Let $u:\mathbb{R}^N \to \mathbb{R}^N$, $u \in BV(\mathbb{R}^N)$, be a function of bounded variation. 
We have that the following holds



> 
> $$(\ast) \qquad \frac{1}{|B\_r(0)|}\int\_{B\_r(0)} \frac{|u(x+z)-u(x)-Az|}{|z|} dz =0$$
>  for the points where $Du$ is not singular, where $A$ is the approximate differential of $u$. 
> 
> 
> 


What happens in the points where $Du$ is singular? What should replace $A$ to still get convergence of that integral to $0$?



Difference quotient for functions of bounded variation
In fact the following is true:



> 
> **Theorem.** If $u\in BV(\mathbb{R}^N)$, then for almost all $x\in\mathbb{R}^n$: $$
>  \frac{1}{|B\_r(0)|}\int\_{B\_r(0)}\left|\frac{u(x+z)-u(x)-[Du(x)]\_{ac}z}{r}\right|^{\frac{N}{N-1}}\,
>  dz\to 0 \quad \text{as $r\to 0$.} $$
> 
> 
> 


Here $[Du(x)]\_{ac}$ is the density of the absolutely continuous part of the distributional derivative $Du$ which, by the definition of $BV$ is a Radon measure.


In fact the above convergence holds for all $x$ such that:


1. $x$ is a Lebesgue point of $u$,
2. Lebesgue point of $Du$,
3. $$\lim\_{r\to 0}\frac{[Du]\_s(B(x,r))}{r^n}=0$$


where $[Du]\_s$ is the singular part of $Du$. 


For other points you need not have concergence no matter what linear operator $A$ you take. 


Moreover,
$[Du]\_{ac}$ equals almost everywhere to the approximate derivative of $u$ at all points at which the above convergence to zero holds. 


The above results is proven in **Evans and Gariepy**, *Measure theory and fine properties of functions.*, Theorems 1 and 4 in Chapter 6.


