Derivative same as the weak derivative Are there any conditions under which the derivative almost everywhere is the same as the weak derivative (in the context of the Sobolev space $W^{1,p}(U)$)?
 A: I guess no conditions are required besides the existence of the (strong) derivative and [Edit: it being in $L_{loc}^1(U)$]. 
Recalling the definition (e.g. book Evans): Let $u,v\in L_{loc}^1(U)$.
$v$ is called the $\alpha$th weak partial derivative of $u$ provided
$$\int_U uD^\alpha\phi dx = (-1)^{|\alpha|}\int_U v\phi dx$$ for all test functions $\phi \in C_c^\infty(U)$. 
The (strong) derivative satisfies this definition if it exists and provided that it is in $L_{loc}^1(U)$. We conclude since a weak derivative is uniquely defined up to a set of measure zero (see e.g. Evans for a proof).
Edit ctd: the question then remains what if the strong derivative is not in $L_{loc}^1(U)$? Is this even possible? The above integrals are to be understood as Lesbegue integral. Moreover, $\phi$ being in $C_c^\infty(U)$ is of bounded variation. This all makes the above integration by parts formula valid for functions which are not necessarily Riemann integrable. In this sense, I do not know of an example for which the strong derivative exists which fails the above integration by parts formula.
