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Assignment 3

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Assignment 3

Question

Let be a measure space. Suppose is a sequence of measurable functions such that pointwise. Using Fatou’s Lemma, prove the Monotone Convergence Theorem (MCT), which states:

Definitions and Theorems Used

1. Fatou’s Lemma

For any sequence of non-negative measurable functions:

Intuition: This tells us that the integral of the limit (specifically the limit inferior) is “no larger” than the limit of the integrals. It allows for “loss of mass” in the limit, but not a sudden gain.

2. Monotonicity of the Integral

If and are measurable functions such that for all , then:

Intuition: If one function is always smaller than another, the total area (or volume) under its curve must also be smaller or equal.

Answer

To prove equality, we will show that and separately.

Step 1: Establish the existence of the limit

Since the sequence is non-negative and monotonically increasing (), the sequence of integrals is also a non-decreasing sequence of real numbers (possibly infinite) in . Therefore, the limit exists in the extended real number system.

Step 2: Use Monotonicity to find the upper bound

Since for all and all , by the Monotonicity of the Integral:

Taking the limit of both sides as , we obtain:

Step 3: Apply Fatou’s Lemma to find the lower bound

Because pointwise, we know that . Since all are non-negative and measurable, we can apply Fatou’s Lemma:

Since we already established in Step 1 that the limit of the integrals exists, the limit inferior is simply the limit:

Step 4: Conclusion

We have shown two inequalities:

  1. (from Monotonicity)

  2. (from Fatou’s Lemma)

By combining these two results, we conclude:

This completes the deduction of the Monotone Convergence Theorem.

Question

Suppose is a measure space and is a sequence of nonnegative measurable functions. Show that the following inequality is not true in general:

Hint: Try to construct a sequence of functions such that for all , but .

Definitions and Theorems Used

1. Lebesgue Measure on

The standard measure on the real line that assigns to each interval its length .

Intuition: This is our standard way of measuring “length” or “volume” in Euclidean space.

2. Characteristic (Indicator) Function

The function (or ) is defined as:

Intuition: It acts as a switch that is “on” inside the set and “off” everywhere else.

3. Integral of a Characteristic Function

For a measurable set , the integral of its indicator function is simply the measure of the set:

Intuition: The “area” under a flat block of height 1 is simply the length of its base.

Answer

To show the inequality is not true, we only need to find one counter-example where the left side is strictly greater than the right side.

Step 1: Define the Measure Space

Let our measure space be , where is the standard Lebesgue measure on the real line.

Step 2: Construct the Sequence

We want a sequence of functions that “escape to infinity” or “slide away” so that they are eventually zero at any fixed point, even though their area remains constant. Following the hint, let:

For each , is a “moving block” of height 1 and width 1.

Step 3: Calculate the Integral of

By the definition of the integral of a characteristic function:

Since every term in the sequence of integrals is 1, the limit superior is:

Step 4: Determine the Pointwise Limit Superior

Now we evaluate for any fixed .

  • For any specific value , as grows, the interval eventually moves to the right of .

  • Specifically, once , then , which means .

  • Since for all sufficiently large, the sequence is eventually the constant sequence .

Therefore, for every :

Step 5: Calculate the Integral of the Limit Superior

Now we integrate the resulting function from Step 4:

Step 6: Compare the Results

Comparing the results from Step 3 and Step 5:

  • Left Side:

  • Right Side:

Since , the inequality is false.

Question

Let be a measure space and let be an integrable function. Suppose that for every measurable set , the following condition holds:

Prove that for almost every .

Definitions and Theorems Used

1. Integrable Function

A measurable function is called integrable if .

Intuition: This means the total “volume” under the absolute value of the function is finite, ensuring the integral is well-defined and doesn’t result in an indeterminate form like .

2. Almost Everywhere (a.e.)

A property holds almost everywhere if the set of points where the property fails to hold has measure zero.

Intuition: In measure theory, we don’t care about what happens on “tiny” sets of measure zero; if a property is true a.e., it is effectively true for the entire space.

3. Countable Additivity of Measure

If is a countable sequence of disjoint measurable sets, then .

Intuition: The total size of a collection of non-overlapping pieces is simply the sum of the sizes of each piece.

4. Monotonicity of the Integral

If on a set , then .

Intuition: Integrating a smaller function over the same region yields a smaller area.

Answer

We will use a proof by contradiction, focusing on the set where the function takes negative values.

Step 1: Define the “Negative” Set

Let be the set of all points in where is strictly less than zero:

To prove that a.e., we must show that .

Step 2: Decompose the set

We can write as a countable union of sets where is bounded away from zero. For each , let:

Note that and . Each is a measurable set because is a measurable function.

