I read the following theorem in my lecture material stating as follows: For any probability measure $\mu$ on $(\mathbf{R},\,\mathcal{B})$ with a density function $f(x)$, the following condition holds for almost all $x$ $$\lim_{h\downarrow 0}\frac{1}{2h}\mu(\{x-h,\,x+h\})=f(x). $$
My question is, by stating a measure 'with a density', does that mean $\mu\ll Leb$, and $\mu$ does not have singular or discrete part? Besides, what is the name of above theorem, where can I find the proof of it?
Assume that $\mu$ is absolutely continuous w.r.t. the Lebesgue measure. This implies the existence of a measurable and Lebesgue integrable function $f$ such that \begin{align*} \mu(-\infty,x] = \int_{-\infty}^x f(y) \, \mathrm{d}y. \end{align*} This shows that the map $F(x) = \mu(-\infty,x]$ is absolutely continuous. (The classic link between probability densities and Radon-Nikodym derivatives.) In particular, $F$ is almost everywhere differentiable with derivative $f$. (I suppose this result is known to you. Otherwise see Royden's or Rudin's.) Can you take it from here?