Absolutely continuous representative in $W^{1,p}(\Omega,X)$ Let $\Omega\subset\mathbb{R}^N$ be a domain and consider the Sobolev space $W^{1,p}(\Omega)$. It is know that if $u\in W^{1,p}(\Omega)$, then for almost every segment of line parallel to the coordinate axes, $u$ restricted to this segment can be viewed as a absolutely continuous function, i.e. there is a function $v$ defined in the segment which is equivalently to $u$ in the segment and $v$ is absolutely continuous.
My question is: if instead  of $W^{1,p}(\Omega,\mathbb{R})$, we consider $W^{1,p}(\Omega,X)$, where $X$ is a separable Banach space and $W^{1,p}(\Omega,X)$ is viewed in the Bochner sense, still we have the same result? or at least in the case where $\Omega=(0,1)$, is true that if $u\in W^{1,p}((0,1),X)$, then we can assume without loss of generality that $u$ is continuous?
Any reference is appreciated.
Thank you

The answer to your second question is affirmative. If $u \in W^{1,p}(0,T;X)$, then you can write $u(t) = \int_0^t \dot u(\tilde t) \, d\tilde t$. Now, it's not too hard to show that $u$ is absolutely continuous.