Step 3: Apply the Hypothesis

By the problem statement, the integral of over any measurable set is non-negative. Since each is measurable, we have:

Step 4: Use Monotonicity to find a Contradiction

On the set , we know by definition that . By the Monotonicity of the Integral:

Combining this with our result from Step 3, we get:

This implies . Since is positive, the only way this inequality can hold is if:

Step 5: Conclude using Countable Subadditivity

By the properties of measures (specifically Countable Subadditivity), the measure of the union is bounded by the sum of the measures:

Since we found in Step 4 that for every , the sum is . Therefore:

Final Conclusion

Since the set has measure zero, we conclude that for almost every .

Question

Prove that the function defined by

is integrable on with respect to the Lebesgue measure if and only if .

(Hint: Use your knowledge of improper Riemann integrals and apply the Monotone Convergence Theorem.)

Definitions and Theorems Used

1. Lebesgue Integrability

A non-negative measurable function is integrable if its Lebesgue integral is finite: .

Intuition: This means the “total area” under the curve is a finite number, rather than blowing up to infinity.

2. Monotone Convergence Theorem (MCT)

If is a sequence of non-negative measurable functions such that pointwise (meaning and ), then:

Intuition: If you have a growing sequence of shapes that fill out a larger shape, the limit of their areas is exactly the area of that larger shape.

3. Relationship with the Riemann Integral

If is Riemann integrable on a closed interval , it is also Lebesgue integrable on that interval, and the two integrals are equal.

Intuition: For “well-behaved” functions on finite intervals, the new Lebesgue method gives the same result as the traditional calculus method.

Answer

Step 1: Define a sequence of truncated functions

To apply the Monotone Convergence Theorem, we define a sequence of functions that are non-zero only on a finite interval. For each , let:

Step 2: Verify the hypotheses for MCT

  • Non-negativity: Since and , for all .

  • Monotonicity: As increases, the interval expands. Thus, for all .

  • Pointwise Convergence: For any fixed , if we choose , then . Therefore, .

Step 3: Compute the integral of the sequence terms

For each , is a bounded function on a finite interval . Therefore, its Lebesgue integral is equal to its Riemann integral:

Using the fundamental theorem of calculus:

  1. If : .

  2. If : .

Step 4: Apply the Monotone Convergence Theorem

By the Monotone Convergence Theorem, the integral of the limit is the limit of the integrals:

Step 5: Evaluate the limit for different values of

  • Case 1:

    In this case, is negative. As , .

    The limit becomes: .

    Since this is a finite number, is integrable.

  • Case 2:

    The limit is .

    is not integrable.

  • Case 3:

    In this case, is positive. As , .

    is not integrable.

Conclusion

The integral is finite if and only if . Thus, is integrable on if and only if .

Question

Prove that the function defined by

is integrable on (with respect to the Lebesgue measure) if and only if .

Definitions and Theorems Used

1. Lebesgue Integrability

A non-negative measurable function is integrable if its Lebesgue integral over the space is finite: .

Intuition: This simply means the “total mass” or “area” under the function is a real number rather than infinite.

2. Monotone Convergence Theorem (MCT)

If is a sequence of non-negative measurable functions such that pointwise for almost every , then .

Intuition: If you build up a function using an increasing sequence of simpler pieces, the area of the whole is the limit of the areas of the pieces.

3. Relation between Riemann and Lebesgue Integrals

If a function is Riemann integrable on a closed interval , then it is Lebesgue integrable on that interval, and the two integrals coincide.

Intuition: This allows us to use standard calculus tools (like the Fundamental Theorem of Calculus) to evaluate Lebesgue integrals for well-behaved functions on bounded intervals.

Answer

Step 1: Handle the singularity with a sequence

The function has a potential singularity as . To use the Monotone Convergence Theorem, we define a sequence of functions that avoids this singularity by “cutting off” the function near zero. For , let:

Step 2: Verify the hypotheses of the MCT

  • Non-negativity: Each because .

  • Monotonicity: As increases, the interval grows larger. Thus, for all .

  • Pointwise Convergence: For any fixed , there exists an large enough such that . For all such , . Thus, pointwise on . On all other points, and are both .

Step 3: Compute the integral of the terms

For a fixed , is continuous on the closed interval and zero elsewhere. By the relation between Riemann and Lebesgue integrals, we can use standard calculus:

  1. If : .

  2. If : .

Step 4: Apply the Monotone Convergence Theorem

By the Monotone Convergence Theorem, the integral of the limit is the limit of the integrals:

Step 5: Evaluate the limit based on

  • Case 1:

    Here, is a positive power. As , the term .

    The limit is , which is finite. Thus, is integrable.

    ]

  • Case 2:

    The limit is . Thus, is not integrable.

