Let $\Omega \in \mathbb{R^n}$, (open set). and $f\in L^2(\Omega)$.
If we had the equality in distribution sense : $u=f$ : $(1)$
Does it implies that $u\in L^2(\Omega)$ ?
This is what I did :
Let $\phi \in D(\Omega)$ (test function), from $(1)$ we have
$\int_{\Omega}u.\phi .d\lambda=\int_{\Omega}f.\phi .d\lambda$
$\int_{\Omega}(u-f).\phi .d\lambda=0$ ;$\forall \phi \in D(\Omega)$
Then $u=f \quad$a.e. then $u\in L^2(\Omega)$