MAT 601 Assignment

Homework 3

ROLAND APERE AMBE

SC24P163

January 16, 2025

Problem 1
Let (X, A, µ) be a measure space, and letR h : X → R be a non-negative measurable
function. Define ν on (X, A) by ν(A) = A h dµ for A ∈ A. We need to show that for
every non-negative measurable function g : X → R, we have:
Z Z
g dν = gh dµ.
X X

Proof:
Pn
1. For Simple Functions: Let g = i=1 ci χAi be a simple function, where Ai ∈ A
and ci ≥ 0. Then:

Z n n Z Z n
! Z
X X X
g dν = ci ν(Ai ) = ci h dµ = ci χ A i h dµ = gh dµ.
X i=1 i=1 Ai X i=1 X

2. For Non-negative Measurable Functions: Let g be a non-negative measurable

function. There exists a sequence of simple functions {gn } such that gn ↑ g. By the
Monotone Convergence Theorem (MCT):
Z Z Z Z
g dν = lim gn dν = lim gn h dµ = gh dµ.
X n→∞ X n→∞ X X

3. For g ∈ L1 (X, A, ν, R): If g is integrable with respect to ν, then g + and g − are

integrable. Applying the above result to g + and g − and subtracting gives:
Z Z Z Z Z Z
+ − + −
g dν = g dν − g dν = g h dµ − g h dµ = gh dµ.
X X X X X X

Thus, the result holds for all non-negative measurable functions and for functions in
L1 (X, A, ν, R).

Problem 2
We need to prove that a measurable function f is Lebesgue integrable on (0, 1) if and
only if:
∞
X
2n λ({x ∈ [0, 1] : |f (x)| ≥ 2n }) < ∞.
n=1

Proof:
1. Necessity: Assume f is Lebesgue integrable on (0, 1). Let An = {x ∈ [0, 1] :
|f (x)| ≥ 2n }. Then:
Z 1 Z
|f | dλ ≥ |f | dλ ≥ 2n λ(An ).
0 An

1
 Summing over n:

∞
X Z 1
n
2 λ(An ) ≤ |f | dλ < ∞.
n=1 0

P∞
2. Sufficiency: Assume n=1 2n λ(An ) < ∞. Let Bn = {x ∈ [0, 1] : 2n−1 ≤ |f (x)| <
2n }. Then:
Z 1 ∞ Z
X ∞
X ∞
X
|f | dλ = |f | dλ ≤ 2n λ(Bn ) ≤ 2n λ(An−1 ) < ∞.
0 n=1 Bn n=1 n=1

Thus, f is Lebesgue integrable on (0, 1) if and only if the given series converges.

Problem 3(a)
P∞
Assume 0 ≤ xn ≤ 1, n = 1, 2, . . ., and an ≥ 0 for all n. Suppose further that n=1 an <
∞. We need to prove that:
∞
X a
p n
n=1 |x − xn |
converges almost everywhere on R.

Proof:
P∞
Let f (x) = n=1
√ an . We need to show that f (x) < ∞ almost everywhere.
|x−xn |

1. Integral Test: Consider the integral:

Z 1 ∞ Z 1
X 1
f (x) dx = an p dx.
0 n=1 0 |x − xn |
R1 R1
Since 0 ≤ xn ≤ 1 =⇒ x − 1 ≤ x − xn ≤ x =⇒ 0
√ 1
dx ≤ 0
√1 dx
|x−xn | |x−1|

Simplify the Integrand

The integrand √ 1 can be simplified because x ∈ [0, 1], and |x − 1| = 1 − x.
|x−1|
Thus, the integral becomes:
Z 1
1
√ dx.
0 1−x
—

2
 Substitution
Let u = 1 − x. Then:

• du = −dx,
• When x = 0, u = 1,
• When x = 1, u = 0.