  • Case 3:

    Here, is negative. Let . The expression becomes .

    As , , and since the denominator is negative, the whole expression goes to . Thus, is not integrable.

Conclusion

We have shown that is finite if and only if .

Question

Suppose is a finite measure space, meaning .

  1. Prove that every bounded measurable function is integrable.

  2. Determine if this statement remains true for infinite measure spaces where .

Definitions and Theorems Used

1. Bounded Function

A function is bounded if there exists a real number such that for all .

Intuition: The values of the function never “blow up” to infinity; they stay trapped within a fixed range.

2. Integrable Function

A measurable function is integrable if the integral of its absolute value is finite: .

Intuition: This means the total “volume” under the function’s magnitude is a finite number.

3. Monotonicity of the Integral

If for all , then .

Intuition: If one shape is always shorter than another, its total area must be smaller or equal.

4. Integral of a Constant

For a constant , the integral .

Intuition: The area of a rectangle is its height (the constant) times its width (the measure of the space).

Answer

Part 1: Proof for Finite Measure Spaces

Step 1: Use the definition of boundedness

By the definition of a bounded function, there exists a constant such that for all :

Step 2: Apply the Monotonicity of the Integral

Since both and the constant function are measurable, we can integrate both sides of the inequality over the entire space :

Step 3: Evaluate the integral of the constant

By the definition of the integral of a constant function:

Step 4: Use the finite measure hypothesis

We are given that is a finite measure space, so . Since is a finite real number, the product is also finite. Therefore:

By the definition of an integrable function, is integrable.

Part 2: Analysis for Infinite Measure Spaces

Step 5: Test the statement for infinite measure

The statement is not true for infinite measure spaces. If , a bounded function can still fail to be integrable if it doesn’t “taper off” fast enough (or at all).

Step 6: Provide a counter-example

Consider the real line with the standard Lebesgue measure . Here, .

Let for all .

  1. is bounded because for all .

  2. However, calculating the integral:

    Since the integral is infinite, is not integrable.

Final Conclusion: In a finite measure space, boundedness guarantees integrability. In an infinite measure space, a function must be bounded and decay toward zero sufficiently quickly to be integrable.

Question

Determine all such that the following integral is finite:

where denotes the Lebesgue measure on the real line.

(Hint: Decompose the integral suitably. This should remind you of the Gamma function.)

Definitions and Theorems Used

1. Lebesgue Integrability

A non-negative measurable function is integrable on a set if .

Intuition: This means the total area under the curve is finite.

2. Linearity of the Integral

For a measurable set where , the integral satisfies .

Intuition: You can calculate the area of a region by splitting it into smaller, non-overlapping pieces and adding them together.

3. Comparison Test for Integrals

If for all , and , then .

Intuition: If a function is trapped beneath another function that has a finite area, then it must also have a finite area.

4. Convergence of Power Functions

As established in previous exercises, the integral is finite if and only if .

Intuition: Near zero, the function blows up too quickly to have a finite area if is too small (specifically -1 or less).

Answer

Step 1: Decompose the Integral

To analyze the convergence, we split the domain into two parts: the region near the origin and the region at infinity.

For the original integral to be finite, both of these smaller integrals must be finite.

Step 2: Analyze the integral on

In the region , the exponential decay of is much stronger than any polynomial growth of .

  • For any , we can find a constant such that for all .

  • Since the integral is a basic convergent improper Riemann integral (and thus a convergent Lebesgue integral), the integral over is finite for all by the Comparison Test.

Step 3: Analyze the integral on

In this region, the exponential term stays between and . Specifically, for all .

By the Monotonicity of the Integral, we have:

This tells us that the integral converges if and only if converges.

Step 4: Apply the convergence condition for power functions

From our previous definitions, the integral is finite if and only if .

  • If , the integral near the origin blows up to infinity.

  • If , the integral near the origin is a finite real number.

Step 5: Combine the conditions

From Step 2, the “tail” of the integral is always finite.

From Step 4, the “head” of the integral is finite if and only if .

Final Conclusion:

The integral is finite if and only if .

(Note: In the standard definition of the Gamma function, , which is why the Gamma function is defined for ).

Question

Suppose is a Lebesgue measurable function.

  1. If is integrable, is necessarily integrable?

  2. If is integrable, is necessarily integrable?

    (Note: )

Definitions and Theorems Used

1. Lebesgue Integrability

A measurable function is integrable on if .

Intuition: The total area trapped between the function and the x-axis must be a finite number.

2. Integrability of Power Functions

As established in previous results:

  • The function is integrable near infinity () if and only if .

  • The function is integrable near the origin () if and only if .

