Suppose that $f \in BV([a,b])$: may $f$ have non-Lebesgue points which are not jumps?
In my opinion no, because $BV$ functions have only jump discontinuities (and Lebesgue points are essentially points where $f$ does not oscillate too much).
If I'm wrong, could you show me a counterexample?
By definition we have that $f\in \mathrm{BV} (a,b)$ if and only if $f'\in\mathcal{M}_b((a,b))$.
What is a Lebesgue point? It is an $x\in (a,b)$ such that $$ \lim_{\varepsilon\to0}\frac{2}{\varepsilon}\int_{[x-\varepsilon,x+\varepsilon]} |f(y)-f(x)|dy=0 $$ and comes into play when considering measurable functions or almost everywhere defined functions for which we are able to say that non Lebesgue points are negligible if we have local integrability.
We are asking here: can a bounded variation function have non Lebesgue points on points where it doesn't jump?
The answer is no.
Because if we consider a bounded variation function $f$ we can separate its derivative into two mutually singular parts $\mu_j$ and $\mu_c$ such that $\mu_j$ contains all the atomic components of $f'$ and $\mu_c$ is the remainder.
Now we define $f_c(x)=\mu_c((a,x))$ and $f_j(x)=\mu_j((a,x))$. $f_c$ is a continuous bounded variation function and thus all of its points are Lebesgue points. $f_j$ is a locally constant function outside of the points of the support of $\mu_j$ and thus all of the points outside the support of its derivative are Lebesgue points.
Outside of the support of $\mu_j$ all the points of both $f_j$ and $f_c$ are Lebesgue points and thus their sum will be a Lebesgue point. We can conclude.