Math 516
Professor Lieberman
April 10, 2009
HOMEWORK #6 SOLUTIONS
Chapter 11
5. (a) We just have to check countable S
additivity. If (En ) is a disjoint sequence of
measurable sets, then we set E = En . It follows that
λE = µE + νE
X
X
X
=
µEn +
νEn =
(µEn + νEn )
X
=
λEn .
(b) First, if νE < ∞, we set λE = µE − νE. If νE = ∞, then we need to be more
careful. Specifically, we set
λE = sup{µ(E 0 ) − ν(E 0 ) : E 0 ⊂ E and νE 0 < ∞}.
Again, we just have to check that λ is countably additive,
so let (En ) be a
S
sequence
of
disjoint
measurable
sets
and
set
E
=
E
.
Suppose
first that
n
P
λEn = ∞ for all n. Then for any M > 0, there is a positive integer N (M )
such that
N (M )
X
λEn > M,
n=1
and hence, for each n ≤ N (M ), there is a subset En0 of En with νEn < ∞ and
N (M )
X
SN (M )
µ(En0 ) − ν(En0 ) > M.
n=1
Setting E 0 = n=1 En0 , we see that ν(E 0 ) < ∞ and µ(E 0 ) − ν(E 0 ) > M , so
λE = ∞.
P
Next, suppose that
λEn = A < ∞. Then, for any ε > 0, there is a positive
integer N (ε) such that
N (ε)
X
ε
λEn > A − .
2
n=1
Then, for each n ≤ N (ε), there is a set En0 ⊂ En such that νEn0 < ∞ and
ε
λEn ≥ λEn0 +
.
2N (ε)
It follows that
N (ε)
X
λE ≥
λEn0 ≥ A − ε.
n=1
1
2
P
Hence λE ≥
λEn .
0
0
On thePother hand, if E 0 ⊂ E and νE 0 < ∞, we set E
n = E ∩ En . Then
P
νE 0 =
νEn0 so this sum converges. In addition, µE 0 =
µEn0 , so we have
X
λE 0 =
λEn0 .
P
0
0
From the definition
λEn ,
P of λ, we know that λEn ≤ λEn , so we have λE ≤
and hence λE ≤
λEn . Combined with our previous inequality, we conclude
that
X
λE =
λEn ,
and hence λ is a measure.
(c) If ν is σ-finite, let (Xn ) be a sequence of disjoint sets with X = ∪Xn . If λ is
a measure such that µ = ν + λ, then we must have λ(E ∩ Xn ) = µ(E ∩ Xn ) −
ν(E ∩ Xn ) for any measurable set E. Then countable additivity implies that
λE =
∞
X
µ(E ∩ Xn ) − ν(E ∩ Xn ).
n=1
(d) To show that λ is not unique, take B to be the σ-algebra {∅, X} and
(
0 if E = ∅,
µE = νE =
∞ if E = X.
We then note that any λ with λ∅ = 0 and λX ≥ 0 satisfies µ = λ + ν.
14. Suppose first that f is measurable with respect to B0 . For each α ∈ R, we set
Aα = {x ∈ X : f (x) ≥ α}. From property (iii) from Proposition 11.4, there are sets
Bα , Cα , and Dα with Bα and Cα in B, Dα ⊂ Cα , µCα = 0 and Aα = Bα ∪ Dα . Since
Q is a dense subset of R, Lemma 11.9 gives a function g, which is measurable with
respect to B such that g ≥ α a.e. in Bα and g ≤ α a.e. in X \ Bα for all α ∈ Q. If
we take
[
E=
Cα ,
α∈Q
it follows that f = g off of E.
Conversely, suppose there are g which is measurable with respect to B and E ∈ B
such that µE = 0 and f = g off of E. Then g is measurable with respect to B0
(because it’s a larger σ-algebra) and f = g a.e. [µ0 ]. Proposition 11.8 implies that f
is measurable with respect to B0 .
21. (a) For each positive integer n, Set An = {x : |f (x)| ≥ n1 }. Then
Z
1
|f | ≥ µAn ,
n
so An has finite measure. On the other hand, {x : f (x) 6= 0} is the union of the
An ’s, so it is of σ-finite measure.
3
(b) From part (a), we have that E = {x : f (x) 6= 0} is σ-finite, so Proposition
11.7 gives a sequence (ϕn ) of simple functions which increases to f such that ϕn
vanishes outside a set of finite measure.
(c) From the definition of the integral, there are simple functions ϕ1 and ϕ2 such
that 0 ≤ ϕ1 ≤ f + , 0 ≤ ϕ2 ≤ f − and
Z
Z
ε
ε
+
(f − ϕ1 ) dµ < ,
(f − − ϕ2 ) dµ < .
