Problem 1
R
R
Suppose
that f = limn→∞ fn < ∞. Then for large enough n, we have that
R
fn < ∞. So we can apply the result of Problem 3 here: fn χE is bounded
above by fn , and the sequence fn eventually
satisfies
the conditions of Problem
R
R
3. Therefore, we have that limn→∞ fn χE = f χE , as desired.
If this condition fails to hold, we can run into some serious trouble. The idea
here is that there is now enough room for things to “escape”. For example, if
we define
1
fn (x) := χ[1/n,1] (x) + nχ[1 − 1/n, 1](x)
x
this function clearly approaches x1 pointwise as n → ∞, and the integrals of the
fn clearly go to ∞, but the integral of fn (x) on [ǫ, 1] will always be larger than
the integral of x1 by 1 for any ǫ > 0 and any n. The problem is that we can have
an arbitrarily thin and tall rectangular box of fixed area added to our function
without violating the condition in the problem.
Problem 2
R
f dµ for each measurable set E. Clearly λ(∅) = ∅ f = 0. We
S
P
also need to check that λ
j λ(Aj ) whenever the Aj ’s are
j Aj agrees with
S
disjoint. Let A = j Aj . I will first treat the case of finite unions. Using χAj to
RP
P R
denote the indicator of Aj , we have that λ(A) =
j Aj f dµ =
j f χAj dµ =
P
j λ(Aj ) by linearity of the integral. If the union is infinite, we have that
R
Pk
λ(A) = limk→∞ j=1 f χAj dµ, so by the monotone convergence theorem, this
Pk
P∞
is equal to limk→∞ j=1 λ(Aj ) = j=1 λ(Aj ) as desired.
R
P
P
Let g = j aj χAj be a simple function. Then gdλ is equal to j aj λ(Aj ) =
R
R
P R
P
aj χAj f dµ = f gdµ. Now, apply the monotone converj
j aj Aj f dµ =
gence theorem: if gj is a sequence of simple measurable functions converging to
g, then by the nonnegativity
R sequence
R of f we Rhave that f gj is an increasing
f gj dµ =
of functions.
Therefore,
f
g
dµ
=
lim
f
g
dµ
=
lim
j→∞
j
j→∞
R
R
limj→∞ gj dλ = gdλ, using the Lebesgue monotone convergence theorem
twice.
Define λ(E) =
R
E
Problem 3
My friends and I always joke about this problem: you just take the proof of
Lebesgue’s dominated convergence theorem and put n’s on all of the g’s.
We have that f ∈ L1 by assumption. So we apply Fatou’s lemma:
Z
Z
Z
Z
Z
g + f ≤ lim inf gn + fn = g + lim inf fn
Z
Z
Z
Z
Z
g − f ≤ lim inf gn − fn = g − lim sup fn
1
R
R
R
So
f is trapped between Rlim sup fn and lim inf fn . Therefore the limit of
R
fn exists and is equal to f .
Problem 4
R
R
R
Suppose first that |fn − f | → 0. We would like to show that |fn | → |f |.
Note that the functions |fn | are bounded above by |fn − f | + |f | ∈ L1 by the
triangle inequality. But the integral of |fn − f | + |f | converges to the integral of
|f | by assumption, so the generalized dominated
theorem applies
R convergence
R
to the functions |fRn |, and weRget that limn→∞ |fn | = |f |.
Conversely, if |fn | → R |f |, then we use the triangle
inequality to get
R
|fn − f | ≤ |fn | + |f |. But |fn | + |f | converges to 2|f | by assumption, so
we
R can use
R the generalized dominated convergence theorem to conclude that
|fn | → |f |.
Problem 5
Part a
Define fn (x) := ae−nax − be−nbx . Notice that this function is negative for small
x and positive for large values of x. Let’s find the place where fn (x) = 0:
ae−nax = be−nbx
log a − nax = log b − nbx
log a − log b = (na − nb)x
log b − loga
.
x=
nb − na
Therefore, we have that
Z
0
∞
|fn (x)| dx ≤
Z
log b−log a
nb−na
be−nbx − ae−nax .
0
We compute the right hand antiderivative to be − n1 e−nbx + n1 e−nax . This is
1
−nax
− e−nbx . Now, we plug in the value for x to get that the integral is
n e
a
b
1
(a/b) b−a − (a/b) b−a .
n
In particular, this is a constant times
1
n
so the sum diverges.
Part b
This is easier: Using The antiderivative from part a is equal to 0 at both 0 and
∞, so each summand in this sum is zero.
2
Part c
We use the geometric series formula to compute that the sum of ae−nax is
ae−ax
1−e−ax for any x > 0, with a similar formula for the other term. So the function
ae
be
in question is 1−e
−ax − 1−e−bx .
1−bx
). At
This function has antiderivative log(1 − bx) − log(1 − ax) = log( 1−ax
0, this is zero, and at ∞ this is log(b/a).
ae−ax
be−bx
1
The only thing that remains to be shown is that 1−e
−ax − 1−e−bx is in L .
−ax
Again, we need to find the zero of this function: This occurs when ae
(1 −
e−bx ) = be−bx (1 − e−ax ). Expanding, this gives ae−ax − ae−(a+b)x = be−bx −
be−(a+b)x . So we get be−ax − ae−bx = (b − a)e−(a+b) x. Multiply both sides
by e(a+b)x to get bebx − aeax = b − a. The left hand side is increasing and is
ae−ax
be−bx
therefore only equal to b−a when a = 0. Therefore, the function 1−e
−ax − 1−e−bx
is nonnegative on (0, ∞). Because it is nonnegative and Riemann integrable, it
follows that it is in L1 .
−ax
−bx
3