# 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

## 1 Answer

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.

• Thank you. Two clarifications: A) a countably infinite support (your first example) is an open subset with strictly positive Lebesgue measure? B) which cases is (1) ruling out? – TEX May 26 '18 at 15:43
• My first example has closed support (the strictly positive integers) with zero Lebesgue measure. In my opinion (1) is not ruling out anything but purely discrete distributions (such as that example). You always have the option to add in a mixture of a discrete distribution to make condition (1) fail to be without atoms. – dioid May 26 '18 at 16:20
• 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
• Yes, I can confirm that. – dioid May 26 '18 at 16:37
• Could you help me with this other question on a similar topic? math.stackexchange.com/questions/2797048/… – TEX May 27 '18 at 11:35