Mathematical expectation is inside convex hull of support

Let $\xi$ be a random variable supported in some set $A \in \mathbb{R}^n$: $\xi \in A$ a.e. How to show that $\mathsf E \xi \in \mathop{\mathrm{conv}}{A}$?

Let $s(x)$ be a support function of set $A$: $s(x) = \sup\limits_{y \in A} \langle x,y\rangle$, $x \in \mathbb{R}^n$. Then
$$y \in \mathop{\overline{\mathrm{conv}}}A \Leftrightarrow \langle x,y \rangle \leqslant s(x) \, \forall x$$
So $\langle x, \mathsf E \xi \rangle = \mathsf E \langle x, \xi\rangle \leqslant \mathsf E s(x) = s(x)$ since $\xi \in A \subseteq \mathop{\overline{\mathrm{conv}}}A$ a.e. This shows that $\mathsf E\xi \in \mathop{\overline{\mathrm{conv}}}A$. But how to show that we have stronger result: $\mathsf E \xi \in \mathop{\mathrm{conv}}{A}$?

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If $\mathbb E \zeta$ is on the boundary $\overline{\text{conv}} A\setminus \text{conv} A$, there is a hyperplane $H=\{x\mid \langle x,y\rangle=0\}$ containing $\mathbb E \zeta$, such that $A$ lies on one side, $\{x\mid \langle x,y\rangle\geq 0\}$. But $\langle \zeta,y\rangle$ is non-negative and has expectation zero, so must be zero almost surely: $\zeta\in H$ with probability $1$. The result then follows by induction on dimension and considering $A\cap H$.

Note that the result isn’t true in infinite dimensions: the (finitary) convex hull of unit vectors $(1,0,0,0,\dots),(0,1,0,0,\dots),\dots\in\mathbb R^{\infty}$ does not contain $(1/2,1/4,1/8,\dots)$ for example.