Does Newton-Leibnitz apply to Sobolev space

Question

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:

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

Answer 1

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

Derived how? There is no fault in your proof, just non-trivial detailsare missing.

The formula is however, true. The best way to think about Sobolev functions in $W^{1,p}$ is through the absolute continuity on lines:

Since we can rotate the coordinate directions you have absolute continuity on lines parallel to any coordinate direction. since the absolutely continuous functions satisfy the integration by parts formula, your formula for $u(x)-u(y)$ follows.

Theorem 2.23 quoted above is the same as Theorem 2 in Section 4.9.2 in the book aby Evans and Gariepy, although the statement there is less intuitive that the above one. For the above statement see "Cortona lectures" available on my webpage.