The topological support $\sigma(\mu)$ of a Borel measure $\mu$ can be defined as the smallest closed Borel set $X$ that supports $\mu$, in the sense that $\mu(\mathbb{R}\setminus X)=0$. One can show that the topological support exists and is given by the set of its growth points, i.e.
\begin{equation*}
\sigma(\mu)=\left\{x\in\mathbb{R}:\forall\delta>0,\,\mu(I_\delta(x))>0\right\},
\end{equation*}
with $I_\delta(x)=(x-\delta,x+\delta)$.
Let $\mu$, $\nu$ be two Borel measures and suppose that $\mu\ll\nu$ and $\nu\ll\mu$, i.e. each of them is absolutely continuous with respect to the other one; in this sense, I will say that $\mu$ and $\nu$ are equivalent.
I would like to show $\sigma(\mu)=\sigma(\nu)$. If I am not wrong, this equality can be shown by noticing that, for a given $\mu$, $\sigma(\mu)$ coincides with the spectrum of the self-adjoint multiplication operator $(A_\mu f)(x)=xf(x)$ on the Hilbert space $L^2_\mu(\mathbb{R})$ (when defined on the proper domain); if $\mu$ and $\nu$ are equivalent, the operators $A_\mu$ and $A_\nu$ can be readily shown to be unitarily equivalent and thus isospectral. However, I expect a more immediate argument to show this equality to exist.
Best Answer
That was easy. Let $x\in\sigma(\mu)$ and suppose $x\notin\sigma(\nu)$. Then there is $\delta_0>0$ such that $\nu(I_{\delta_0}(x))=0$. But, since $\mu\ll\nu$, this implies $\mu(I_{\delta_0}(x))=0$, contradicting the initial assumption. Hence $\mu\ll\nu$ implies $\sigma(\mu)\subset\sigma(\nu)$. Repeating the argument exchanging $\mu$ and $\nu$ shows the claim.