Math 318 HW #9 Solutions
1. Let A be a bounded, measurable set and let (fn ) be a sequence of measurable functions on A
converging to f . Suppose ϕ ∈ L(A) and that |fn (x)| ≤ ϕ(x) for all x ∈ A and all n = 1, 2, . . ..
Show that
Z
Z
f g dm
fn g dm =
lim
n→∞ A
A
if g is measurable and essentially bounded on A, meaning that there exists M > 0 such that
|g(x)| ≤ M a.e.
Proof. By assumption, there exists M > 0 such that {x ∈ A : |g(x)| > M } has measure zero.
Therefore,
|fn (x)g(x)| = |fn (x)| |g(x)| ≤ M ϕ(x) a.e.
Since ϕ ∈ L(A), Theorem 25.4(3) implies that M ϕ ∈ L(A). Also, since fn is measurable for
all n = 1, 2, . . . and g is measurable, Proposition 18.4 implies that fn g is measurable for all
n = 1, 2, . . .. Finally, limn→∞ fn g = f g by the Algebraic Limit Theorem.
Thus, the hypotheses of Corollary 28.10 are satisfied by (fn g), f g, and M ϕ, and so we see
that
Z
Z
lim
fn g dm =
f g dm,
n→∞ A
A
as desired.
2. Exercise 31.16. Give an example of an unbounded function (hence non-Riemann integrable)
which is continuous a.e. on [0, 1].
Answer: Define the function f : [0, 1] → R by
(
1/x
f (x) :=
0
if x 6= 0
if x = 0.
Then f (x) = 1/x on (0, 1], and so f is continuous on (0, 1]. The function f is certainly not
continuous at x = 0, but this is the only point of discontinuity. Hence, the set of discontinuity
is {0}, which clearly has measure zero, so f is continuous a.e.
3. Exercise 31.33. Let (hn ) be an increasing
sequence
of functions in L(A), with hn ≤ h for all
R
R
n and some h ∈ L(A). If limn→∞ A hn dm = A h dm, prove that h = limn→∞ hn a.e. on A.
Proof. Since (hn (x)) is an increasing sequence that is bounded above by h(x) for each x ∈ A,
we know that limn→∞ hn (x) exists for each x, so we can define
g(x) := lim hn (x).
n→∞
Also, define f (x) := |h(x)| + |h1 (x)|. Then, since
h1 (x) ≤ hn (x) ≤ h(x)
1
for each x ∈ A and each n = 1, 2, . . ., it follows that
|hn (x)| ≤ f (x) for all n = 1, 2, . . . .
Moreover, since h1 , h ∈ L(A), Theorem 25.4(4)&(5) imply that f ∈ L(A). Therefore,
Z
Z
g dm
lim
hn dm =
n→∞ A
A
by the Lebesgue Dominated Convergence Theorem.
R
R
Combining this with the assumption that limn→∞ A hn dm = A h dm, we see that
Z
Z
g dm.
h dm =
A
A
Using Theorem 25.4(4) again, we have that
Z
Z
Z
0=
h dm −
g dm = (h − g)dm.
A
A
A
Since hn ≤ h for all n, the Order Limit Theorem implies that g ≤ h, so the function h − g is
non-negative. Therefore, h − g = |h − g| and Problem 1(b) from HW 8 implies that h − g = 0
a.e. Since g(x) = limn→∞ hn (x) for all x ∈ A, we conclude that
h = lim hn a.e.,
n→∞
as desired.
2