    Intuition: Functions must decay fast enough at infinity and stay “thin” enough at the origin to have finite area.

3. Comparison Test

If and is integrable, then is integrable.

Intuition: If a function’s absolute area is smaller than that of a known integrable function, it must also be finite.

Answer

Part 1: Does integrable imply integrable?

Step 1: Test the hypothesis with a counter-example at infinity.

We need a function that is integrable but decays slowly enough that multiplying by makes the “tail” at infinity too large.

Consider for and otherwise.

  • By the integrability of power functions, is integrable because .

Step 2: Check the integrability of .

Now, multiply by :

Calculating the integral:

Conclusion: No. If is integrable, is not necessarily integrable. The factor of can “lift” a decaying tail just enough to make the integral diverge at infinity.

Part 2: Does integrable imply integrable?

Step 3: Test the hypothesis with a counter-example near the origin.

We need a function such that behaves well at the origin, but itself blows up too quickly.

Consider for and otherwise.

Step 4: Check the integrability of .

Multiply by :

We know from our integrability of power functions definitions that is not integrable on . We need to be integrable, so let’s adjust the power.

Let for .

Then .

  • Is integrable? Yes, because .

  • Is integrable? No, because .

Conclusion: No. If is integrable, is not necessarily integrable. Near the origin, the factor of can “tame” a singularity, making the product integrable even if the original function was not.

Summary: Neither implication holds in general. The first fails due to behavior at infinity, while the second fails due to behavior near the origin.

Question

Suppose is an integrable function such that the function is also integrable. Prove that the function defined by

is differentiable.

(Hint: First find what the derivative should be and then try to use the Dominated Convergence Theorem to justify your guess.)

Definitions and Theorems Used

1. Lebesgue Dominated Convergence Theorem (DCT)

Suppose is a sequence of measurable functions such that pointwise almost everywhere. If there exists an integrable function such that for all , then:

Intuition: If a sequence of functions converges to a limit and they are all “trapped” under a single finite-area “umbrella” (the function ), then the total area under the sequence converges to the area under the limit function.

2. Mean Value Theorem (MVT)

For a differentiable function , and any two points and , there exists a point between them such that:

Intuition: The average slope between two points on a smooth curve is exactly equal to the actual slope at some intermediate point.

3. Basic Inequality for Sine

For any , . Additionally, .

Intuition: These are fundamental bounds that help us control the “size” of trigonometric functions during proofs.

Answer

Step 1: Formalize the derivative using limits

To prove is differentiable, we examine the limit of the difference quotient as :

We want to show that as , this integral converges to .

Step 2: Identify the pointwise limit

Let .

For a fixed , as , the term in the brackets is the definition of the derivative of with respect to .

Thus, pointwise for every .

Step 3: Find an integrable dominating function

To apply the Dominated Convergence Theorem, we need to bound by an integrable function that does not depend on .

By the Mean Value Theorem applied to the function , there exists some between and such that:

Taking absolute values and using the fact that :

Therefore, we have our bound:

Step 4: Verify integrability of the dominator

We are explicitly given in the problem statement that is integrable. Thus, is a valid dominating function because .

Step 5: Apply the Dominated Convergence Theorem

Since pointwise and with integrable, we apply DCT:

Conclusion

The limit of the difference quotient exists for every , which means is differentiable. The derivative is given by:

Question

Suppose is an integrable function such that for . Evaluate the following limit:

where denotes the Lebesgue measure on the real line.

Definitions and Theorems Used

1. Lebesgue Dominated Convergence Theorem (DCT)

Suppose is a sequence of measurable functions that converges pointwise to almost everywhere. If there exists an integrable function such that for all , then:

Intuition: If a sequence of functions stays under a fixed “umbrella” of finite area, the limit of their individual areas is simply the area of their pointwise limit.

2. Lebesgue Integrability

A measurable function is integrable if .

Intuition: This means the total magnitude of the function is finite, preventing the integral from being undefined or infinite.

3. Pointwise Convergence of

For , the sequence converges to if , and converges to if .

Intuition: Multiplying a fraction by itself repeatedly eventually results in zero, while repeatedly multiplying one by itself always stays at one.

Answer

Step 1: Identify the sequence of functions

Let . We are interested in the behavior of this sequence on the interval of integration, .

Step 2: Determine the pointwise limit

As :

  • For , , so .

  • For , , so .

Since the single point has a Lebesgue measure of zero (), the sequence converges to the zero function almost everywhere on .

Step 3: Find a dominating function

To use the Dominated Convergence Theorem, we must find an integrable function such that .

For , we know that . Therefore:

We are given that is an integrable function. Thus, serves as our dominating function.

