Radon-Nikodem Derivative of a purely nonatomic Borel Measure

Question

If $\mu$ is a purely non-atomic Borel measure on a topological space $X$ then must its density be a continous function to $\mathbb{R}$?

My intuition says yes because all my counterexamples are not diffuse....

Important definitions:

Density: If $\mu$ is a Borel measure on $X$ then $\mu$'s density is a measurable function $g:X \rightarrow \mathbb{R}$ satisfying: for every Borel subset $B$ of $X$ $\mu(B)=\int I_B g dm$ (where $m$ is Lebesgue measure).

Answer 1

(a)
In $\mathbb R$, let $\mu$ be Lebesgue measure restricted to $[0,1]$. Then the density $$ h(x) = \begin{cases} 1,\qquad 0 \le x \le 1\\ 0,\qquad \text{otherwise} \end{cases} $$ is not continuous.

(b)
In $\mathbb R^2$, let $\mu$ be "arc length" measure on a circle. It is nonatomic. But it has no density (continuous or not) with respect to Lebesgue measure.