Does Sigma Algebra Necessarily Induce a Measure? I am wondering if we have a space $X$ and an outer measure $\mu^*$ defined on $P(X)$, is it always true that the restriction $\mu^*|_\mathcal{F}$ is in fact a measure for an arbitrary $\sigma$-algebra $\mathcal{F}$ on $X$ even without necessarily satisfying the Catheodory’s Criterion? If yes, why? If no, why do we call sets in $\mathcal{F}$ as $\mathcal{F}$-measurable?
 A: No, this is not true. For example, this doesn't hold for $\mathcal{F}=\mathcal{P}(\mathbb{R})$ when $\mu^*$ is the Lebesgue outer measure. However, the restriction of $\mu^*$ to the sets satisfying the Caratheodory criterion is a measure.
A: As d.k.o. has indicated, it is not the case that one can extend arbitrary outer measures to arbitrary $\sigma$-algebras.  For example, $\mathscr{P}(\mathbb{R})$, the power set of $\mathbb{R}$, is a $\sigma$-algebra.  But any outer measure $\mu^*$ on $\mathscr{P}(\mathbb{R})$ extends to $\mathcal{F}=\mathscr{P}(\mathbb{R})$ iff $\mu^*$ was a measure in the first place.  And it is easy to construct outer measures that are not measures; this is Vitali's paradox.
However, every $\sigma$-algebra admits a whole family of measures.  For example, consider the well-known Dirac delta: $$\delta_x(S)=\begin{cases}
1 & x\in S\\
0&x\notin S
\end{cases}$$  For any $x\in X$ and any $\sigma$-algebra $\mathcal{F}\subseteq\mathscr{P}(X)$, it is easy to show that the rule defining $\delta_x$ gives a measure on $\mathcal{F}$.  (Of course, $\mathcal{F}$ need not be complete w.r.t. $\delta_x$ — indeed, it probably is not.)
If you don't know what measure you plan to use or will be changing measures frequently, but have a common underlying $\sigma$-algebra, it makes sense to "lift" the concept of measurability to the $\sigma$-algebra.  (Probabilists and $C^*$-algebraists generalize that sort of reasoning much further.)  That's probably why your professor did so.
