Sheet 10
Analysis III, HS15
Output Friday, 20. November.
Delivery Thursday, 26. November. Mailbox in building Y27 on floor J for groups 1
and 2 floor G for group 3. Please write the name of your exercise instructor on your
homework!
Exercise 1.
Consider a measurable space (Ω, A, µ). Let (fn )n≥1 be a sequence of measurable functions fn : Ω → R such that fn ≥ −g for all n ≥ 1, and where g : Ω → [0, ∞] is an
integrable measurable function. Prove that
Z lim inf fn dµ ≤ lim inf
n→∞
Z
fn dµ.
n→∞
Solution 1.
Define the sequences hn = fn + g ≥ 0 and gn = inf k≥n hk for n ≥ 1. Then (gn )n≥1 is an
increasing sequence of non-negative measurable functions. We have
lim inf fn + g = lim inf (fn + g) = lim inf hn = lim gn .
n→∞
n→∞
n→∞
n→∞
R
R
Additionally, by monotone convergence (Satz 3.23), (limn→∞ gn )dµ = limn→∞ gn dµ,
so that
Z lim inf fn dµ =
n→∞
Z lim gn − g dµ = lim
n→∞
≤ lim inf
n→∞
Z
Z
n→∞
(hn − g)dµ = lim inf
n→∞
(gn − g)dµ
Z
fn dµ.
Remark 1. Alternatively, the result follows by applying Fatou’s lemma to the sequence
(hn )n≥1 .
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Exercise 2.
Consider a measurable space (Ω, A, µ). Let (fn )n≥1 be a sequence of non-negative
meaR
surable functions
which converges pointwise to a measurable function f , and f dµ =
R
limn→∞ fn dµ < ∞. Prove that
Z
Z
f dµ = lim
fn dµ
n→∞ E
E
for all E ∈ A.
Solution 2.
R
R
R
R
Let E ∈ A.
ObserveR that f dµ = (f χE + f χE c )dµ = E f dµ + E c f dµ. Similarly,
R
R
fn dµ = E fn dµ + E c fn dµ. Fatou’s lemma (Lemma 3.25) implies that
Z
f dµ ≤ lim inf
Z
n→∞
Ec
Ec
fn dµ.
Thus
Z
Z
n→∞
fn dµ −
fn dµ = lim sup
lim sup
n→∞
E
Z
f dµ − lim inf
Z
Ec
fn dµ
fn dµ
Z
Z
f dµ −
Ec
Z
n→∞
Z
fn dµ
n→∞
Zn→∞
≤
Ec
fn dµ − lim inf
= lim sup
=
Z
f dµ =
Ec
f dµ.
E
Another application of Fatou’s lemma gives
Z
f dµ ≤ lim inf
Z
n→∞
E
fn dµ,
E
whence
Z
lim sup
n→∞
fn dµ ≤
E
Z
f dµ ≤ lim inf
n→∞
E
Z
fn dµ.
E
The result follows.
Exercise 3.
Consider a measurable space (Ω, A, µ) such that µ(Ω) < ∞. Let (fn )n≥1 be a sequence
of integrable measurable functions fn : Ω → [−∞, ∞] which converges uniformly on Ω
to a function f . Prove that
Z
Z
f dµ = lim
n→∞
2
fn dµ.
Solution 3.
Since (fn )n≥1 converges uniformly on Ω to a function f , we can consider N ∈ N such
that |fn (ω) − f (ω)| ≤ 1 when n ≥ N and ω ∈ Ω. On the other hand, we observe that
g = |f1 | + · · · + |fN | is an integrable measurable function. Then, for ω ∈ Ω
|fn (ω)| ≤ |fn (ω) − f (ω)| + |fN (ω) − f (ω)| + |fN (ω)| ≤ 2 + g(ω).
Therefore, the result follows by the dominated convergence theorem (Satz 3.26) and the
fact that µ(Ω) < ∞.
Remark 2. Alternatively, the result follows by observing that
Z
Z
Z
f dµ − fn dµ ≤ |f − fn | dµ ≤ sup |f (ω) − fn (ω)| µ(Ω).
ω∈Ω
Exercise 4.
Compute the following:
(i)
lim
n
Z n
1 − nx
√
n→∞ 0
x
dx.
(ii)
Z 1
lim
n→∞ 0
(1 + nx2 )(1 + x2 )−n dx.
Solution 4.
(i) Recall the inequality 1 + x ≤ ex for x ∈ R and observe that for n ≥ 1
n
1− x
e−x
fn (x) := χ[0,n] (x) √ n ≤ √
for x ≥ 0.
x
x
√
√
In addition, limn→∞ fn (x) = e−x / x. On the other hand, f (x) = e−x / x is a
non-negative measurable function with respect B(R), and moreover
Z ∞
f (x)dx =
√
π.
0
Then by Satz 3.27, f is Lebesgue integrable. Since fn ≥ 0, by the dominated
convergence theorem (Satz 3.26) we get
lim
n
Z n
1 − nx
n→∞ 0
√
x
Z ∞
lim fn (x)dx =
dx =
0
3
n→∞
√
π.
(ii) Note that (1 + x2 )n ≥ 1 + nx2 for x ∈ [0, 1] and n ≥ 1. Then for n ≥ 1,
fn (x) := χ[0,1] (x)
1 + nx2
≤1
(1 + x2 )n
for x ≥ 0.
Moreover, limn→∞ fn (x) = 0. Then by Satz 3.27 and the dominated convergence
theorem (Satz 3.26) we get
Z 1
lim
n→∞ 0
2 −n
2
(1 + nx )(1 + x )
Z ∞
lim fn (x)dx = 0.
dx =
0
n→∞
Exercise 5.
Give an example that shows that we cannot remove the hypothesis that the function f
be non-negative from the Tonelli theorem (Satz 3.34).
Solution 5.
Consider two measurable spaces (N, P(N), µ) and (N, P(N), ν). We define
f (n, m) =

−n

 2−2
−2 +
2−n
if
if

 0
n = m,
n = m + 1,
otherwise.
This function is supported on the diagonal and the sub-diagonal of N × N. On the
diagonal f takes positive values and on the sub-diagonal f takes negative values. So
non-negativity is violated. Note that the counting measures are defined on the whole
powerset of N, so everything is measurable with respect to them. Note also that the
product measure µ × ν is the counting measure on N × N. We then take a look at why
the statement of Tonelli theorem does not hold, i.e. why the order of integration can
not be exchanged. Note first that
Z Z
f (n, m)dµdν =
N N
3
f (n, m) = 2 − 2−1 = .
2
n≥1 m≥1
X X
On the other hand taking the integrals in reverse order gives
Z Z
f (n, m)dνdµ =
N N
X X
f (n, m) =
m≥1 n≥1
X
(2 − 2m − 2 + 2−(m+1) )
m≥1
=
1
(2−m−1 − 2m ) = − .
2
m≥1
X
So we conclude that we can not exchange the order of integration.
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