Let $\Omega \subset \mathbb{R}$, bounded open set.
And $f\in L^2(\Omega)$, if we have that for $g\in D'(\Omega)$.
$$\int_\Omega f.\phi dx=\int_\Omega g.\phi dx \; \text{for all } \phi \in D(\Omega)$$
do we have $f=g$ a.e, or can we just say $g\in L^2(\Omega)$?.
$g$ and $f$ are defined on totally different spaces so $g=f$ a.e. doesn't make sense. Strictly speaking $g \in L^{2}(\Omega)$ also doesn't make sense. A precise statemenet is the following: $f$ corresponds to a distribution, call it $\Lambda_g$, and we have $\Lambda_g=g$. Sometimes people express this by saying $g \in L^{2}(\Omega)$.