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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!

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    $\begingroup$ When it comes to the support of a function, it is actually a matter of preference, as I have seen both practice (taking closure or leaving it as it is). But the support of measure, defined as in Definition 2.5, is always a closed set regardless of the lack of closure operation. So it makes sense to have closure operations in Definition 2.6. $\endgroup$ – Sangchul Lee Sep 27 '19 at 20:28
  • $\begingroup$ I'm happy to see you again @SangchulLee :) Could you write your comment as an answer so that I can accept it? $\endgroup$ – LAD Sep 27 '19 at 20:32
  • $\begingroup$ @SangchulLee I'm quite surprised that you have retyped the images in 3 of my questions in LaTex. This job takes much time. I wonder why you spent time doing so ^o^ $\endgroup$ – LAD Sep 27 '19 at 20:33
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    $\begingroup$ No problem, I am just out of my PC now, so I will try to make an answer when I go back. Anyway, if you get used to $\LaTeX$, it actually takes not much time to write in it. For instance, it took me less than 5 minutes to do so. And why I am doing so is to make your posting better exposed to search engines, so other people have better access to it. I also encourage you to do so whenever available :) $\endgroup$ – Sangchul Lee Sep 27 '19 at 20:38
  • $\begingroup$ @SangchulLee I really appreciate your kindness and contribution to the community. Honestly, I'm having a high workload with study, so I'm able to type LaTex when the image is not a True-HD PDF ;( $\endgroup$ – LAD Sep 27 '19 at 20:40
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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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