# $u\in W^{1,p}(0,1)$ is equal a.e. to an absolutely continuous function?

I have a simple question on Sobolev space theory. Let $1\le p \le \infty$. How can one prove that every $u\in W^{1,p}(0,1)$ is equal a.e. to an absolutely continuous function and that $u'$ exists a.e. and belongs to $L^p(0,1)$?

• What does equal s.e. mean? Also don't you mean $W^{1,p}(0,1)$? If I'm guessing right, it looks like a basic theorem in Sobolev space theory... Dec 2, 2012 at 1:25
• I mean the function can be represented by a function that is a.e. equal to an absolutely continuous function. Dec 9, 2012 at 9:16

Consider the case $p=1$. Take $u\in W^{1,1}(0,1)$ and put $v(t)=u(0)+\int_0^tu'(s)ds$, then $v\in W^{1,1}(0,1)$ and is absolutely continuous. We have $v'=u'$ a.e. so $u=v+c$ a.e.
• Thank you for the answer. The $p=1$ case can be proven by your argument. I still don't know how to prove in general case. Dec 9, 2012 at 9:17
• @Pooya: Just notice that for $p>1$ we have $W^{1,p}\subset W^{1,1}$ and $L^p\subset L^1$ since $(0,1)$ has finite measure. Dec 9, 2012 at 22:40
• @Jose27 Can you elaborate why $W^{1,p} \subset W^{1,1}$ is true for all $p > 1$? I apologize if this question is fundamental, but I too am studying Sobolev spaces but have little knowledge in measure theory. Jan 10, 2015 at 23:09
• @dragon: This is jsut Holder's inequality: $$\int_0^1 |f|dx \leq \left( \int_0^1 |f|^p dx \right)^{1/p}.$$ Now if $u\in W^{1,p}(0,1)$ then using $f=u$ and $f=u'$ we arrive at $u\in W^{1,1}(0,1)$. Jan 11, 2015 at 1:49
• @Jose27 Hi Jose. I think we should just define $v$ by $v(t)=\int_0^tu'(s)ds$, i.e. without adding the $u(0)$ term. The reason is that, $u$ is a $L^1$ function and defined almost everywhere, so it makes no sense to say $u(0).$ However, though we define $v(t)$ in this slightly different way, it is still absolutely continuous. And so your reasoning still applies. Mar 17, 2020 at 15:05
We want to show $$u(x)-u(a) = \int_a^{x}u'(t)dt$$ a.e., then consider the mollifier and convolve it with $$u$$, then the formal integration by parts yields: $$u_{\epsilon}(x)-u_{\epsilon}(a) = \int_a^{x}u'_\epsilon(t)dt$$ and by $$L^1$$ convergence of $$u_\epsilon$$ and $$u'_\epsilon$$: $$\int_a^{x}u'_\epsilon(t)dt \to \int_a^{x}u'(t)dt$$. By picking a subsequence of $$u_\epsilon$$, we may assume $$u_\epsilon \to u$$ a.e., then $$u(x)-u(a) = \int_a^{x}u'(t)dt$$ a.e., thus $$u(x)$$ can be identified as an absolutely continuous function $$v$$ such that $$v(x) = u(a) + \int_a^{x}u'(t)dt$$.