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Exercise 1.2.12 (Monotonicity and subadditivity as a consequense of additivity)
Let be the set of all Lebesgue measurable sets on , and let be the map that obeys empty set and countable additivity axioms. Then also obeys monotonicity and countable subadditivity.
Answers
- (i)
- (monotonicity)
Let . Then, we have the disjoint union . Therefore, by additivity (since by definition). - (ii)
- (countable subadditivity)
Let be a countable sequence of sets in . Notice that we can write the union of these sets in a disjoint formThis is not hard to demonstrate.
- Pick an arbitrary . Then for some we have .
- Pick an arbitrary . Then for some . Set . We then have ; it thus follows , and we are done.
- For all we obviously have .
Using this, we obtain
2020-05-30 00:00