MA2224 (Lebesgue integral) Tutorial sheet 9
[April 1, 2016]
Name: Solutions
1. For f (x) = max
integrals.]
1
2
R
− |x|, 0 , plot y = f (x) and compute R f dµ. [Hint: use Riemann
Solution:
Since f is continuous and nonzero only on [−1/2, 1.2], the Lebesgue integral agrees with
the Riemann integral:
Z
Z
f dµ
f dµ =
[−1/2,1/2]
R
Z
=
1/2f (x) dx
−1/2
0
Z 1/2
1
=
+ x dx +
−1/2 2
0
2 0
x x
x
=
+
+
−
2
2 −1/2
2
1 1 1 1
1
=
− + − =
4 8 4 8
4
Z
1
− x dx
2
1/2
x2
2 0
2. With f as in Q1, let fn (x) = f (x − n). Plot f1 , f2 , f3 .
Solution:
Let
Fn =
n
X
(−1)j+1
j
j=1
fj
and compute
Z
Fn dµ
R
Solution: Since integrals are linear and
Z
Fn dµ =
R
R
R
fn dµ =
Z
n
X
(−1)j+1
j=1
R
R
f dµ = 1/4,
n
1 X (−1)j+1
fj dµ =
4 j=1
j
R
j
2
P∞ (−1)n+1
3. With fn as in the previous question, F (x) =
fn show that the improper
n=1
n
R∞
Rb
Riemann
integral 0 F (x) dx = limb→∞ 0 F (x) dx exists while the Lebesgue integral
R
F dµ does not exist (i.e. that F = F χ[0,∞) is not Lebesgue integrable).
[0,∞)
Solution: If b = n + (1/2), then (because if x ≤ n + 1/2, then fj (x) = 0 for j > n)
Z n+(1/2)
Z b
n
∞
1 X (−1)j+1
1 X (−1)j+1
Fn (x) dx =
→
F (x) dx =
4 j=1
j
4 j=1
j
0
0
as n → ∞ (and the limit exists by the alternating series test — it is the sum of an alternating
series with terms having absolute values that decrease monotonically to 0 — in fact the
limit is (1/4) ln 2 but we don’t need that value).
If b > 2 is not n + 1/2 for any n, then it is between n + 1/2 and n + 3/2 for some n and
Z b
Z n+(1/2)
Z b
(−1)n+2
fn+1 (x) dx.
F (x) dx =
Fn (x) dx +
0
0
n+(1/2) n + 1
The latter integral has small absolute value. To be precise it has abolute value smaller than
1
4(n + 1)
So (considering b → ∞ continuously we still get the same limit as restricting to b =
n + 1/2)
Z b
∞
1 X (−1)j+1
F (x) dx =
lim
.
b→∞ 0
4 j=1
j
(and the improper Riemann integral exists).
Since the fn have nonoverlapping supports (that is, for each x there is at most one n with
fn (x) 6= 0)
∞
X
1
|F (x)| =
fn
n
n=1
We can see by the Monotone Convergence theorem that
Z
Z X
n
n
X
1
1
|F | dµ = lim
fj dµ = lim
=∞
n→∞ R
n→∞
j
4j
R
j=1
j=1
(as the harmonic series
P∞
1
j=1 j
diverges). Thus F is not Lebesgue integrable.
Aside: It follows from the Monotone Convergence theorem that |F | is measurable. It is in
fact true that F is measurable since F (x) = limn→∞ Fn (x) shows that F is the pointwise
limit of measurable functions Fn — in fact the Fn are
R even continuous. For F to be
integrable it would have to be measurable and to have R |F | dµ < ∞. As the latter fails,
we don’t need to worry about measurability.
Richard M. Timoney
3