Homework 2 Solutions, Real Analysis I, Fall, 2010.
(6) Rudin, chapter 1, problem 5.
(a) Solution: To prove A = {x : f (x) < g(x)} is measurable,
we would like to say that A = (f − g)−1 ([−∞, 0)), which
would be measurable since [−∞, 0) is open and f − g is
measurable. Unfortunately, this doesn’t avoid the possibility that f − g is of ∞ − ∞ type, which must be handled
separately.
To be precise, the remarks in 1.9 show that if f, g are Rvalued functions, we may consider them to be C-valued,
and then f − g = f + (−1)g is measurable. This leaves
out the case of infinite values, however. The worst case
happens when ∞ − ∞ occurs, which is on
E = [f −1 (∞) ∩ g −1 (∞)] ∪ [f −1 (−∞) ∩ g −1 (−∞)].
E is measurable since {∞} and {−∞}, as closed points,
are Borel sets. Also let
F = f −1 ({∞, −∞}) ∪ g −1 ({∞, −∞})
be the set in which either f or g is infinite. Then E ⊂ F
and F is measurable also. On F c , f − g is measurable,
since f, g are finite and measurable there. Then we may
define measurable functions
f˜ = f χF c ,
g̃ = gχF c ,
h = f˜ − g̃.
Note f˜ is measurable since
−1
f ((α, ∞]) − F for α ≥ 0,
−1
f˜ ((α, ∞]) =
f −1 ((α, ∞]) ∪ F for α < 0.
Now we may check
A = [h−1 ((−∞, 0)) ∪ f −1 (−∞) ∪ g −1 (∞)] − E,
which is clearly measurable.
The other set {x : f (x) = g(x)} is also measurable, since
it is equal to
h−1 (0) ∪ E,
and {0} is a Borel set.
(b) Solution: Let fn : X → R be a sequence of measurable
functions. Then we know by Theorem 1.14 that k(x) =
lim sup fn (x) and `(x) = lim inf fn (x) are measurable. The
set of points where {fn (x)} converge to a finite limit are
1
2
then the set of points in which k(x) = `(x) and k(x) is
finite. This is equal to
{x : k(x) = `(x)} − k −1 ({−∞, ∞}).
This is measurable by part (a) and since {−∞, ∞} is a
Borel set.
(7) Rudin, chapter 1, problem 10.
Solution: Let µ(X) < ∞ and let fn be a sequence of bounded
complex measurable functions on X. Assume fn → f uniformly
on X.
Assume |fn | ≤ Cn . The uniform convergence fn → f means
for all > 0, there is an N so that n ≥ N implies |fn (x) −
f (x)| < . Note is independent of x ∈ X. So choose and
find N as above. Then
|f (x)| ≤ |fN (x)| + |fN (x) − f (x)| ≤ CN + for all x ∈ X.
Then for n ≥ N ,
|fn (x)| ≤ |f (x)| + |fn (x) − f (x)| ≤ CN + 2 for all x ∈ X.
So for all n = 1, 2, . . . and x ∈ X, we have |fn (x)| ≤ C for
C = max{C1 , C2 , . . . , CN −1 , CN + 2}.
Now we may apply the Dominated Convergence Theorem,
since
Z
C dµ = Cµ(X) < ∞ =⇒ C ∈ L1 (µ).
X
This proves
lim
n→∞
Z
X
fn dµ =
Z
f dµ.
X
To provide a counterexample when µ(X) = ∞, let fn (x) = n1 .
Then fn converges uniformly to f (x) = 0, but
Z
1
fn dµ = µ(X) = ∞,
n
X
R
while X f dµ = 0.
(8) Rudin, chapter 1, problem 12.
Solution: Let f ∈ L1R(µ). We want to show that for all > 0,
there is δ > 0 so that E |f | dµ < whenever µ(E) < δ.
We prove the statement by contradiction. If this is not true,
then there
R is an > 0 and measurable sets En with µ(En ) <
−n
2 , but En |f | dµ ≥ .
3
(1)
We will apply Fatou’s Lemma to the sequence of positive
functions {|f |(1 − χEn ) = |f |χEnc }. Fatou’s Lemma states
Z
Z
(lim inf |f |χEnc ) dµ ≤ lim inf
|f |χEnc dµ.
X
n→∞
n→∞
X
First of all, analyze the right hand side. For each n,
Z
Z
Z
Z
|f |χEnc dµ =
|f | dµ −
|f | dµ − .
|f | dµ ≤
X
X
En
X
R
Therefore, the right-hand side of (1) is ≤ X |f | dµ − .
Now analyze the integrand of the left-hand side of (1).
lim inf |f |χEnc = |f | lim inf (1 − χEn ) = |f |(1 − lim sup χEn ).
n→∞
n→∞
n→∞
Recall the definition of lim sup to compute
lim sup χEn (x) = inf sup χEi (x)
k≥1 i≥k
n→∞
= inf χAk (x),
Ak =
k≥1
∞
[
Ei
i=k
= χB (x),
B=
∞
\
k=1
Ak =
∞
∞ [
\
Ei .
k=1 i=k
The Ak and B are measurable, and µ(B) ≤ µ(Ak ) for all k, by
set inclusion. Now the measure of Ak is given by
µ(Ak ) = µ(
∞
[
i=k
Ei ) ≤
∞
X
µ(Ei ) <
i=k
∞
X
2−i = 2−k+1 .
i=k
(This is by assumption on the measure of Ei and Lemma 1
below.) Thus for all k, µ(B) ≤ µ(Ak ) < 2−k+1 . This implies
µ(B) = 0. Putting this together, we have the left-hand side of
(1) satisfies
Z
Z
c
(lim inf |f |χEn )dµ =
|f |(1 − lim sup χEn )dµ
X n→∞
ZX
Zn→∞
=
|f | dµ −
|f |χB dµ
X
X
Z
Z
=
|f | dµ −
|f | dµ
X
B
Z
=
|f | dµ,
X
4
since B has measure 0. Putting this together with the righthand side of (1) then shows
Z
Z
|f | dµ ≤
|f | dµ − ,
X
X
R
which is a contradiction since we assume X |f | dµ < ∞.
Finally, we need the following Lemma:
Lemma 1. Let Ai be measurable sets. Then
∞
∞
[
X
µ( Ai ) ≤
µ(Ai ).
i=1
i=1
Proof. Of course, this is an equality if the Ai are disjoint sets.
Define
B1 = A1 ,
B2 = A2 − A1 ,
B3 = A3 − (A1 ∪ A2 ), . . . .
Then Bi ⊂ Ai implies µ(Bi ) ≤ µ(Ai ), and
∞
∞
[
[
Ai =
Bi .
i=1
i=1
So compute
∞
∞
∞
∞
[
[
X
X
µ( Ai ) = µ( Bi ) =
µ(Bi ) ≤
µ(Ai ).
i=1
i=1
i=1
i=1