MA2224 (Lebesgue integral) Tutorial sheet 2
[January 29, 2016]
Name: Solutions
1. Let f : A → B be a function and suppose E, E1 , E2 ⊆ B. Show that
(a) χf −1 (E) = χE ◦ f .
Solution: For a ∈ A we have either a ∈ f −1 (E) or a ∈
/ f −1 (E). If a ∈ f −1 (E), then
f (a) ∈ E and so χE (f (a)) = 1 = χf −1 (E) (a). If a ∈
/ f −1 (E), then f (a) ∈
/ E and so
χE (f (a)) = 0 = χf −1 (E) (a).
Thus χf −1 (E) (a) = χE (f (a)) = (χE ◦ f )(a) in both cases.
(b) f −1 (E1 ∩ E2 ) = f −1 (E1 ) ∩ f −1 (E2 )
Solution:
a ∈ f −1 (E1 ∩ E2 )
⇐⇒
⇐⇒
⇐⇒
⇐⇒
f (a) ∈ E1 ∩ E2
f (a) ∈ E1 and f (a) ∈ E2
a ∈ f −1 (E1 ) and a ∈ f −1 (E2 )
a ∈ f −1 (E1 ) ∩ f −1 (E2 ).
2. Let I1 = (3, 5], I2 = (−1, 4], I3 = (−4, −2], I4 = (4, ∞), I5 = (9, 10].
For each of the follwing sets E, find the ‘standard’ representation of the set (as a union of
intervals in the interval algebra which do not overlap or share end points and are arranged
left to right). Also find the measure m of the set.
(a) E = I5 ∪ I2
Solution: ‘standard form’ E = I2 ∪ I5 = (−1, 4] ∪ (9, 10]
(Note: disjoint intervals, no shared end points, ordered to increase left to right)
m(E) = (4 − (−1)) + (10 − 9) = 6
(b) E = I1 ∪ I2 ∪ I4
Solution: Solution: ‘standard form’ E = (−1, ∞)
m(E) = ∞
(c) E = I1 ∪ I2 ∪ I3
Solution: ‘standard form’ E == I3 ∪ (−1, 5] = (−4, −2] ∪ (−1, 5]
m(E) = (−2 + 4) + (5 + 1) = 8
(d) E = I1 ∪ I2 ∪ I3 ∪ I5
Solution: ‘standard form’ E = I3 ∪ (−1, 5] ∪ I5 = (−4, −2] ∪ (−1, 5] ∪ (9, 10]
m(E) = 2 + 6 + 1 = 9
3. Let A denote the collection of all subsets of R that are either finite or complements of finite
subsets. (Subsets with finite complement are called ‘co-finite’ subsets.) Show that A is an
algebra.
Solution:
(a) ∅ ∈ A since it is finite.
(b) (Finite unions.) If E1 , E2 ∈ A, then we can consider two cases. In the first case both
E1 and E2 are finite and then so is E1 ∪ E2 finite — so E1 ∪ E2 ∈ A. In the second
case at least one of E1 or E2 is infinite and then must have a finite complement. In
that case
(E1 ∪ E2 )c = E1c ∩ E2c
must be finite (since at least one of E1c and E2c is finite). So E1 ∪ E2 ∈ A (in all
cases).
(c) (complements.) If E ∈ A, then either E is finite or E c is finite. If E is finite then E c
is co-finite (has finite complement E) and so E c ∈ A in that case. If E c is finite, then
also E c ∈ A.
Richard M. Timoney
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