Download ANALYSIS 3 PROBLEMS FOR WEEK 5 SOLUTIONS Problem 1

ANALYSIS 3
PROBLEMS FOR WEEK 5
SOLUTIONS
Problem 1. Recall that for a function f : X → R the following three statements are equivalent:
(i) f is measurable,
(ii) {f < a} is measurable for each a ∈ R,
(iii) {f > a} is measurable for each a ∈ R. First of all let us prove that |f | is measurable, indeed
this follows from
{|f | < a} = ∅, if a < 0
and
{|f | < a} = {f < a} ∩ {f > −a}, if a ≥ 0
combined with (ii) and (iii) and the measurability of f .
Next assume that p > 0. We have
{|f |p > a} = X,
and
if a ≤ 0
1
{|f |p > a} = {|f | > a p }, if a > 0,
and so by (iii) and the measurability of |f | it follows that |f |p is measurable for every p > 0.
Finally assume that f (x) 6= 0 for every x ∈ X. Note that
o
n1
> 0 = {f > 0}
f
n1
o n
1o
>a = f <
, if a > 0
f
a
and
n1
o n
1o
>a = f <
∪ {f > 0}, if a < 0
f
a
so that using (iii) again we conclude.
Problem 2. You have seen in class that B(R2 ) = B(R) ⊗ B(R), so if we prove that A is
open in R2 with the euclidean topology we are done. This is trivial because the function
R2 3 (x, y) 7→ f (x, y) := |x − y| ∈ R is continuous, indeed
|f (x, y) − f (x0 , y 0 )| = ||x − y| − |x0 − y 0 || ≤ |x − y − x0 + y 0 | ≤ |x − x0 | + |y − y 0 |
and the last term is a norm on R2 and so, since all norms are equivalent on a finite dimensional
space, it induces the usual Euclidean topology on R2 .
Problem 3. You have seen the implication (⇒) in class. Let us prove (⇐). So let (fn )n be a
sequence of bounded simple functions and assume
(1)
lim sup |fn (x) − f (x)| = 0
n→∞ x∈X
In particular this implies that fn (x) → f (x) for every x ∈ X, which implies the measurability
of f (the pointwise limit of measurable functions is measurable, as you have seen in class).
For the boundedness of f simply notice that by (1) there exists N such that
sup |fN (x) − f (x)| ≤ ε
x∈X
and moreover supx∈X |fN (x)| ≤ C < ∞, therefore
|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ ε + C < ∞
1
2
ANALYSIS 3 PROBLEMS FOR WEEK 5 SOLUTIONS
for every x ∈ X.
Problem 4. Without loss of generality we can assume that the the decimal representation of
every rational number of the form k10−m ends with infinite zeros (rather than infinite nines),
because the set of all such rationals has measure zero. With this assumption each real number
has a uniquely defined decimal expansion.
For each n ≥ 1 let us define
En = {x ∈ [0, 1] | the nth decimal place of x is zero}
Then we have
A=
∞
\
Enc
n=1
and claim that m(A) = 0. It clearly suffices to show (by the decreasing monotone convergence
theorem for measures) that
N
\
m(
Enc ) → 0
n=1
as N → ∞. We shall prove that
N
\
m(
Enc ) = (
n=1
9 N
)
10
by induction. In fact, given this for N − 1, we consider
c
EN
∩
N
−1
\
Enc .
n=1
For any y ∈
TN −1
n=1
Enc we have
y=
N
−1
X
−n
yn 10
n=1
+
∞
X
yn 10−n
n=N
0
where each yn 6= 0 for n ≤ N − 1. For each fixed sequence y10 , . . . , yN
−1 , the collection of all y
in this set with that fixed sequence is in one-to-one correspondence with the interval [0, 1] by
the affine map
N
−1
X
yn0 10−n 10−N +1
y 7→ y −
n=1
By the scaling property of Lebesgue measure, this set has measure 10−N +1 . But then the interc
section of EN
and the set of y with this fixed sequence has measure (0.9)10−N +1 (independent
of the fixed sequence) so the result follows.
Problem 5. There are many possible constructions. For example, let (rn )∞
n=1 be an enumeration of Q ∩ [0, 1] (since the latter set is countable) and let
∞
[
U=
(rn − 2−n , rn + 2−n )
n=1
for some small > 0. Then we have
m(U ) ≤
∞
X
2−n+1 = .
n=1
Furthermore, U is a dense set in [0, 1] (since it contains the dense set Q ∩ [0, 1]) so the closure
U = [0, 1] (in the restriction topology). Thus
m(U \ U ) = m([0, 1] \ U ) = 1 − and the result follows for any < 1.
ANALYSIS 3
PROBLEMS FOR WEEK 5
SOLUTIONS
3
Problem 6. If f is bounded the result trivially holds by choosing δ < µ(E)εsup f . If f is not
bounded consider the sequence fn := min{f, n}, then fn (x) ≤ n for every x ∈ X, (fn )n is an
increasing sequence of positive functions and limn→∞ fn (x) = f (x). Therefore we can apply the
monotone convergence theorem to conclude
Z
Z
f dµ = lim
fn dµ
n→∞ E
E
R
for every E ⊂ X. Note that by assumption X f dµ < ∞, hence both sides of the equality
above are finite. Now let ε > 0 be given, then there exists N such that
Z
Z
ε
f
dµ
−
f
dµ
< .
N
2
E
E
Take δ = ε/2N , then we have
Z
Z
Z
Z
ε
f dµ ≤ f dµ −
fN dµ +
fN dµ ≤ + N µ(E) < ε
2
E
E
E
E
whenever E ⊂ X with µ(E) < δ.