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.
Bugs Bunny
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