinuous function s.t. $f(x + y) = f(x) + f(y)$.

Answer 3

Let $r_1,r_2, \dots$ be the rationals. Let $f(x) = x^{-1/2}, x\in (0,1),$ $f=0$ elsewhere. For $x\in \mathbb R,$ define

$$g(x) = \sum_{n=1}^{\infty}\frac{f(x-r_n)}{2^n}.$$

Then $g$ is a Borel measurable function from $\mathbb R$ to $[0,\infty].$ Note that $\lim_{x\to r_n^+} g(x)=\infty$ for all $n.$

Now $\int_{\mathbb R} g <\infty$ by the monotone convergence theorem. Thus $g<\infty$ a.e. Define $h=g$ wherever $g$ is finite, $h=0$ where $g=\infty.$ The function $h$ fits the bill.

To be sure, there are some things to check, but I'll leave it here for now