For a function $u\in W^{1,p}$, we always use a sequence of $C^1$ function to approach it and derive the consequences. However, can we just claim for a.e. x, y:

$$u(x)-u(y)= \int_0^1 Du(y+t(x-y))\cdot (x-y) dt$$

I don’t think it’s right but I just can not figure out a counter example. So I tried to prove that it’s right in the following way:

Consider a $C^1$ sequence $u_n$ approaching u in $W^{1,p}$, the key here is to prove:

$$\int_0^1 D(u-u_n)(y+t(x-y))\cdot (x-y) dt$$

Tends to 0 as n tend to infinity, which can be derived from the definition.

Is there any fault in this proof? I still don’t think it’s right but I can’t tell it. Maybe I need some help here.