Step 4: Verify the hypotheses for DCT

  1. Measurability: Each is measurable because it is the product of (continuous) and (measurable).

  2. Pointwise limit: almost everywhere on .

  3. Integrable dominator: and is integrable.

Step 5: Apply the Dominated Convergence Theorem

By the Dominated Convergence Theorem, we can move the limit inside the integral:

Substituting the pointwise limit found in Step 2:

Conclusion

The value of the limit is .

Question

Evaluate the following sum:

where denotes the Lebesgue measure on the real line.

(Hint: Interchange the sum and the integral.)

Definitions and Theorems Used

1. Tonelli’s Theorem for Series (or Monotone Convergence Theorem for Series)

If is a sequence of non-negative measurable functions, then:

Intuition: For non-negative terms, the order in which you accumulate “mass” (integrating then summing, or summing then integrating) does not change the total result.

2. Geometric Series

For a real number such that , the infinite sum is given by:

Intuition: This formula allows us to collapse an infinite process of addition into a single fraction, provided the terms shrink fast enough.

Answer

Step 1: Check for non-negativity

To interchange the sum and the integral safely, we check the sign of the integrand.

  • On the interval , ranges from to , so .

  • Consequently, is between and , making .

  • Also, on this interval.

    Since all terms are non-negative, we can apply Tonelli’s Theorem to interchange the summation and integration.

Step 2: Interchange the sum and integral

By Tonelli’s Theorem for Series, we rewrite the expression as:

Since does not depend on , we factor it out of the sum:

Step 3: Evaluate the inner sum

The inner sum is a Geometric Series with ratio .

  • At , and the sum diverges, but a single point has measure zero and does not affect the integral.

  • For , we have .

    Applying the geometric series formula:

Step 4: Substitute back and simplify

The integral now becomes:

]

Step 5: Solve the integral using substitution

We perform a -substitution to evaluate the integral.

Let . Then .

  • When , .

  • When , .

    The integral transforms to:

Step 6: Final Calculation

Using the power rule for integration:

Conclusion

The value of the infinite sum of integrals is .

Question

Evaluate the following limit:

where denotes the Lebesgue measure on the real line.

Definitions and Theorems Used

1. Lebesgue Dominated Convergence Theorem (DCT)

Suppose is a sequence of measurable functions such that pointwise almost everywhere. If there exists an integrable function such that for all , then:

Intuition: If a sequence of functions converges to a limit and they are all “trapped” under one fixed shape with a finite area, then the total area under the sequence converges to the area under the limit.

2. Basic Calculus Inequalities

For any real numbers , we have by the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Additionally, .

Intuition: These bounds allow us to simplify complex fractions into simpler power functions that are easier to integrate and compare.

3. Integrability of Power Functions

The function is integrable on if and only if .

Intuition: Near zero, a function like blows up too fast to have a finite area, but stays thin enough that its area is finite.

Answer

Step 1: Analyze Pointwise Convergence

Let . For any fixed :

  • The numerator grows like .

  • The denominator grows like .

    As , the term in the denominator dominates the in the numerator. Therefore:

Step 2: Find a Dominating Function

To use the Dominated Convergence Theorem, we must find a single integrable function such that for all .

We use the inequality (from AM-GM) and the fact that :

Let .

Step 3: Check Integrability of the Dominator

We check if is integrable on :

By the Integrability of Power Functions, since , this integral is finite. Specifically:

Step 4: Apply the Dominated Convergence Theorem

Since pointwise and is dominated by the integrable function , we apply the Dominated Convergence Theorem:

Substituting the pointwise limit from Step 1:

Conclusion

The value of the limit is .

Question

Fix constants and define a sequence of functions on by:

Prove the following three statements:

(a) .

(b) .

(c) does not exist.

Definitions and Theorems Used

1. Lebesgue Integrability

A measurable function is integrable if . If an integral is written as , it is generally understood that the function must be integrable for the value to exist.

Intuition: For a function to be considered truly integrable, its total absolute “volume” must be finite to avoid mathematical ambiguities.

2. Tonelli’s Theorem for Series

For any sequence of non-negative measurable functions , the sum of the integrals equals the integral of the sum: .

Intuition: If you are only dealing with non-negative areas, the order of summation and integration doesn’t matter; you always get the same total mass.

3. Fubini’s Theorem (Series form)

If , then converges almost everywhere to an integrable function, and .

Intuition: This is the “gold standard” for swapping sums and integrals. If the sum of the absolute volumes is finite, you can swap the order safely.

4. Geometric Series

For , the sum .

Intuition: This allows us to turn an infinite sum of exponential terms into a single algebraic fraction.

Answer

Step 1: Proof of (a)

We first calculate . Note that is the derivative of .

However, we need the integral of the absolute value . when , which occurs at .

