Homepage Solution manuals Terence Tao An Introduction to Measure Theory Exercise 1.2.12 (Monotonicity and subadditivity as a consequense of additivity)

Exercise 1.2.12 (Monotonicity and subadditivity as a consequense of additivity)

Let L(d) be the set of all Lebesgue measurable sets on d, and let m : L [0,+] be the map that obeys empty set and countable additivity axioms. Then m also obeys monotonicity and countable subadditivity.

Answers

(i)
(monotonicity)
Let A,B : A B. Then, we have the disjoint union B = A (BA). Therefore, by additivity m(B) = m(A) + m(BA) m(A) (since m(BA) 0 by definition).
(ii)
(countable subadditivity)
Let E1,E2 be a countable sequence of sets in d. Notice that we can write the union of these sets in a disjoint form n=1E n = n=1E n ( i=1n1E i) .

This is not hard to demonstrate.

  • Pick an arbitrary x n=1En ( i=1n1Ei). Then for some n we have x En ( i=1n1Ei) En En.
  • Pick an arbitrary x n=1En. Then x En for some n . Set n0 := min {n : x En}. We then have n0 < n; it thus follows x En0 ( i=1n01Ei), and we are done.
  • For all n,n : n > n we obviously have [En ( i=1n1Ei)] [En ( i=1n1 Ei)] = .

Using this, we obtain

m ( n=1E n) = n=1m (E n ( i=1n1E i)) n=1m (E n) .

User profile picture
2020-05-30 00:00
Comments