Let $\mu$ be a probability measure on $\mathbb{R}$ with the property that the set \begin{align} S= \{x\in \mathbb{R}: \mu(V)>0 \text{ for every open set } V \text{ containing } x \} \end{align} is equal to $\mathbb{R}$.
Is $\mu$ absolutely continuous with respect to Lebesgue measure $\lambda$?
Where absolutely continuous means that for any measurable set $A$ we have that $\lambda(A)=0$ implies that $\mu(A)=0$.
Not necessarily: let $\{r_n\}_{n=1}^{\infty}$ be a fixed enumeration of $\mathbb{Q}$, and define a probability measure $\mu$ on $\mathbb{R}$ by $$ \mu=\sum_{n=1}^{\infty}2^{-n}\delta_{r_n}$$
If $x$ is a real number and $V$ is an open set containing $x$, then $V$ contains a rational number $r$, and so $\mu(V)\geq \mu(\{r\})>0$. Therefore $S=\mathbb{R}$. But $\mu$ is not absolutely continuous with respect to Lebesgue measure because $\mu(\mathbb{Q})=1$.