Real Analysis HW 6 Solutions
Problem 33: Let {fn } be a sequence of integrable functions
R on E for which fn converges
to f a.e.R on E and
f
is
integrable
over
E.
Show
that
|f − fn | → 0 if and only if
E
R
limn→∞ E |fn | = E |f |.
R
Solution: If E |f − fn | → 0 then by the reverse triangle inequality
Z
Z
Z
0 ≤ |f | −
|fn | ≤
|f − fn |.
E
E
E
R
R
Therefore limn→∞ E |fn | = E |f |.
R
R
If limn→∞ E |fn | = E |f |, since |fn −f |+|f |−|fn | → 0 a.e., and 0 ≤ |fn −f |+|f |−|fn | ≤ 2|f |,
we have by Lebesgue Dominated Convergence
Z
Z
Z
Z
|fn − f |
|fn | = lim
|f | −
0 = lim
|fn − f | +
n→∞
n→∞
E
E
E
E
Problem 34: Let f be a nonnegative measurable function on R. Show that
Z n
Z
lim
f=
f.
n→∞
−n
R
Solution: Let fn = f χ[−n,n] , then fn is nonnegative, monotone and fn → f pointwise. By
monotone convergence
Z
Z n
Z
f.
lim
f = lim
fn =
n→∞
−n
n→∞
R
R
Problem 35: Let f be a real-valued function of two variables (x, y) that is defined on the
square Q = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} and is a measurable function of x for each fixed
value of y. Suppose for each fixed value of x, limy→0 f (x, y) = f (x) and that for all y, we
have |f (x, y)| ≤ g(x), where g is integrable over [0, 1]. Show that
Z 1
Z 1
lim
f (x, y)dx =
f (x)dx.
y→0
0
0
1
Also show that if the function f (x, y) is continuous in y for each x, then
Z 1
f (x, y)dx
h(y) =
0
is a continuous function of y.
Solution: It suiffces to consider a sequence {yn } ⊂ [0, 1] such that yn → 0. Define fn (x) =
f (x, yn ). Then since fn → f pointwise and |fn | ≤ g, Lebesgue dominated convergence gives
Z 1
Z 1
Z 1
f (x, yn )dx = lim
fn (x)dx =
f (x)dx.
lim
n→∞
n→∞
0
0
0
Since this is true for every such sequence {yn }, we have
Z 1
Z 1
lim
f (x, y)dx =
f (x)dx.
y→0
0
0
Now fix y ∈ [0, 1] and suppose that {yn } ⊆ [0, 1] such that yn → y. Define fn (x) = f (x, yn ).
Since f (x, y) is continuous, fn (x) → f (x, y) for all x ∈ [0, 1]. Once again, since |fn (x)| ≤ g(x)
dominated convergence gives
Z 1
Z 1
fn (x)dx =
f (x, y)dx = h(y).
lim h(yn ) = lim
n→∞
n→∞
0
0
Therefore h is continuous.
Problem 36: Let f be a real-valued function of two variables (x, y) that is defined on the
square Q = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} and is a measurable function on x for each fixed
value of y. For each (x, y) ∈ Q let the partial derivative ∂f /∂y exist. Suppose there is a
function g that is integrable over [0, 1] and such that
∂f
(x, y) ≤ g(y) for all (x, y) ∈ Q.
∂y
Prove that
Z 1
Z 1
∂f
d
f (x, y)dx =
(x, y)dx for all y ∈ [0, 1].
dy 0
0 ∂y
Solution: Fix y0 ∈ [0, 1) and define for n ≥ 1,
Dn f (x) = n(f (x, y0 + 1/n) − f (x, y0 )).
Since the derivative exists, Dn f (x) → ∂f
(x, y0 ) pointwise. Also by the mean value theorem,
∂y
there is a ζ ∈ [y0 , y0 + 1/n] such that |Dn f (x)| = | ∂f
(x, ζ)| ≤ g(x). We conclude by the
∂y
Lebesgue Dominated Convergence Theorem that
Z 1
Z 1
Z 1
Z 1
∂f
d
(x, y0 )dx = lim
Dn f (x)dx = lim Dn
f (x, y0 )dx =
f (x, y0 )dx .
n→∞ 0
n→∞
dy 0
0 ∂y
0
Since this is true for all y0 ∈ [0, 1) we have the result.
R n
Problem 38: For each of the two functions f on [1, ∞) defined below, show that limn→∞ 1 f
exists while f is not integrable over [1, ∞). Does this contradict the continuity of integration?
2
(i) Define f (x) = (−1)n /n for n ≤ x ≤ n + 1.
(ii) Define f (x) = (sin(x))/x for 1 ≤ x < ∞.
Solution: Neither of these cases contradict the continuity of integration since continuity
assumes that f integrable over a set E containing each of the sets in the collection {Ek },
which we do not have in either of these cases.
(i) We can easily see that
n
Z
lim
n→∞
f=
1
∞
X
(−1)k
k=1
converges. However
n
Z
|f | =
lim
n→∞
1
k
∞
X
1
k=1
k
diverges. So f is not integrable.
(ii) Note that since sin (x)/x is continuous and bounded so we may use the Riemann
integral. It follows from the fundamental theorem of calculus for Riemann integrals
Z n
Z n
sin x
d cos x cos x
dx = −
+ 2
dx
x
dx x
x
1
1
Z n
cos x
cos n
−
dx
= cos 1 −
n
x2
1
It follows that if n ≥ m > 0,
Z n
Z n
sin x cos m cos n | cos x|
dx ≤ −
dx
+
x
m
n
x2
m
m
Z n
1
2
≤ +
dx
2
n
m x
4
≤ →0
n
R n sin x
as m, n goes to infinity. Therefore 1 x dx is Cauchy and converges. In particular we
have the well known result
Z ∞
sin x
π
dx = ,
x
2
0
which can be gotten by complex integration or the Laplace transform.
sin (x)/x is not Lebesgue integrable since
Z
(k+1)π
kπ
1
| sin x|
dx ≥
x
(k + 1)π
3
Z
(k+1)π
| sin x|dx =
kπ
2
(k + 1)π
However
and therefore
Z
lim
n→∞
1
n
∞
| sin x|
2X1
dx ≥
=∞
x
π k=2 k
diverges.
Problem 39: Prove the theorem regarding continuity of integration.
Solution: Let {En }∞
n=1 be an ascending collection of subsets of E. Define fn = f χEn , then
fn converges pointwise to f χS∞
. Since |fn | ≤ |f |, it follows by Lebesgue dominated
k=1 Ek
convergence theorem that
Z
Z
Z
f.
lim
f = lim
fn = S
n→∞
En
n→∞
E
∞
k=1
Ek
∞
In the case that {En }∞
n=1 are descending, choose {fn }n=1 as above, and note that fn converges
T∞
pointwise to f χ k=1 Ek . The result again follows from dominated convergence.
.
4