Is a measure, whose distributional derivative is a measure, absolutely continuous wrt Lebesgue?

Question

Suppose $\mu$ is a finite Borel measure on $\mathbb{R}^n$ with the property that it's distributional gradient $\nabla\mu$ is a vector-valued finite Borel measure. Does it follows then that $\mu$ itself is absolutely continuous with respect to Lebesgue measure?

Answer 1

Yes, it is, and it is even more in $L^p$ for $p<\frac{n}{n-1}$.

Take $\varphi\in C^\infty_c$ and $\psi = (1-\Delta)^{-1}\varphi$ which is a Schwartz function. Then $$ \langle\mu,\varphi\rangle = \langle\mu,(1-\Delta)\psi\rangle = \langle\mu,\psi\rangle + \langle\nabla\mu,\nabla\psi\rangle $$ where there is a scalar product in the second duality product. Since $\mu$ and $\nabla\mu$ are finite measures, $$ |\langle\mu,\varphi\rangle| \leq C\left(\|\psi\|_{L^\infty} + \|\nabla\psi\|_{L^\infty}\right) \leq C\,\|\psi\|_{W^{1,\infty}}. $$ By Sobolev embeddings, if $q>n$, $$ \|\psi\|_{W^{1,\infty}} \leq \|\psi\|_{W^{2,q}} = \|(1-\Delta)^{-1}\varphi\|_{W^{2,q}} \leq C\,\|\varphi\|_{L^q} $$ where the last inequality follows by elliptic regularity (for example Calderon-Zygmund). Hence $$ \mu \in L^{q'}(\Bbb R^n) $$