No, if $X$ is finite and $|X|>1$. As $\overline{\mu}(\{x\})=\mu(x)$, you use up all the measure on 1-element subsets. Then, if $\mu(x)>0$ and $x\neq y$, there is no measure left for $\overline{\mu}(\{x,y\})$ to be non-zero.

If $|X|>2$, then there is a single solution: $\overline{\mu}(\{x\})=\mu(x)$ and $\overline{\mu}(S)=0$ for any other set $S$.

Indeed, let $a=\overline{\mu}(\emptyset)$. By the condition for $S=\{x\}$, $\overline{\mu}(\{x\}) =\mu (x)-a$. By the condition for $S=\{x,y\}$ with $x\neq y$, $\overline{\mu}(\{x,y\}) =a$.

Now if $|X|>2$, there are at least $1+|X|$ two-element subsets. Thus, $\sum_{|S|\leq 2} \overline{\mu}(S) \geq a+ \sum_{x\in X}\mu (x) =1+a$. This forces $a=0$.

Now it is easy to show that for all larger subsets $\overline{\mu}(S)=0$.

No, of $X$ is finite and $|X|>1$. As $\overline{\mu}(\{x\})=\mu(x)$, you use up all the measure on 1-element subsets. Then, if $\mu(x)>0$ and $x\neq y$, there is no measure left for $\overline{\mu}(\{x,y\})$ to be non-zero.

No, if $X$ is finite and $|X|>1$. As $\overline{\mu}(\{x\})=\mu(x)$, you use up all the measure on 1-element subsets. Then, if $\mu(x)>0$ and $x\neq y$, there is no measure left for $\overline{\mu}(\{x,y\})$ to be non-zero.