Math 515
Professor Lieberman
October 4, 2004
HOMEWORK #6 SOLUTIONS
Chapter 4
10. (b) Recall that there are multiple types of improper Riemann integrals. This solution is
specifically written for the case
Z b
Z ∞
f (x) dx,
f (x) dx = lim
b→∞ 0
0
but the same argument works for all cases. So suppose f is Lebesgue integrable on [0, ∞)
and that the improper Riemann integral of f exists. Set
(
f (x) if x ≤ n,
fn (x) =
0
if x > n.
Then hfn i is a sequence of measurable functions that converges a.e. to f and |fn | ≤ |f | a.e.
Since |f | is integrable, it follows from the Lebesgue Convergence Theorem that
Z
Z
fn → f.
But Proposition 40 implies that
Z
n
Z
fn =
f (x) dx,
0
so
Z
Z
∞
f=
f (x) dx.
0
14. (a) Let hn = gn + g and Fn = |fn − f |. Then |Fn | ≤ hn , Fn → 0 Ra.e., and
R hn → 2g, so
Theorem 17 (which is Theorem 49 in class numbering) implies that Fn to 0 = 0.
(b) =⇒: First, we note that, for any functions fn and f , we have
Z
Z
Z
Z
|fn | − |f | = (|fn | − |f |) ≤ ||fn | − |f ||
Z
≤ |fn − f |.
R
R
R
Hence, if |fn − f | → 0, then |fn | → |f |.
⇐=: Use part (a) with gn = |fn |.
1
2
Chapter 5
4. We use the hint. Let ε > 0 and suppose that g is a continuous function such that D+ g ≥ ε.
Then, for every x ∈ [a, b), there is a number η(x) > 0 such that
g(x + h) − g(x)
ε
≥ε−
h
2
for 0 < h < η(x) by Problem 2.49(b) and hence g(x + h) ≥ g(x) + εh/2 for 0 < h < η(x).
Since g is continuous, it follows that g(x + h) > g(x) for 0 < h ≤ η(x). It follows that the
maximum of g on any interval [x, y] ⊂ [a, b] cannot occur at a point in [x, y). Hence, the
maximum occurs at y so g(y) > g(x) if y > x.
If we only know that D+ f ≥ 0, then let ε > 0 and set g(x) = f (x) + εx. Then D+ g ≥ ε,
so x ≤ y implies that
f (x) < f (y) + ε[y − x].
Since this inequality is true for every ε > 0, it follows that f (x) ≤ f (y).
Chapter 11
22. (a) First, ν is nonnegative and defined on B. In addition,
Z
Z
Z
ν(∅) = g dµ = χ∅ g dµ = 0 dm = 0.
∅
Finally, if hEi i is a sequence of disjoint measurable sets, then we write E = ∪Ei . It follows
that
Z
Z
Z X
∞
ν(E) =
g dµ = χE g dµ =
χEi g dµ.
E
i=1
From Corollary 46, we know that
Z X
∞
∞ Z
∞
X
X
χEi g dµ =
χEi g dµ =
ν(Ei ),
i=1
i=1
i=1
so ν is countably additive and hence a measure.
(b) Suppose f is simple. Then there are finitely many nonnegative numbers a1 , . . . , aN and
finitely many disjoint measurable open sets E1 , . . . , EN such that
N
X
f=
an χEn .
n=1
Then
Z
f dν =
X
an ν(En ) =
X
Z
an
g dµ =
X
Z
an
Z
χEn g dµ =
f g dµ.
En
For a general f , let hϕn i be an increasing sequence of simple functions which converges
to f . Then
Z
Z
fn dν = fn g dµ,
so
Z
Z
f dν = lim
Z
fn dν = lim
by the Monotone Convergence Theorem.
Z
fn g dµ =
f g dµ,