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 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, as we have, for a.e. x:

$$u_n(x) \to u(x)$$

Therefore, we have, for a.e. x, y:

$$ u(x)-u(y)=\lim_{n\to\infty} u_n(x)-u_n(y)=\lim_{n\to\infty} \int_0^1 Du_n(y+t(x-y))\cdot (x-y) dt= \int_0^1 Du(y+t(x-y))\cdot (x-y) dt$$