I have some trouble about the definition of the space $L^1((0,T);W^{1,\infty}(\mathbb R)$, on one side the formal definition given by Bochner consists in the set of measurable functions $u$ from $(0,T)$ to $W^{1,\infty}(\mathbb R)$ (with the Borel $\sigma$-algebra) such that
\begin{equation} \int_0^T \|u(t)\|_{W^{1,\infty}}<+\infty. \qquad (1) \end{equation}
Measurability is equivalent (Pettis' theorem) to weak measurability and almost surely (almost everywhere?) separably valued. Namely $t\mapsto G u(t)$ is measurable from $(0,T)$ to $\mathbb R$ for all $G\in W^{1,\infty}(\mathbb R)'$ (topological dual space) and there exists a negligeable subset $N$ of $(0,T)$ such that $\{f(t)\, |\, t\in(0,T)\setminus N\}$ is a separable subset of $W^{1,\infty}$.
On the other side I read sometimes it is the space of functions $u$ from $(0,T)$ to $W^{1,\infty}(\mathbb R)$ such that $u(t) \in W^{1,\infty}$ for almost all $t\in(0,T)$ and $(1)$ holds.
But I have some doubt on the consistency of the second definition. Considering such a function, I would verify the first definition is fullfilled, but the dual space of $W^{1,\infty}$ is somewhat misterious to me and moreover $W^{1,\infty}$ is not separable so proving almost surely separated valued seems tricky.
Is the second definition is false or at least a sufficient criterion of integrability?
Thanks.
The second definition is bogus. In fact, if you only assure that $u(t) \in W^{1,\infty}$ for (almost) all $t \in (0,T)$, the function $$ t \mapsto \|u(t)\|_{W^{1,\infty}}$$ might fail to be measurable, hence the left-hand side of (1) is not even well defined.
Moreover, the separability condition is crucial. There are weakly measurable $u$, for which (1) holds, but which fail to be almost surely separably valued. I think that $$ u(t) := \max(t, \cdot) \in W^{1,\infty}$$ defines such a function.