MA2224 (Lebesgue integral) Tutorial sheet 7
[March 11, 2016]
Name: Solutions
1. For f = 2χ(0,1)∩Q + 3χ[1/2,9/2]\Q + 8χ[5,6] compute
R
R
f dµ and
R
[1,2]
f dµ.
Solution:
Since integrals are additive (for non-negative simple functions) and we can pull out positive
scalar factors
Z
Z
Z
Z
f dµ = 2 χ(0,1)∩Q dµ + 3 χ[1/2,9/2]\Q dµ + 8 χ[5,6] dµ
R
R
R
R
R
We know R χE dµ = µ(E) (for E ∈ L ), and so we need the measures of the 3 sets.
OSince (0, 1) ∩ Q is a countable set has measure 0.
Since [1/2, 9/2] \ Q = [1/2, 9/2] \ ([1/2, 9/2] ∩ Q) is all of the intervale [1/2, 9/2] but
a countable set, it has the same measure as [1/2, 9/2]. An argument that is perhaps more
watertight is that we can write
[1/2, 9/2] = ([1/2, 9/2] \ Q) ∪ ([1/2, 9/2] ∩ Q),
which is a disjoint union of measurable sets. So
µ([1/2, 9/2]) = µ([1/2, 9/2] \ Q) + µ([1/2, 9/2] ∩ Q), = µ([1/2, 9/2] \ Q) + 0.
In short µ([1/2, 9/2] \ Q) = 9/2 − 1/2 = 4.
Ans µ([5, 6]) = 1.
So
Z
f dµ = 2(0) + 3(4) + 8(1) = 20.
R
For the other integral
R
[1,2]
f dµ, we have
Z
Z
f dµ =
χ[1,2] f dµ
[1,2]
R
and
χ[1,2] f = 2χ[1,2] χ(0,1)∩Q + 3χ[1,2] χ[1/2,9/2]\Q + 8χ[1,2] χ[5,6]
= 2χ[1,2]∩(0,1)∩Q + 3χ[1,2]∩[1/2,9/2]\Q + 8χ[1,2]∩[5,6]
= 0 + 3χ[1,2]\Q + 0
So
Z
Z
χ[1,2] f dµ = 3µ([1, 2] \ Q) = 3µ([1, 2]) = 3
f dµ =
[1,2]
R
(since µ([1, 2] ∩ Q) = 0 and using the same argument as before).
2. For f (x) = |x|χ[−5/6,5/6] (x), let En,j = f −1
j = 1, 2, 3, 4.
j−1
2n
, 2jn
. Compute E1,1 , E1,2 and E2,j for
Solution:
E1,1 =
=
=
E1,2 =
=
E2,1 =
=
=
E2,2 =
=
E2,3 =
=
E2,4 =
=
1
f
0,
2
5 5
1
5
5
x∈ − ,
: |x| ∈ 0,
∪ −∞, −
∪
,∞
6 6
2
6
6
1 1
5
5
− ,
∪ −∞, −
∪
,∞
2 2
6
6
1
1 5
−1
,1
f
= {x ∈ R : |x| ∈ ,
}
2
2 6
1 5
5 1
− ,− ∪ ,
6 2
2 6
−1
1
f
0, 2
2
1
5
5
5 5
,∞
: |x| ∈ 0,
∪ −∞, −
∪
x∈ − ,
6 6
4
6
6
1 1
5
5
− ,
,∞
∪ −∞, −
∪
4 4
6
6
1 2
1 1
−1
f
,
}
= {x ∈ R : |x| ∈ ,
22 22
4 2
1 1
1 1
− ,− ∪ ,
2 4
4 2
2 3
1 3
−1
f
,
= {x ∈ R : |x| ∈ ,
}
22 22
2 4
3 1
1 3
− ,− ∪ ,
4 2
2 4
3 5
3 4
−1
f
,
= {x ∈ R : |x| ∈ , }
22 22
4 6
5 3
3 5
− ,− ∪ ,
6 4
4 6
−1
Compute
Z X
4
j−1
χE2,j dµ
22
R j=1
2
Solution: We have
Z X
Z X
4
4
j−1
j−1
χ
dµ
=
χE2,j dµ
E
2,j
22
22
R j=1
R j=2
Z
4
X
j−1
χE2,j dµ
=
22
R
j=2
=
4
X
j−1
j=2
22
µ(E2,j )
2
3
1
µ(E2,2 ) + µ(E2,3 ) + µ(E2,4 )
=
4 4 4 2 1
3
5 3
1 1
+
+
2
−
=
4 2
4 2
4
6 4
1
=
2
Richard M. Timoney
3