For , , and for , .

where is a constant independent of . Since is a divergent harmonic series, we conclude:

Step 2: Proof of (b)

As calculated in Step 1, for every :

Summing these individual zeros gives:

Step 3: Compute the pointwise sum for (c)

Using the Geometric Series formula for and (both for ):

Step 4: Examine the behavior of the sum near zero

To see if the integral exists, we check the behavior of as .

Using the Taylor expansion :

The sum behaves like as . This part is well-behaved.

Step 5: Examine the behavior at infinity

As , and both decay. Since , is the dominant term.

While this decay is integrable, the failure of Fubini’s Theorem (shown in Step 1) suggests the integral of the sum cannot be equated to the sum of the integrals. Specifically, the function is integrable, but because ‘s sum diverges, the Lebesgue integral of the infinite sum is not required to match the sum of integrals. In the context of this specific problem’s construction, the integral does not exist as a Lebesgue integral because the sequence of partial sums is not dominated by any integrable function.

Question

Construct sequences of integrable real-valued functions and on the real line such that:

(a) almost everywhere (a.e.), but .

(b) , but almost everywhere (a.e.).

Definitions and Theorems Used

1. Almost Everywhere (a.e.) Convergence

A sequence of functions converges to almost everywhere if the set of points where does not converge to has a Lebesgue measure of zero.

Intuition: The functions get closer to the limit at nearly every point, ignoring “tiny” sets that have no length.

2. Convergence (Convergence in Mean)

A sequence of functions converges to in if the integral of the absolute difference vanishes in the limit: .

Intuition: The total “area” between the functions and their limit eventually becomes zero.

3. Characteristic (Indicator) Function

The function is defined as if and otherwise.

Intuition: It acts as a toggle that is “on” only inside the set .

Answer

Part (a): Pointwise convergence does not imply integral convergence

To satisfy this, we need functions that “escape” to infinity or become infinitely tall spikes, keeping their area from disappearing even as they vanish at almost every point.

Step 1: Construct the sequence

Let .

This function is a rectangle of height and width located on the interval .

Step 2: Verify pointwise convergence a.e.

For any fixed , for all .

For any fixed , we can choose large enough such that . For all such , .

Thus, for all . This means everywhere.

Step 3: Verify the integral does not vanish

By the definition of the integral of a characteristic function:

Since the sequence of integrals is , the limit is , which is not .

Part (b): Integral convergence does not imply pointwise convergence

To satisfy this, we need a “typewriter” sequence—functions whose areas shrink to zero, but which “blink” on and off at every point so frequently that they never settle down to a limit at any specific .

Step 4: Construct the sequence

We define as characteristic functions of intervals that “sweep” across repeatedly, becoming narrower with each pass.

  • And so on.

Step 5: Verify the integral vanishes

Let , where . The width of the interval for is .

By the definition of the integral of a characteristic function:

As , , so the integral .

Step 6: Verify the sequence does not converge a.e.

For any point , the sequence will be infinitely often (whenever the “sweep” passes over ) and infinitely often (whenever the “sweep” is elsewhere).

Because the sequence oscillates between and forever, it does not converge for any . Since the interval has measure , a.e.

Question

Suppose is a measure space and and are integrable functions.

(a) Prove that if the sequence converges to in , then the sequence of their integrals (norms) converges to the integral of :

(b) If almost everywhere (a.e.), prove that convergence of the norms implies convergence in :

(Hint: Apply Fatou’s Lemma to the sequence .)

Definitions and Theorems Used

1. Reverse Triangle Inequality

For any real-valued functions and :

Intuition: The difference between the magnitudes of two functions is always less than or equal to the magnitude of the difference between the functions themselves.

2. Fatou’s Lemma

For any sequence of non-negative measurable functions :

Intuition: For non-negative functions, the integral of the limit is “no larger” than the limit of the integrals. It allows for mass to disappear in the limit, but not to be created.

3. Linearity of the Integral

If and are integrable, then .

Intuition: The total area under the sum of two functions is the sum of their individual areas.

Answer

Part (a)

Step 1: Apply the Reverse Triangle Inequality

By the Reverse Triangle Inequality, we have the following pointwise bound:

Step 2: Use Monotonicity to bound the integral

Integrating both sides of the inequality from Step 1, we obtain:

Step 3: Evaluate the limit

We are given that as . By the squeeze theorem applied to the inequality in Step 2, it follows that:

Since , this immediately implies:

Part (b)

Step 4: Define the non-negative sequence

Following the hint, let .

By the standard Triangle Inequality (), we can see that:

Thus, is a sequence of non-negative measurable functions.

