Measures with a continuous density function

Question

For simplicity, let's just consider the measure space $\mathbb R$ with the lebesgue measure $\lambda$. If a measure $\mu <<\lambda$ (abosolutely continuous), then by Radon–Nikodym theorem, $\mu$ has a density $f:\mathbb R\to \mathbb R$, which is locally integrable.

I am wondering if there is a property $\mathcal P$ such that if the measure $\mu$ satisfies $\mathcal P$, then $f=d\mu/d\lambda$ is a continuous function. (Sorry for being too "formal" with the word "property" here.)

I am asking this because I came across this answer by Ian. In the answer, it is claimed that if a radon measure $\mu<< \lambda$, then $f$ is continuous. So here, the property $\mathcal P$ is "being a radon measure". However, if we consider the measure $\mu$ on $\mathbb R$ with density being the Heaviside function, $$ H(x)=1,x\ge 0\\ H(x)=0, x<0 $$ then $\mu$ is a radon measure, but the density $H$ is not continuous. (The antiderivative of $H$ is indeed continuous: $[xH(x)]'=H(x)$.)

So, I am thinking about a property $\mathcal P$ described in the language of measure theory that could imply the continuity of $f$.

Answer 1

One known sufficient conidition is the following: If $\mu$ is a finite measure whose Fourier transform is integrable then $\mu$ has continuous density (which is given by the inversion theorem).

Answer 2

I think the real problem is with the cited answer, which I think is a wrong-headed. The other answer, by Ilya, is more to the point.

In short, the premise of this problem is false: Radon-Nikodym derivatives need not be continuous, and the "property" the OP wants to generalize is false.

Consider the pair of measures $\sigma$ and $\tau$ which are the restrictions of Lebesgue measure to the intervals $[0,1]$ and $[0,2]$, respectively, so for each measurable $A$ we have $\sigma(A)=\lambda(A\cap[0,1])$ and $\tau(A)=\lambda(A\cap[0,2])$. They are both Radon measures, and it is easy to check that $\sigma\ll\tau$ and $\tau\ll\lambda$, and that the Radon Nikodym derivative of $\sigma$ with respect to $\tau$ is the function $f=\chi_{[0,1]}$, or at least is $\tau$-a.e. equal to that. No measure $0$ modification of this function is continuous.

The original question brings out the problem caused by the multiple meanings of the word "continuous" in this context. It is true that if $\tau\ll\lambda$ the function $x\mapsto \tau([a,x])=\int_a^x f\,d\lambda$ is absolutely continuous, but it is not true that $f$ is necessarily continuous. Add to this the fact that all measures are continuous in the sense that they preserve monotone convergence. The accepted answer to that original question does not help.