In Theorem 8.2. of Brezis's book of Functional Analysis, says that a function $u \in W^{1,p}(I)$ has a continuous representante $\tilde{u}$ such that
$$
\int_y^x u'(t) dt = \tilde{u}(x) - \tilde{u}(y).
$$
I'm trying to verify the following remark:
I would like to know if there is a straightforward demonstration, or whether is necessary a more sophisticated argument.
My attempt: Let $v$ be the continuous representative of $u'$. It is clear that $\tilde{u}$ has $v$ as its weak derivative, because given $\varphi \in C^\infty_0(I)$ $$ \int \tilde{u} \varphi' = \int u \varphi' = - \int u' \varphi = -\int v \varphi. $$ I'm trying to prove the limite $$ \lim_{h \rightarrow 0} \frac{\tilde{u}(x + h) - \tilde{u}(x)}{h} $$ exists and equals to $v(x)$, for all $x \in I$, however it seems not to be the right way to go.
In your formula of Theorem 8.2 it should be $\int_y^x u'(t) \, dt = \tilde{u}(x) - \tilde{u}(y)$. Then we have $$ \lim_{h\searrow 0} \frac{\tilde{u}(x+h) -\tilde{u}(x)}{h} = \frac{1}{h} \int_x^{x+h} u(t) \, dt = u'(x)$$ for a.e. $x\in \overline{I}$ by Lebesgue's differentiation theorem, and likewise $$\lim_{h\searrow 0} \frac{\tilde{u}(x) -\tilde{u}(x-h)}{h} = u'(x)$$ for a.e. $x\in \overline{I}$. But as $u'$ is supposed to be continuous, this hold for all $x\in \overline{I}$.