Consider a random variable $V$ defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that
1) The support of $V$ is an open subset $\mathcal{V}$ of $\mathbb{R}^K$ with strictly positive Lebesgue measure.
2) The distribution of $V$ is absolutely continuous on $\mathcal{V}$ with respect to Lebsgue measure.
Question: which of the two assumptions is sufficient for having $\forall v\in \mathcal{V}$ $$ \mathbb{P}(V=v)=0 $$ ?
My thoughts: I'm tempted to say that 1) is sufficient for the desired conclusion as 1) implies that the support of $\mathcal{V}$ is non-finite. 2) adds more by implying that the cdf of $V$ is continuous and there is a pdf. Could you say whether I'm right or wrong and why?
Your argument that 1) is sufficient doesn’t work since you may have a discrete variable with countably infinite support, e.g. X with support on strictly positive integers with probability that X=k being $2^{-k}$. Even when support is uncountable you may have a mixture of a continuous and discrete variable, e.g. with B bernoulli with p = 0.5 and Z standard gaussian the random variable $Y = BX + (1-B)Z$ will have support $R$ but atoms (i.e. non-zero probability) at strictly positive integers.
However 2) is sufficient since by definition of absolute continuity each point has probability less than Lebesgue measure of set with single point which is zero.
As a general counterexample to 1) being sufficient, any V satisfying 1) and not having atom you may pick any point v of support of V and make a mixture with probability 0.5 of V and a point mass at v which still satisfies 1) but has an atom at v.