Consider a measurable space $\{\mathcal{I},\mathcal{B}\}$, where $\mathcal{I} = [0,1]$ and $\mathcal{B}$ are the Borel sets on $\mathcal{I}$. And also, denote $\mathcal{C}$ as the cantor set on $\mathcal{I}$.
Define an order $\mathcal{R}$ on $\mathcal{I}$ as: if both $s,t$ lie in $\mathcal{C}$, or both $s,t$ lie in the complement of $\mathcal{C}$, then $s\mathcal{R}t$ iff $s>t$; if $t \in \mathcal{C}$, while $s$ not, then $s\mathcal{R}t$. And define initial segment as $I(s,\mathcal{R})=\{t:s\mathcal{R}t\}$.
Now, define a set function $v=f\circ\lambda$, where $f$ is canter function, and $\lambda$ is the Lebesgue measure function.
Question, is there a measure function $\varphi^{\mathcal{R}}v$ on measurable space $\{\mathcal{I},\mathcal{B}\}$, which is induced by $\varphi^{\mathcal{R}}v(I(s,\mathcal{R})) = v(I(s,\mathcal{R})), \forall s \in \mathcal{I}$? If there is, is it atomic?
This is a statement on Page 95 of "Values of Non-Atomic Games 1972 by Aumann and Shapley". Could anyone can explain in detail? Thank you.
P.S. Measure function $\varphi^{\mathcal{R}}v$ is Non-Atomic on $\{\mathcal{I},\mathcal{B}\}$ iff $\forall S \in \mathcal{B}$, if $\varphi^{\mathcal{R}}v(S)>0$, then there must exist $T \in \mathcal{B}, T \subset S$ and $\varphi^{\mathcal{R}}v(T)>0$. And the Cantor function is exactly Devils Staircase function.
I don't see how such a measure (call it $\mu$ for brevity) could exist. The definition of order implies that $I(1,\mathcal R)$ is precisely $C\setminus \{0\}$. Since $\lambda(C)=0$, we must have $\mu(C\setminus \{1\})=0$. Let $\alpha=\mu(\{1\}) = \mu(C)$.
Next, for $s\notin C$ the set $I(s,\mathcal R)$ is the union of $C$ and the interval $[0,s)$. Therefore, $\mu([0,s))=f(s) - \alpha$. If $\alpha>0$ than we can pick $s\notin C$ such that $f(s)<\alpha/2$, and arrive at a contradiction: $\mu([0,s))<0$. Thus, $\alpha=0$.
For all $s\notin C$ we have $\mu([0,s))=f(s)$. Also, $f$ is constant on every interval in the complement of $C$. Therefore, each such interval has zero $\mu$-measure. So the measure is just zero, which obviously does not work.