Extension of Measure

measure-theoryprobability

Consider the following steps of extension of a measure:

(1) Define $\mathbb{P}$, a probability measure on $F$, a field of subsets of a set $\Omega$.

(2) First extension of $\mathbb{P}$ goes as follows. Establish $G$, which is a collection of all limits of increasing sequence of sets in $F$. This is the collection of all countable unions of sets in $F$. Define $\mu$ on $G$, where $\mu=\mathbb{P}$ on $F$.

(3) Now, we try to extend $\mu$ to the class of all subsets of $\Omega$.

My question: in (3), why is it that the $\mu$ will not be countably additive on all subsets but only on a smaller $\sigma$-field? The book defines the outer measure $\mu^*$ on $\Omega$ after (3) due to this reason. AND THEN, the author identifies a $\sigma$-field on which the outer measure $\mu^*$ is countably additive.

Best Answer

It can be answered with an example.

Take for instance $\Omega=[0,1]$ and $P=m^*$ Lebesgue outer measure.

Take $V\subseteq [0,1]$ a non measurable set.

Then $m^*$ defined on the powerset of $[0,1]$ is not even finitely additive.

This is because $m^*(V)+m^*([0,1] \setminus V)>m^*(([0,1] \setminus V) \cup V)$

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