Is this a typo in the definition of support of a random variable

definitionprobability theory

Good evening, I'm reading about the support of a random variable in my lecture notes:

Definition 2.5(src) (Support) Let $X$ be a real-valued random variable on $(\Omega,\mathcal{A},\mathbb{P})$. The support of $X$, denoted $\mathit{Supp}(X)$, is defined as follows:
$$ \mathit{Supp}(X) = \{x\in\mathbb{R}; \ \forall N_x, \ \mathbb{P}(X \in N_x) \neq 0 \} $$
where $N_x$ is an open neighborhood of $x$.

Definition 2.6(src) (Support: discrete and continuous case). Let $X$ be a real-valued random variable on $(\Omega, \mathcal{A}, \mathbb{P})$.

  • If $X$ is discrete (see definition 2.9), then
    $$ \mathit{Supp}(X) = \overline{\{ x \in \mathbb{R}; \ \mathbb{P}(X = x) \neq 0 \}}.$$

  • If $X$ is absolutely continuous with respect to the Lebesgue measure (see definition-theorem 2.1) and $f_X$ is a p.d.f. of $X$, does not have any isolated point, then
    $$ \mathit{Supp}(X) = \overline{\{ x \in \mathbb{R}; \ f(x) \neq 0 \}},$$
    where $f$ is a density of $X$.

Whereas the definition of support is given by Wikipedia's page as follows:

In practice, support of a discrete random variable $X$ is often defined as the set
$$ R_{X} = \{ x \in \mathbb{R} : P(X=x) > 0 \}. $$
And support of a continuous random variable $X$ is defined as the set
$$ R_{X} = \{ x \in \mathbb{R} : f_{X}(x) > 0 \}, $$
where $f_{X}(x)$ is a probability density function of $X$.(src)

Clearly, the support in my lecture note is the closure of that from Wikipedia.

My question: Is it a typo in my lecture note?

Thank you so much for your clarification!

Best Answer

Closure operations appearing Definition 2.6 are simple consequences of the fact that the support, defined in Definition 2.5, is always a closed set. In fact, Definition 2.6 is rather a consequence of Definition 2.5 applied to special cases.

I would like to remark that Definition 2.5 is quite natural and extends to even broader class of objects like distributions and so forth, I would consider it more standard than what is discussed in the Wikipedia article.

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