Is the set $Unif(0,\frac{1}{n})$ for odd and even $n$ a 2-alternating capacity?

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Let $\Omega$ be a complete metrizable splace $\mathscr A$ its Borel-$\sigma$ algebra and $\mathscr M$ the set of all probability measures on $\Omega$. Every non-empty sub-set $\mathscr P \subset \mathscr M$ defines an upper probability $$v(A)=\sup\{P(A)|P\in\mathscr{P}\},\quad A\in\mathscr A.$$

Then, a $2$-alternating capacity is defined with the following conditions:

$1.$ $v(\emptyset)=0$ and $v(\Omega)=1$

$2.$ $A\subset B\Longrightarrow v(A)\leq v(B)$

$3.$ $A_n\uparrow A\Longrightarrow v(A_n)\uparrow v(A)$

$4.$ If $\mathscr P$ is weakly compact, then we also have $F_n\downarrow F,\,\, F_n\, \text closed\Longrightarrow v(F_n)\downarrow v(F)$

$5.$ $v(A\cup B)+v(A\cap B)\leq v(A)+v(B)$

That is also to say that the set $$P_v=\{P\in\mathscr M|P(A)\leq v(A)\quad\text{for all}\quad A\in\mathscr A\}$$ is not larger than the closed convex hull of the set $\mathscr P$ determining $v$.

Example $1:$ Let $\Omega=\{1,2,3\}$, $P_0=\{\frac{1}{2},\frac{1}{2},0\}$, $P_1=\{\frac{4}{6},\frac{1}{6},\frac{1}{6}\}$ and let $v$ be the upper probability determined by $\mathscr P=\{P_0,P_1\}$. Then, $$P_v=\Bigg\{\frac{3+t}{6},\frac{3-t-s}{6},\frac{s}{6}\Bigg|0\leq s,t\leq 1\Bigg\}$$ whereas the convex closure of $P$ is the proper subset of $P_v$ determined by $s=t$. In this example, $v$ is $2$-alternating.

Example $2:$ Let $\Omega=\{1,2,3,4\}$, $P_0=\{\frac{5}{10},\frac{2}{10},\frac{2}{10},\frac{1}{10}\}$, $P_1=\{\frac{6}{10},\frac{1}{10},\frac{1}{10},\frac{2}{10}\}$ and let $\mathscr P=\{P_0,P_1\}$. Here $v$ is not $2$-alternating. Let $A=\{1,2\}$ and $B=\{1,3\}$. Then,

$$v(A\cup B)+v(A\cap B)=\frac{15}{10}> v(A)+v(B)=\frac{14}{10}$$

Question: Let $U^n=Unif(0,\frac{1}{n})$ the uniform measure on unit interval. And $\mathscr{P}_0$ corresponds to those with odd $n$, $\mathscr{P}_1$ contains exactly those with even $n$. Is $\mathscr{P}_0$, $\mathscr{P}_1$ or $\{\mathscr{P}_0,\mathscr{P}_1\}$ a $2$-alternating capacity? If we change the definition with "odd and even $n$ less than $10^{10}$", does $\mathscr{P}_0$, $\mathscr{P}_1$ or $\{\mathscr{P}_0,\mathscr{P}_1\}$ become a $2$-alternating capacity?

The reference of this question is this paper. Examples are from page $254$.