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?

Solution 1:

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.

Solution 2:

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.