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Let us stick to $\mathbb{R}$ and $\mathcal{B}$, the Borel Sigma field. Given that $\mathbb{F}$ is a sigma field of $\Omega$, consider a random variable $X$

$X:\begin{bmatrix} \Omega \\ \mathbb{F} \\ P \end{bmatrix} \to \begin{bmatrix} \mathbb{R}\\\mathcal{B}\\ P_X \end{bmatrix}$

Definition: $S_X:=$ smallest closed subset of $\mathbb{R}$ such that $P_X(S_X)=1$. Call this support.

Definition: $U_X:=\{x\in\mathbb{R} \ | \ \forall \ \varepsilon>0, \ P_X(V_{\varepsilon}(x))>0 \}$ (where $V_\varepsilon(x)$ is the $\varepsilon$ neighbourhood of $x$)

I was able to show that the definition $S_X$ is equivalent to the definition $U_X$.

Consider now, that $X$ is also a continuous random variable, which means: $$P_X(a,b)=\int_a^b p_d(t)dt$$ where $p_d(t)$ is the density function.

this answer, at the very end, defines the support $G_X$, for this case, as follows: $$G_X=\text{closure}\{x\in\mathbb{R}\ | \ p_d(x)>0 \}$$

Now I want to show that, for the case of continuous random variable, $G_X \equiv S_X\equiv U_X$.

It was rather easy to show that $ S_X \subseteq G_X$, but the reverse assertion is proving to be trouble some. I suspect some application of Lagrange Mean value theorem or something of that sort. Any ideas?

nickbros123
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    The definition of $G_X$ does not even meke sense since pdf's are defined only a.e. in general. You can use this definition if it is assumed that $p_d$ is a continuous function. – Kavi Rama Murthy Apr 07 '24 at 07:16

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Proof of $G_X \subseteq S_X$ under the assumption that $p_d$ is continuous:

Suppose $x \in G_X$. If $x \notin S_X$ then $x \notin U_X$, so there exists $\epsilon >0$ such that $P_X((x-\epsilon, x+\epsilon))=0$. By definition of $G_X$, the interval $(x-\epsilon, x+\epsilon)$ intersects $\{x: p_d(x)>0\}$. So there exists $y $ in this interval with $p_X(y)>0$. But $P_X((x-\epsilon, x+\epsilon))=0$ says that $\int_{x-\epsilon}^{x+\epsilon} p_d(t)dt=0$. By continuity and non-negativity of $p_d$ this implies that $p_d(t) =0$ for all $t$ in $(x-\epsilon, x+\epsilon)$. In particular, $p_d(y)=0$, a contradction.

Kavi Rama Murthy
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  • In the linked answer, I didn't notice the hypothesis that $p_d(x)$ is also continuous, which was mentioned, my bad. Thanks for the correction and proof – nickbros123 Apr 07 '24 at 08:54