Let be a set and an outer measure on , we call a subset to be -measurable if
for all . This should remind the reader of a full measure since this is similar to disjoint additivity. Thus, the natural question arises of the significance of these measurable sets. Caratheodory's Theorem provides and answer by proving that if is the collection of measurable sets, then is indeed a Measure Space.
Consider an outer measure on .
Then is a -algebra and is a measure on .
First, observe that is closed under complements due to symmetry in the meaning of -measurability. Now, we show if then  to conlcude that is an algebra.
Suppose . Then
and by subadditivity
But certainly, since the inequality in the other direction also holds, and we conclude
hence and we have shown that is an algebra.
Now, suppose are disjoint. Then
so is finitely additive.
Next, we show is closed under countable disjoint unions. Given a disjoint sequence of sets , for all , by countable subadditivity,
It remains to show the other inequality direction. Since is closed under finite unions, . By using the definition of -measurability,
and by monotonicity,
Since this holds for all ,
Similarly as before, the other inequality direction is proved by monotonicity, so we conclude equality and we have that .
The only thing that remains to be shown is closure under countable unions. Consider a sequence of sets not necessarily disjoint. Define
Their unions are equal and is disjoint, hence contained in , proving it to be a algebra.
This finishes all parts to the proof.
- ↑ Craig, Katy. MATH 201A Lectures 6,7. UC Santa Barbara, Fall 2020.
- ↑ Folland, Gerald B., Real Analysis: Modern Techniques and Their Applications, second edition, §1.4