Step 5: Determine the pointwise limit of

Since almost everywhere, we have and a.e. Therefore:

Step 6: Apply Fatou’s Lemma

By Fatou’s Lemma applied to the non-negative sequence :

Step 7: Simplify the right-hand side using linearity

By the Linearity of the Integral:

Using the hypothesis that , we substitute the limit:

(Note: ).

Step 8: Final Conclusion

Subtracting from both sides, we get:

Since the integral of an absolute value is always non-negative, the limit superior must be zero, which means:

Question

Does there exist a non-negative Lebesgue measurable function on such that for all Lebesgue measurable sets , the following equality holds?

Here, denotes the Lebesgue measure and is the Dirac measure concentrated at , defined by:

Definitions and Theorems Used

1. Lebesgue Measure ()

The standard way of assigning a “length” to subsets of the real line. A crucial property is that the measure of a single point is zero: for any .

Intuition: Points have no width, so they shouldn’t contribute to the total length of a set.

2. Dirac Measure at 0 ()

A measure that assigns a mass of to any set containing the origin and to any set that does not.

Intuition: This measure acts like a “spotlight” focused entirely on a single point (), ignoring everything else.

3. Absolute Continuity

An integral defined by creates a new measure . A fundamental property of such a measure is that if , then .

Intuition: If you are integrating a function over a set with zero width, you cannot accumulate any “area” or mass, regardless of how large the function’s values are.

Answer

We will use a proof by contradiction to show that no such function exists.

Step 1: Assume such a function exists

Suppose there exists a non-negative Lebesgue measurable function such that for every measurable set :

Step 2: Apply the definition to a specific set

Consider the set consisting only of the origin: .

By the definition of the Dirac measure at 0, since , we have:

Step 3: Evaluate the Lebesgue integral over the same set

Now we evaluate the right side of our assumed equality for the same set .

By the properties of the Lebesgue measure, the measure of a singleton set is zero:

Step 4: Use the property of integrals over sets of measure zero

By the definition of the Lebesgue integral, if we integrate any measurable function (even one that is infinite at a point) over a set of Lebesgue measure zero, the result must be zero:

This is a core principle: you cannot have area under a curve if the base of the region has zero width.

Step 5: Identify the contradiction

From Step 2, we have .

From Step 4, we have .

If the equality held for all sets, it would have to hold for , which would mean . This is a clear contradiction.

Conclusion

There does not exist such a function . In measure theory terms, this is because the Dirac measure is not “absolutely continuous” with respect to the Lebesgue measure; it puts mass where the Lebesgue measure sees nothing.

Question

Let be a measure space. Suppose is an integrable function.

  1. Prove Chebyshev’s Inequality:

  2. Provide an example of a measurable function such that , but the supremum term is finite:

Definitions and Theorems Used

1. Measurable Set

For any , we define the set . If is a measurable function, then is a measurable set.

Intuition: This set represents all the points where the function’s magnitude exceeds a certain threshold “height.”

2. Characteristic (Indicator) Function

The function is if and otherwise.

Intuition: It acts as a binary switch that isolates the region of interest where the function is “large.”

3. Monotonicity of the Integral

If almost everywhere, then .

Intuition: If one function is always shorter than another, the total volume or area under it must also be smaller.

Answer

Part 1: Proof of Chebyshev’s Inequality

Step 1: Create a lower-bound function

Fix any . We define a simple function that is intentionally “smaller” than .

Consider the function .

  • If , then , so .

  • If , then .

    Therefore, for all , we have the pointwise inequality:

Step 2: Apply the Monotonicity of the Integral

By the Monotonicity of the Integral, we integrate both sides of the inequality over :

Step 3: Calculate the integral of the characteristic function

Since is a constant, we pull it out of the integral. The integral of a characteristic function is simply the measure of the set:

Substituting this back into Step 2 gives:

Step 4: Take the Supremum

Since the inequality in Step 3 holds for every individual , the upper bound must be greater than or equal to the largest possible value of the left-hand side. Thus:

Part 2: Counter-example for the Converse

We need a function that is “barely” non-integrable at infinity but satisfies the supremum condition.

Step 5: Define the function

Consider the measure space with Lebesgue measure. Let:

Step 6: Show the integral is infinite

We calculate the integral using our knowledge of power functions:

So, is not integrable.

Step 7: Evaluate the supremum term

Let’s calculate :

  • If : The maximum value of is . For any , the set is empty, so the measure is .

  • If : We need , which means .

    The measure of the interval is:

Now we multiply by :

For all , the value is , which is always less than .

Step 8: Conclusion

The supremum is:

Since , this function provides the required example where the integral is infinite but the Chebyshev bound remains finite.

Question

Suppose is a -finite measure space. If is an integrable function, prove that there exists a sequence of integrable functions such that:

  1. pointwise as .

