Absolutely continuous representative of Sobolev function

Question

I'm following the proof of this result on Leoni, A First Course in Sobolev Spaces (theorem 7.13). Let $I\subset \textbf R$ be and interval and take $u \in W^{1,p}(I)$ where $1 \leq p < \infty$. Take $x_0 \in I$ a Lebesgue point of $u$ and define $$ \bar u(x)=u(x_0)+\int_{x_0}^x u'(t) \;\mathrm dt\,,\;\; x \in I $$ Since $u'\in L^p(I)\subset L^1_{\mathrm{loc}}(I)$, it follows that $\bar u$ is locally absolutely continuous. More precisely (using other kind results that give sufficient conditions in order to have absolute continuity like continuity and differentiability a.e. and the so called condition N of Lusin), $\bar u$ is absolutely continuous. Moreover $\bar u'=u'$ a.e and integrating against every test function $\varphi$ and using integration by parts, one has $$ \int_I (\bar u - u) \;\varphi ' \mathrm d x=0 $$ so that there exits a constant $c \in \textbf R$ such that $u - \bar u=c$ a.e.

Here there is the point that I can't get: now the text says that since $x_0$ is a Lebesgue point of $u$ and $\bar u(x_0)=u(x_0)$ then $c$ have to be zero.

What I know is that there exists a (measurable) null set $A$ such that for every $x \in I \setminus A$ the equality $u(x) - \bar u(x)=c$ holds. Then the fact that $c=0$ is proved if I know that $x_0 \notin A$. I know that there exists at least one Lebesgue point outside $A$ (the set of Lebesgue point of a locally integrable functions is such that its complement is a null set)

I know I could take the function $\bar u + c$ as continuous representative of $u$, by I'd like to understand the last step.

Answer 1

Let $L$ be the set of all Lebesgue points of $u$. You have proved that, given $x_0\in L$, there exist an absolutely continuous function $\overline{u}$, a constant $c$ and a null set $A$ such that $$u(x)-\overline{u}(x)=c,\quad\forall \ x\in I \setminus A.\tag{1}$$

It remains to show that $c=0$, which should be a consequence of the following two facts: (i) $\boldsymbol{x_0}$ is a Lebesgue point of $\boldsymbol{u}$ and (ii) $\boldsymbol{\overline{u}(x_0)=u(x_0)}$.

You said "there exists at least one $x\in L\cap (I\setminus A)$". Actually, almost all $x\in I$ belongs to $L\cap ( I\setminus A)$. Therefore there exists a sequence $(x_n)$ in $L\cap (I\setminus A)$ such that $$x_n\overset{n\to\infty}{\longrightarrow} x_0$$ and thus, by $(1)$, $$u(x_n)-\overline{u}(x_n)=c,\quad\forall \ n\in\mathbb N.\tag{2}$$ Since $\boldsymbol{x_0}$ is a Lebesgue point of $\boldsymbol{u}$, $u$ is approximately continuous at $x_0$ and thus $$u(x_n)\overset{n\to\infty}{\longrightarrow} u(x_0).$$ Taking the limit in $(2)$, we obtain $$u(x_0)-\overline{u}(x_0)=c.$$ Since $\boldsymbol{\overline{u}(x_0)=u(x_0)}$ we conclude that $c=0$.