Math 516
Professor Lieberman
March 28, 2005
HOMEWORK #8 SOLUTIONS
Chapter 11
30. First, note that, because ν1 and ν2 are finite, so is αν1 + βν2 . To see that it’s a measure,
let hEn i be a sequence of measurable sets. Then
X
X
(αν1 + βν2 )(∪En ) = αν1 (∪En ) + βν2 (∪En ) = α
ν1 (En ) + β
ν2 (En )
X
=
(αν1 (En ) + βν2 (En )) ,
so αν1 + βν2 is countably additive. It’s clear that αν1 + βν2 (∅) = 0.
To check that |αν| = |α||ν|, let A, B be a Hahn decomposition for ν. If α > 0, then this
is a Hahn decomposition for αν, so
|αν(E)| = αν(E ∩ A) − αν(E ∩ B) = α|ν|(E).
If α < 0, then B, A is a Hahn decomposition for αν, so
|αν(E)| = −αν(E ∩ A) + αν(E ∩ B) = −α|ν|(E).
finally, if α = 0, then αν is the signed measure which is zero on every measurable set, so
|αν| = |α||ν| in this case as well.
Now we give a quick solution to problem 29: if ν is a signed measure and E is a measurable
set, then
νE = ν + E − ν − E ≤ ν + E ≤ ν + E + µ− E = |ν|E,
and
νE = ν + E − ν − E ≥ −ν − E ≤ −ν − E − µ+ E = −|ν|E.
Therefore |νE| ≤ |ν|(E).
To see that |ν1 + ν2 | ≤ |ν1 | + |ν2 |,let A and B be a Hahn decomposition for ν1 + ν2 . Then
(ν1 + ν2 )(E ∩ A) = ν1 (E ∩ A) + ν2 (E ∩ A) ≤ |ν1 |(E ∩ A) + |ν2 |(E ∩ A)
(by Problem 29) and similarly
(ν1 + ν2 )(E ∩ B) ≥ −|ν1 |(E ∩ B) − |ν2 |(E ∩ B).
Subtracting these two inequalities gives
|ν1 + ν2 |(E) ≤ |ν1 |(E ∩ A) + |ν2 |(E ∩ A) + |ν1 |(E ∩ B) + |ν2 |(E ∩ B)
= |ν1 |(E) + |ν2 |(E).
32. (a) From Problem 29, we have
1
1
µ ∧ ν = (µ + ν − |µ − ν|) ≤ (µ + ν − (µ − ν)) = ν
2
2
and
1
1
µ ∧ ν = (µ + ν − |µ − ν|) ≤ (µ + ν − (ν − µ)) = µ.
2
2
1
2
If λ is smaller than µ and ν, then let A, B be a Hahn decomposition for µ − ν. Then for
any measurable set E, we have
λ(E) = λ(E ∩ A) + λ(E ∩ B) ≤ µ(E ∩ A) + ν(E ∩ B)
and
µ ∧ ν(E ∩ A) = µ(E ∩ A), µ ∧ ν(E ∩ B) = ν(E ∩ A),
because |µ − ν|(E ∩ A) = (µ − ν)(E ∩ A) and |µ − ν|(E ∩ B) = (ν − µ)(E ∩ B). It follows
that
λ(E) ≤ µ ∧ ν(E ∩ A) + µ ∧ ν(E ∩ B) = µ ∧ ν(E).
(b) We define
1
µ ∨ ν = (µ + ν + |µ − ν|).
2
The argument from part (a) shows that µ ∨ ν is larger than µ and ν but smaller than any
other measure which is larger than µ and ν. The formulae show that µ ∨ ν + µ ∧ ν = µ + ν.
(c) If µ and ν are mutually singular, let A and B be disjoint sets with A ∪ B = X and
µ(A) = ν(B) = 0. Then for any set E, we have
µ ∧ ν(E) = µ ∧ ν(E ∩ A) + µ ∧ ν(E ∩ B) ≤ µ(E ∩ A) + ν(E ∩ B) = 0.
But the zero measure is smaller than µ and ν, so µ ∧ ν(E) ≥ 0. Hence µ ∧ ν = 0.
Conversely, if µ∧ν = 0, let A, B be a Hahn decomposition for µ−ν. Then µ+ν = |µ−ν|,
so (µ + ν)(A) = (µ − ν)(A), which implies that ν(A) = 0 and (µ + ν)(B) = (ν − µ)(B),
which implies that µ(B) = 0. Therefore µ and ν are mutually singular.
33. (a) Using the hint, we write X = ∪Xi with each Xi having finite µ-measure and Xi ∩Xj = ∅
if i 6= j. Then, for each i, there is a nonnegative measurable function fi defined on Xi such
that
Z
ν(Ei ) =
fi dµ
Ei
P
for any subset Ei of Xi . We extend fi to be zero outside of Xi and define f =
fi .
Since only one term in this sum is nonzero for any given x ∈ X, there are no convergence
problems. Since f is a sum of measurable functions, it is measurable and nonnegative. For
any measurable set E, we set Ei = E ∩ Xi . Then
Z
X
XZ
ν(E) =
ν(Ei ) =
fi dµ =
f dµ.
Ei
E
(b) Let ε > 0 and let E be the subset of X on which f − g ≥ ε. Then
Z
0 = νE − νE =
f − g dµ ≥ εµ(E).
E
It follows that the set on which f − g ≥ 1/n has zero measure for any positive integer n.
Hence f ≥ g a.e., and a similar argument shows that f ≤ g a.e. as well.
36. Here’s the version that we prove: Let (X, B, µ) be a σ-finite measure space and let ν be a
signed measure defined on B which is absolutely continuous. Then there is a measurable
function f such that for each E ∈ B, we have
Z
νE =
f dµ.
E
The function f is unique in the same sense as in the Radon-Nikodym theorem.
3
For the proof, let f1 and f2 be the nonnegative measurable functions, given by the
Radon-Nikodym theorem, such that
Z
Z
+
−
ν E=
f1 dµ, ν E =
f2 dµ.
E
E
Then f = f1 − f2 is the desired function. It’s unique because, if A, B is a Hahn decomposition for ν, then f1 = 0 a.e. on B and f2 = 0 a.e. on A.