I posted this question first in Math.StackExchange one week ago here, but I didn't get an answer or a helpful comment so I repost it here:

Let $d=3$ and $\Omega\subset \mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function. A sufficient condition for the integral $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ is that $u\in L^{6/5}(\Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(\Omega)\hookrightarrow L^6(\Omega)$.

**Question:** Is the opposite true, i.e is it true that
$$\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$$ implies $u\in L^{6/5}(\Omega)$ or at least $u\in L^1(\Omega)$ ?

**My thoughts:** It is easy to see that $u\in L^1_{loc}(\Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $\Omega$ and $0$ in a neighborhood of the boundary $\partial \Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $u\in H_0^1(\Omega)$ such that $f(u)\in L^{6/5}(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

or

$(2)$ Find $u\in H_0^1(\Omega)$ such that $\int\limits_{\Omega}{f(u)vdx}<\infty,\forall v\in H_0^1(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.