MA2224 (Lebesgue integral) Tutorial sheet 8
[March 18, 2016]
Name: Solutions
P∞
R
χ[n,n+1/2n ] , use the monotone convergence theorem to calculate R f dµ
P
Solution: Let fn = nj=1 χ[j,j+1/2j ] . Since characteristic functions are nonnegative, fn ≥ 0
and fn ≤ fn+1 .
1. For f =
n=1
Of course the fn are measurable, since they are [finite] sums of characteristic functions of
measurable sets (intervals).
P
By definition f (x) = ∞
n=1 χ[n,n+1/2n ] (x) = limn→∞ fn (x) (pointwise limit of the partial
sums of the series).
By the monotone convergence theorem
Z
Z
f dµ = lim
fn dµ
n→∞
R
R
We can easily compute
Z
fn dµ =
R
(as n → ∞). So
n Z
X
j=1
R
R
n
X
1 1 − 1/2n+1
1
=
→1
χ[j,j+1/2j ] dµ =
j
2
2
1
−
1/2
R
j=1
f dµ = 1.
∞
X
(−1)n+1
2. For f =
χ[n,n+1) , calculate f + , f − , show that f is integrable and calculate
n
3
n=1
R
f
dµ
(giving
a jutsification for the calculation).
R
Solution: f + (x) = max(f (x), 0) is nonzero on the intervals [n, n+1) with n odd. (f (x) <
0 on [n, n + 1) if n is even and f (x) = 0 for x < 1.) So
f + (x) =
∞
X
k=1
1
32k−1
χ[2k−1,2k)
For f − (x) = max(−f (x), 0), it is nonzero on [2k, 2k + 1) (n = 2k even.) So
∞
X
1
f (x) =
χ[2k,2k+1)
32k
k=1
−
By use of the Monotone convergence theorem is a very similar way to Q1, we have
Z
+
f dµ =
R
=
=
lim
n→∞
lim
n→∞
1
32k−1
Z
1
R k=1
n
X
k=1
∞
X
k=1
=
Z X
n
32k−1
χ[2k−1,2k) dµ
χ[2k−1,2k) dµ
R
1
32k−1
1
1
3
=
3 1 − 1/9
8
By a similar argument again
∞
X
1
1
1
1
f dµ =
=
=
2k
3
9 1 − 1/9
8
R
k=1
Z
−
As f + and f − are both (measurable and) integrable (ie have finite integrals), so is f =
f + − f − integrable (the definition of f being integrable) and
Z
Z
Z
1
3 1
+
f dµ =
f dµ − f − dµ = − =
8 8
4
R
R
R
Richard M. Timoney
2