Let $\Omega$ be a domain in $\mathbb R^n$. A classical result tells us that if a distribution $T \in \mathcal{D}'(\Omega)$ has partial derivatives (say gradient) in $L^2(\Omega)$ then $T \in L^2_{loc}(\mathbb R^n)$. Is there any known result for distributions satisfying $\operatorname{div} T \in L^2(\Omega)$ or $\operatorname{curl} T \in L^2(\Omega)^3$?
If the answer is affirmative, I expect the conclusion to be be weaker (e.g. $T \in L^1_{loc}$) since here we assume summability only for a linear combination of derivatives.
Thank you in advance for any kind of reference or insight.
EDIT
Recall that if $T \in \mathcal{D}'(\Omega; \mathbb R^3)$, then its distributional curl and divergence are defined as:
$$ \langle \operatorname{div} T, \phi \rangle := - \langle T, \nabla \phi \rangle, \ \phi \in \mathcal{D}(\Omega), \quad \text{and } \quad \langle \operatorname{curl} T, \psi \rangle := \langle T, \operatorname{curl} \psi \rangle, \ \psi \in \mathcal{D}(\Omega; \mathbb R^3). $$
Since I came up with a solution, I will answer my own question.
As suggested in comments, distributions with integrable curl or divergence can be in fact irregular. To see this (for the divergence), define:
$$ T : \mathcal{D}(\mathbb{R}^3; \mathbb{R}^3) \rightarrow \mathbb{R}, \ \ \phi \mapsto \sum_{i=1}^3 \delta_{\mathbf{0}}(\phi_i); \qquad H := \operatorname{curl} T \in \mathcal{D}'(\mathbb{R}^3; \mathbb{R}^3). $$
Then $\operatorname{div} H = \operatorname{div} \operatorname{curl} T = 0 \in L^2 $, on the other hand $H$ is not represented by any $L^1_{loc}$ since an explicit computation shows that:
$$ \langle H, \psi \rangle = \langle \operatorname{curl} T, \psi \rangle = \sum_{i=1}^3 (\operatorname{curl} \psi)_i (\mathbf{0}) $$
for $\psi \in \mathcal{D}(\mathbb{R}^3; \mathbb{R}^3) $, which is clearly irregular just like a Dirac mass.