Natural examples of functions $f$ in $L^1([0,1])$ such that any function $g$ in the class $[f]$ is discontinuous everywhere

Question

I need many motivated examples of functions $f$ in $L^1([0,1])$ such that any function $g$ in the equivalence class $[f]$ is discontinuous everywhere.

Thank you in advance!

Answer 1

Not an answer, but here is a reformulation, and some related concepts that may be enlightening.

Define, for a function $f$ on $[0, 1]$, the essential oscillation $O_e f(x)$ at $x \in [0, 1]$ by

$$O_e f(x) := \lim_{r \to 0_+} \text{esssup}_{(y, z) \in B_r (x) \times B_r (x)} |f(y) - f(z)|,$$

where $\text{esssup}$ denotes the essential supremum, which is taken with respect to the product Lebesgue measure.

The essential oscillation measures how far a function is from being continuous, modulo null sets. One may compare this to the usual definition of the oscillation of a function:

$$Of(x) := \lim_{r \to 0_+} \text{sup}_{(y, z) \in B_r (x) \times B_r (x)} |f(y) - f(z)|.$$

With this definition, a function is continuous at $x$ iff $Of(x) = 0$, and is discontinuous at $x$ with magnitude $Of(x)$ otherwise.

Then your question is equivalent to asking for (natural examples of) functions for which $O_e f(x) > 0$ everywhere. This follows from the fact that points of "essential continuity", that is, points for which $O_e f(x) = 0$ can be repaired to solve $O f(x) = 0$, at the cost of modifying $f$ on a null set.

Remark:

Somewhat less obvious is that you can do this repair job for all points of essential continuity at once with the same null set. More precisely, we have the following

Theorem: Let $f: [0, 1] \to \mathbb R$ be measurable. Then there exists a modification of $f$, that is, a function $\tilde f$ that agrees a.e. with $f$ such that $O_e f = O \tilde f$ everywhere.

In particular, we have $\{O_e f = 0\} = \{O \tilde f = 0\}$, i.e. all points of essential continuity can all be repaired to be points of actual continuity at once with the same null set modification.

Answer 2

Here is one example. Let $(q_n)_n$ be a list containing all rationals in $[0,1]$, let $f_n:[0,1]\to\mathbb{R}$ be given by $f_n(x)=1$ if $|x-q_n|<\frac{1}{n!}$ and $f_n(x)=0$ if not. Let $f=\sum_nf_n:[0,1]\to\mathbb{R}$.

If some function $g$ is in $[f]$ then it has to be unbounded in any open interval $(a,b)$ (so it cannot be continuous at any point): this is because inside any interval $(a,b)$ we can find by recursion an increasing sequence $(n_k)_k$ such that $q_{n_k}\in(a,b)$ and $|q_{n_{k+1}}-q_{n_k}|<\frac{1}{2n_k!}$. Thus, for all $k\in\mathbb{N}$, $f(x)\geq k$ for all $x$ in some nhood of $q_{n_k}$.

Answer 3

As suggested in a comment by Pietro Majer: Let $A$ be a Borel set with positive but not full measure in every interval in $(0,1)$ and let $f = 1_A$ be its indicator function. Then if $g = f$ a.e., we have that $g$ takes both values 0 and 1 in every interval, and thus is nowhere continuous.

Answer 4

The derivative of any nowhere continuously differentiable bounded variation (BV) function will work.

If $f$ is BV, then it is differentiable a.e., but if its derivative were anywhere continuous, then $f$ would be continuously differentiable at that point. Thus every $g\in [f']$ must be everywhere discontinuous.

Some fractal curves will have this property, see Mandelbrot and Frames "canopy of a self contacting fractal tree".