Determine the outer measure $\mu^{*}$ induced by $\mu$ and the $\sigma$-algebra of measurable subsets

Suppose $S={\phi, [1,2]}$, $\mu(\phi)=0$, $\mu([1,2])=1$ as you indicated.

By definition of outer measure, for any $E \subset X$, $\mu^*(E)=\inf{\sum \mu(E_k)}$ such that $E \subset \cup E_k$.

Then, for arbitrary $E$: if $E = \phi$, $\mu^*(E)=0$ (since it is covered by $\cup \phi$); if $E\subset[0,1]$, $\mu^*(E)=1$ (since it is covered by $[0,1] \cup \cup_{k=1}^{\infty}\phi$; otherwise $\mu^*(E)=\infty$ (since no infimum exists).

To determine the $\sigma$-algebra of measurable sets, recall that $E$ is measurable iff for any $A \subset \bf{R}$, $\mu^*(A)=\mu^*(A \cap E) + \mu^*(A \cap E^c)$. This statement is trivial if $A = \phi$ or $A$ not contained in $[0,1]$. Therefore consider $A \subset [0,1]$. Then $\mu^*(A) = 1 = \mu^*(A \cap E) + \mu^*(A \cap E^c)$ if and only if $A \cap E = \phi$ or $A \cap E^c = \phi$. Thus, $E$ is measurable if and only if $[0,1] \subset E$ or $E \cap [0,1] = \phi$.