Real Analysis — Problem Set 5 Issued: 2.10.14 Due: 9.10.14, in class

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Jacobs University Bremen
School of Engineering and Science
Peter Oswald
Fall Term 2014
Real Analysis — Problem Set 5
Issued: 2.10.14
Due: 9.10.14, in class
Reading for this week: Chapters 2.2-3 in Folland [F]. Submit solutions for at least 5 problems.
5.1. Prove (for general measure spaces) the following (simple but useful) fact: If fn ∈ L1 (realor complex-valued functions), and
∞
X
kfn kL1 < ∞,
n=1
then
P
converges a.e. on X to a measurable function f (x),
a) the series ∞
n=1 fn (x)P
1
b) f ∈ L with kf kL1 ≤ ∞
n=1 kfn kL1 , and
c)
Z
∞ Z
X
f dµ =
fn dµ.
X
n=1 X
Why do I (maybe not you) like this Lemma?
5.2. Given an arbitrary measure space. Suppose fn , g ∈ L+ and fn → g a.e.. We observe that
Z
lim
fn dµ = c > 0
n→∞ X
exists. What can you say about the value of
R
X
g dµ?
5.3. Do Exercises 34 and 38 (about convergence in measure) of Chapter 2.4 in [F].
5.4. For the counting measure on N, the comparison of the convergence types considered in
class simplifies a lot. Show (using a table) which of the convergence types implies which
other. For each box (but the diagonal ones) give a short argument (or reference to a
theorem in [F] or else) why the implication holds and provide an example if it does not.
5.5. Suppose (X, M, µ) is a probability space, i.e., µ(X) = 1. Let f : X → R be measurable
(such a function is called random variable in probability theory), and assume that |f |2 ∈ L1
(the random variable has finite second moment).
R
a) Show that f ∈ L1 and that E(f ) := X f dµ is finite (Hint: Use the fact that |f (x)| ≤
|f (x)|2 for all x ∈ {x ∈ X : |f (x)| ≥ 1} and that µ(X) = 1).
b) Show that
Z
|f (x) − E(f )|2 dµ(x)
V (f ) :=
X
is finite (E(f ) and V (f ) are called expectation and variance of the random variable f ,
respectively.
c) Prove the Chebyshev-Markov inequality
µ({x ∈ X : |f (x) − E(f )| ≥ λ}) ≤ λ−2 V (f ).
5.6. Show that (assume Lebesgue measure or just work with the (improper) Riemann integral
definition) that
Z ∞
sin(ex )
dx → 0,
n → ∞.
1 + nx2
0
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