Everyone who studied distributions at some point have seen how locally all distribution are basically formal derivatives of continuous functions.
The question is: does there exists a bounded function $f\in L^\infty(\mathbb R^d,\mathbb R^d)$ such that its distributional derivative is not a measure?
The easiest formulation of the question would be to find such a function in $L^\infty (\mathbb R)$.
Since the topology on the distribuitions is the pointwise topology when looking at bounded functions we can just look at characteristic functions.
Given a measurable set $A$ we know that $\chi_A$ can be approximated by above with open sets $O_n$ and from below with compact sets $K_n$, thus creating two converging sequences $$ T_{\chi_{O_n}}\to T_{\chi_{A}}\leftarrow T_{\chi_{K_n}}. $$
We get $$ \partial T_{\chi_A}(\varphi)=- T_{\chi_A}(\varphi')=\lim -T_{\chi_{O_n}}(\varphi')=\sum_k (\varphi(a^n_k)-\varphi(b^n_k)). $$ The inequality $|\partial T_{\chi_A}(\varphi)|\leq\|\varphi'\|_\infty\mathcal{L}^1(A)$ is trivial but if the number of intervals in $O_n$ is a finite number $M_n\in\mathbb N$ for any $n\in\mathbb N$ then we get $$ |\partial T_{\chi_A}(\varphi)|\leq 2 M_n \|\varphi\|_\infty+\varepsilon, $$ thus pointing out that if $M_n\leq M$ uniformly then we would get a measurable set with a measure as a derivative.
Is such a condition necessary? Of course not, because Cantor's functions is an example of a bounded function whose derivative is a measure supported on an uncountable set.
Does this mean that considering the function $$ f=\sum_n(-1)^{n+1}\frac{n}{2}\chi_{[-\frac{1}{n},\frac{1}{n}]} $$ is enough?
It is bounded because the alternated harmonic series converges and its derivative is $$ \int_{\mathbb R}\varphi' f\mathcal L=\sum_n(-1)^{n+1}\frac{\varphi(1/n)-\varphi(-1/n)}{2/n}. $$
Are there more trivial examples? Is there a mistake and this function in reality has a measure as derivative? Is there an example of a continuous function that doesn't have a measure as a derivative? What are some multidimensional examples?