Suppose U is an open set,not necessarily bounded or has Lipschitz boundary, $f\in L^p(U)$ ,define the difference as usual: $$D^h_i f=\frac{f(x+he_i)-f(x)}{h},\ \ \forall x\in U'\subset\subset U$$
Now we're given the condition that:$\lim \inf_{h\rightarrow 0^+}||D^h_i||_{L^p(U')}<M$.
We're required to show that $f\in W^{1,p}(U)$.
To the best of my knowedge,if we have uniform difference quotients bound here,then use the fact that any bounded sequence in a self-reflexive seperable Banach space has a subsequence which is weakly converge,thus we can find the weak derivative of $f$. But here the bound is not uniform,it seems hard to apply the weak convergence.
The second problem is that to let the quotient difference make sense,we always has to find a compact subset in $U$,thus even we overcome the first obstacle,we can only get $f\in W^{1,p}_\mathrm{loc}(U)$, how can we get the globle one?
Let $\phi$ be a smooth function with compact support in $U$. Let $U'$ be as above and contain the support of $\phi$. Then $$ \int_U f(x) \partial_i \phi(x) \, dx = \int_U \lim_{h \to 0} f(x) D^h_i \phi(x) \, dx $$ Note that $|f(x) D^h_i\phi(x)| \le K |f(x)| \chi_{U'}(x)$, where $K$ is the maximum value of $|D\phi|$ in $U$, so that LDCT applies and thus $$ \int_U f(x) \partial_i \phi(x) \, dx = \lim_{h \to 0} \int_U f(x) D^h_i \phi(x) \, dx = - \lim_{h \to 0} \int_U D_i^h f(x) \phi(x) \, dx. $$ Now apply Holder's inequality to conclude $$ \left| \int_U f(x) \partial_i \phi(x) \, dx \right| \le M \|\phi\|_{L^{p'}(U)}. $$ To address the second question, note that $M$ is indepedent of $U'$ so that this inequality holds for all $\phi \in C_0^\infty(U)$. You can use e.g. the Riesz representation theorem to prove the existence in $L^p(U)$ of the weak derivative $\partial_i f$.