The integral becomes:

Z u=0 Z 1
1 1
√ (−du) = √ du.
u=1 u 0 u
—

Evaluate the Integral

R1
The integral √1 du is a standard integral:
0 u

Z 1 Z 1
1
√ du = u−1/2 du.
0 u 0

Using the power rule for integration:

un+1
Z
un du = +C for n ̸= −1,
n+1
we get:
Z 1 1
u−1/2 du = 2u1/2 = 2(1) − 2(0) = 2.
0 0

Final Answer
Z 1
1
p dx = 2.
0 |x − 1|
R1
=⇒ 0
√ 1
dx ≤ 2,
|x−xn |
we have:
Z 1 ∞
X
f (x) dx ≤ 2 an < ∞.
0 n=1
R1
2. Conclusion: Since 0 f (x) dx < ∞, f (x) must be finite almost everywhere on
[0, 1]. By periodicity, the result extends to R.

3
Problem 3(b)
We need to compute the limit:
Z 1 √ 2 x2
lim n xe−n dx.
n→∞ 0

Solution:
Gamma Function Definition
The Gamma function is defined for z > 0 by the improper integral:
Z ∞
Γ(z) = tz−1 e−t dt.
0
It generalizes the factorial function, satisfying the property:

Γ(n) = (n − 1)! for positive integers n.

—

Define the Integral

R 1 √ −n2 x2
Let In = n 0
xe dx. We aim to evaluate the limit of In as n → ∞.
—

Substitution
Make the substitution u = nx, so du = n dx. The limits of integration change as follows:

• When x = 0, u = 0.

• When x = 1, u = n.

Thus, the integral becomes:

Z nr Z n
u −u2 1 √ −u2
In = e du = √ ue du.
0 n n 0
—

Evaluate the Integral

Rn√ 2
We now evaluate the integral 0
ue−u du. As n → ∞, this integral converges to:
Z ∞
√ −u2
ue du.
0
√
To evaluate this integral, make the substitution v = u2 , so dv = 2u du and u = v.
The integral becomes:
Z ∞ Z ∞
√ −u2 1 ∞ −1/4 −v
Z
1/4 −v dv
ue du = v e · 1/2 = v e dv.
0 0 2v 2 0

4
 This is a Gamma function integral. Recall that:
Z ∞
Γ(z) = tz−1 e−t dt.
0
Thus, the integral evaluates to:

1 ∞ −1/4 −v
Z  
1 3
v e dv = Γ .
2 0 2 4
—

Compute the Limit

Substitute the result back into In :
 
1 1 3
In = √ · Γ .
n 2 4
As n → ∞, √1 → 0, so:
n
 
1 1 3
lim In = lim √ · Γ = 0.
n→∞ n→∞ n 2 4
—

Final Answer
The limit of In as n → ∞ is:

lim In = 0.
n→∞

Problem 3(c)
Assume
R fn ,Rgn , f, g are integrable on R, fn → f a.e., gn → g a.e., |fn | ≤ gn a.e., and
g dµ → R g dµ. We need to show that:
R n
Z Z
fn dµ → f dµ.
R R

Proof:
The Dominated Convergence Theorem (DCT) is a fundamental result in measure theory
and integration. It states that if a sequence of measurable functions {fn } converges
pointwise almost everywhere to a function f , and if there exists an integrable function g
such that |fn | ≤ g for all n, then:
Z Z
lim fn dµ = f dµ.
n→∞ R R

In this case: - The sequence {fn } is dominated by {gn }, meaning |fn | ≤ gn for all n.
- The sequence {gn } converges to g in L1 , i.e., ∥gn − g∥L1 → 0 as n → ∞. - Since gn → g

5
in L1 , g is integrable, and {gn } is uniformly integrable.

By the DCT, the pointwise convergence of {fn } to f , combined with the dominance
condition |fn | ≤ gn , ensures that:
Z Z
lim fn dµ = f dµ.
n→∞ R R
Conclusion:
The result follows directly from the Dominated Convergence Theorem. Specifically:
- The dominance condition |fn | ≤ gn ensures that the sequence {fn } is controlled by the
integrable sequence {gn }.
- The convergence of {gn } to g in L1 guarantees that the limit function f is integrable.
- The DCT allows us to interchange the limit and the integral, yielding the desired result:

Z Z
lim fn dµ = f dµ.
n→∞ R R
This completes the proof.