Given two real $BV$ functions $u$ and $v$ in an open interval $(a,b)$ consider the set


$A=\{x: \text{both } u \text{ and } v \text{ are continuous at } x \text{ and } u(x)=v(x)\}$


is it true that $|Du|(A)=|D(v)|(A)$?


At first glance it seems absurd that something like that holds but I find it hard to come up with a counterexample. For example if $A$ is an open set then this is obviously true. If $A$ is a countable set then since both functions are continuous there, we have $|Du|(A)=|Dv|(A)=0$. It seems that the difficulty arises when $A$ is of Cantor type. 


So one could also ask the following version: Let $u$ be the Cantor-Vitali function and $v$ be a $BV$ function which is continuous on the Cantor set $\mathcal{C}$ and coincides with $u$ there.


Is it true that $|Du|(\mathcal{C})=|Dv|(\mathcal{C})$?



Is the variation of two BV functions the same in the set in which they coincide?
Yes, this works. Let me write $\mu$ and $\nu$ for the (signed) measures induced by $u$ and $v$, respectively. I then claim that $\mu|\_A=\nu|\_A$, and this will then imply that the total variations agree also.


To prove my claim, it suffices to discuss $\mu(U), \nu(U)$ for open $U\supseteq B$, where $B$ is a Borel subset of $A$ (by regularity). Let's focus on a component $I=(c,d)$ of $U$. Here we can assume that $c,d$ are in $A$ or there are sequences $a\_n\in (c,d)\cap A$ that converge to the endpoints; if that is not the case, then we can simply make $I$ smaller.


Since $\mu((a,x))=\nu((a,x))$ for all $x\in A$ by the definition of $A$, we now obtain that also $\mu(I)=\nu(I)$, by approximation (use dominated convergence).


