# Sufficient conditions for probability measure of singleton sets being zero

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?

• How do you define support? It seems that the first property is trivial as any measure is supported on $\mathbb{R}^K$. Also is $\Omega=\mathbb{R}^K$? – Yanko May 26 '18 at 14:25
• Anyway, as long as I'm aware, $V$ being absolutely continuous with respect to Lebesgue measure means that $P(V=v)\leq \mathcal{L}(\{v\})=0$. – Yanko May 26 '18 at 14:27
• 1) The support of $V$ is intended as $\{v\in \mathbb{R}^K \text{ s.t. } \mathbb{P}(\{\omega \in \Omega \text{ s.t. } V(\omega) \in B(v,r)\})>0 \text{ }\forall r > 0 \}$, where $B(v,r)$ denotes the ball with center at $v$ and radius $r$. See here math.stackexchange.com/questions/846011/… – TEX May 26 '18 at 14:34
• 2) $\Omega$ is not $\mathbb{R}^K$. $V$ is defined as a function from $\Omega$ to $\mathbb{R}^K$. – TEX May 26 '18 at 14:35
• 3) Why do you say that the first property is trivial? Suppose that $K=1$ and $V$ is a Bernoulli random variable. Then, I don't think that the support of $V$ is an open subset of $\mathbb{R}$ with strictly positive lebesgue measure. What about the case $K>1$? Thank you for your help – TEX May 26 '18 at 14:43

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.
• So just to be sure that I have understood: 1) is JUST ruling out purely discrete distributions, but it is satisfied for some mixtures of continuous and discrete variables (e.g., $Y=BX+(1-B)Z$ with $B$ Bernoulli and $Z$ standard Gaussian - do you confirm that in this mixture example the support is an open subset with strictly positive Lebesgue measure?). Hence 1) does not rule out the presence of mass points. – TEX May 26 '18 at 16:32