2
2
We then take ϕ = ϕ1 − ϕ2 and observe that ϕ1 = 0 wherever f ≤ 0 and ϕ2 = 0
wherever f ≥ 0. Hence
|f − ϕ| = (f + − ϕ1 ) + (f − − ϕ2 ),
It follows that
Z
|f − ϕ| dµ < ε.
24. As usual, we only have to check countable
additivity. We write (En ) for a sequence
S
of disjoint measurable sets and E = En . Then, for any k, we have
X
X
µk E =
µk En ≤
µEn .
Sending k → ∞, we see that
µE ≤
X
µEn .
Now let N be a positive integer. Then, for any k, we have
µk E =
∞
X
µk En ≥
n=1
N
X
µk En .
n=1
Since the sum on the right is finite, we have
N
X
µEn =
n=1
N
X
lim µk En = lim
n=1
N
X
µk E n .
n=1
Hence
µE ≥
N
X
µEn
n=1
for any positive integer N , so
µE ≥
∞
X
µEn .
n=1
(Note that this inequality is true without assuming that (µn ) is increasing.) Combining the two inequalities shows that µ is countably additive.
4
25. Write m for Lebesgue measure on the real line and B for the σ-algebra of all Lebesgue
measurable subsets of R. We then define µn on B by µn E = µ(E ∩ (n, ∞)). Then
µn converges setwise to the function µ defined by
(
∞ if m(E ∩ [0, ∞)) = ∞,
µE =
0
otherwise.
It’s easy to see that (µn ) is decreasing. On the other hand, we can write [1, ∞) =
∪[n, n + 1). Since µ[1, ∞) = ∞ but µ[n, n + 1) = 0, it follows that µ is not countably
additive and hence it isn’t a measure.
32. (a) We use the formula in the hint and set µ ∧ ν = 21 (µ + ν − |µ − ν|). We now
derive an alternative formula for µ ∧ ν. Let A be A is a positive set for µ − ν
such that X \ A is a negative set for µ − ν. (In other words, A and X \ A form
a Hahn Decomposition for µ − ν.) Then, for any measurable set E, we have
|µ − ν|(E) = (µ − ν)(E ∩ A) − (µ − ν)(E \ A),
so
1
(µ ∧ ν)E = (µ(E ∩ A) + µ(E \ A) + ν(E ∩ A) + ν(E \ A)
2
+ (ν − µ)(E ∩ A) + (µ − ν)(E \ A))
= ν(E ∩ A) + µ(E \ A).
Since A is a positive set for µ − ν, it follows that ν(E ∩ A) ≤ µ(E ∩ A), so
(µ ∧ ν)E ≤ µ(E ∩ A) + µ(E \ A) = µE,
and a similar argument shows that (µ ∧ ν)E ≤ νE.
If λ ≤ µ and λ ≤ ν, we have that, for any measurable set E,
λE = λ(E ∩ A) + λ(E \ A) ≤ ν(E ∩ A) + µ(E \ A),
and hence λ ≤ µ ∧ ν.
(b) We define µ ∨ ν = 12 (µ + ν + |µ − ν|). The proof that µ ∨ ν is the smallest
measure larger than µ and ν is similar to part (a), and the formulas show that
µ ∨ ν + µ ∧ ν = µ + ν.
(c) If µ and ν are mutually singular, let A and B be disjoint measurable sets with
A ∪ B = X and µ(A) = ν(B) = 0. Fix a set E and note that µE = µ(E ∩ B)
and νE = ν(E ∩ A). Hence
1
(µ ∧ ν)(E ∩ A) = (µ(E ∩ A) + ν(E ∩ A) − |µ − ν|(E ∩ A)).
2
But E ∩ A is a null set for µ and hence it’s a positive set for ν − µ, which means
that |µ − ν|(E ∩ A) = (ν − µ)(E ∩ A) = ν(E ∩ A). Hence
(µ ∧ ν)(E ∩ A) = 0.
A similar argument shows that (µ ∧ ν)(E ∩ B) = 0, so (µ ∧ ν)E = 0.
5
Conversely, if µ ∧ ν = 0, then µ + ν = |µ − ν|. Taking A as in part (a), we see
that µ(A) + ν(A) = µ(A) − ν(A), so ν(A) = 0. In addition, if we set B = X \ A,
then B is a negative set, so µ(B) + ν(B) = ν(B) − µ(B) and hence µ(B) = 0.
Hence µ and ν are mutually singular.