If $u\in\mathcal D'$ and $\langle u,\phi' \rangle=0$ then there exists $c\in\mathbb C$ such that $\langle u,\phi \rangle= c\int\phi$

Question

From https://math.stackexchange.com/a/2552946/1084278, we know the following property:

If $u\in\mathcal D'$ and $\langle u,\phi' \rangle=0$ for all $\phi\in\mathcal D$ then there exists $c\in\mathbb C$ such that $\langle u,\phi \rangle= c\int\phi$ for all $\phi\in\mathcal D$.

Proof:

Suppose $\phi\in\mathcal D$. There exists $\psi\in\mathcal D$ with $\phi=\psi'$ if and only if $\int\phi=0$.

But the proof of this property in the link only show the case that $u$ can regard as a locally integrable function. How to prove it when $u$ is a non-regular distribution? In other words, does $u'=0$ mean $u=c$ (in the distribution theory)?