So first I was able to show that $T_\mu$ is in fact a distribution. To show their supports are equal, I'll look at the complements, and so I need to show that the largest open set on which $T_\mu =0$ is also the largest open set on which $\mu = 0$.
For any $V \subset U$, if $\mu(V) =0$, then $T_\mu = 0$ on $V$. If I union over all such $V$, then I think that shows that $supp(\mu) \subset supp(T_\mu)$. But how do I show the reverse containment?
Suppose that $x \notin \newcommand{\supp}{\mathrm{supp}}\supp(\mu)$. Then there is a ball $B(x,r)$ with $\mu(B(x,r)) = 0$. If $\phi$ is any function compactly supported in $B(x,r)$ then $T_\mu(\phi) = 0$. Thus $B(x,r) \cap \supp(T_\mu) = \emptyset$, so that $x \notin \supp (T_\mu)$.
On the other hand, suppose that $x \notin \supp (T_\mu)$. Then there exists a ball $B(x,r)$ with the property that $B(x,r) \cap \supp(T_\mu) = \emptyset$ so in particular $T_\mu(\phi) = 0$ for any smooth function $\phi$ compactly supported in $B(x,r)$. In particular, if $\phi$ is compactly supported in $B(x,r)$, $\phi \ge 0$, and $\phi = 1$ on $B(x,r/2)$ you get $$\mu(B(x,r/2)) = \int_{B(x,r/2)} \phi \, d\mu \le \int_U \phi \, d\mu = 0.$$ Thus $\mu(B(x,r/2)) = 0$ so that $x \notin \supp(\mu)$.
p.s. This assumes, of course, that $\mu$ is a positive measure. For signed measures work instead with the variation measure $|\mu|$.