Derivative same as the weak derivative

Question

Are there any conditions under which the derivative almost everywhere is the same as the weak derivative (in the context of the Sobolev space $W^{1,p}(U)$)?

Answer 1

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.