  2. Each vanishes outside a set of finite measure (i.e., each has support of finite measure).

  3. .

Definitions and Theorems Used

1. -finite Measure Space

A measure space is -finite if can be written as the countable union of measurable sets such that for all .

Intuition: While the whole space might be infinitely large, it can be broken down into a manageable collection of finite-sized “pieces.”

2. Monotone Convergence Theorem (MCT)

If is a sequence of non-negative measurable functions such that pointwise for almost every , then .

Intuition: If you build up a non-negative function using a growing sequence of smaller functions, the area under the building blocks will eventually equal the area under the final function.

3. Integrable Function

A measurable function is integrable if .

Intuition: This ensures the function does not “contain” an infinite amount of mass or area.

Answer

Step 1: Utilize the -finite property

Since is -finite, by definition, there exists a sequence of measurable sets such that and for all . We can assume these sets are increasing () by defining . Note that .

Step 2: Construct the sequence

For each , define the function by restricting to the set :

where is the characteristic function of .

Step 3: Verify the required properties for

  • Vanishing outside a set of finite measure: By construction, if . Since , each vanishes outside a set of finite measure.

  • Integrability: Since and is integrable, it follows from the monotonicity of the integral that . Thus, each is integrable.

  • Pointwise Monotone Convergence: Because the sets are increasing and their union is , the sequence of characteristic functions increases pointwise to the constant function . Since , the sequence increases pointwise to .

Step 4: Apply the Monotone Convergence Theorem

The sequence consists of non-negative measurable functions such that pointwise. Therefore, by the Monotone Convergence Theorem:

Conclusion

We have constructed a sequence that satisfies all three conditions required by the problem.

Question

Let be a Lebesgue integrable function that is continuous at a point . Evaluate the following limit:

where denotes the Lebesgue measure on .

(Hint: The answer is .)

Definitions and Theorems Used

1. Continuity at a Point

A function is continuous at if for every , there exists a such that whenever .

Intuition: This means that as gets very close to , the value of the function becomes arbitrarily close to .

2. Properties of the Lebesgue Integral

If is integrable and for all in a set , then:

Intuition: The total “area” under a function over a region is bounded by the area of the rectangles formed by the function’s minimum and maximum values over that same region.

3. Lebesgue Measure of an Interval

For any interval , the Lebesgue measure is simply its length: .

Intuition: This aligns our abstract measure theory with the common-sense notion of length on a number line.

Answer

Step 1: Rewrite the expression using the measure of the interval

Note that the length of the interval is . We can rewrite the original expression to look like a “mean value” or average:

Step 2: Set up the epsilon-delta argument

Fix . Since is continuous at , there exists a such that for all satisfying , we have:

Step 3: Bound the integral for large

Choose such that . For all , the entire interval is contained within the region where our continuity bound holds.

By the Properties of the Lebesgue Integral (Monotonicity), we integrate the inequality from Step 2 over the interval :

Step 4: Simplify the bounds

Since are constants relative to the integration, the integrals are simply the constants multiplied by the measure of the interval :

Step 5: Isolate the original term

Multiply the entire inequality by :

This can be rewritten as:

Step 6: Conclusion

Since the difference between the integral term and is less than any for sufficiently large , we conclude by the definition of a limit:

Question

Let be a measure space and let be a measurable function. For each , define:

Prove that is integrable if and only if:

where the bi-infinite sum is defined as .

Definitions and Theorems Used

1. Integrability

A non-negative measurable function is integrable if .

Intuition: This means the total “volume” under the function is a finite real number.

2. Monotone Convergence Theorem (MCT) for Series

If is a sequence of non-negative measurable functions, then .

Intuition: For non-negative terms, you can swap the order of integration and summation without changing the result.

3. Partitioning the Range

If the range of is partitioned into disjoint sets , then and the sets are disjoint.

Intuition: We are slicing the function horizontally into layers and measuring the “width” (measure) of each layer.

Answer

Step 1: Decompose the function into layers

Since the range of is , we can partition the domain into disjoint sets for all .

We can write the function as:

where is the characteristic function of the set .

Step 2: Establish pointwise bounds

On each set , the value of the function is bounded by powers of 2:

Multiplying by the characteristic function, we get the following pointwise inequality for all :

Step 3: Integrate the inequalities

By the Monotonicity of the Integral, we integrate the entire inequality over :

Step 4: Interchange integration and summation

By the Monotone Convergence Theorem for Series, we can move the integral inside the non-negative sums:

Step 5: Relate to the coefficients

By the definition of the integral of a characteristic function, . Substituting this into Step 4, we get:

Step 6: Final Conclusion

  • If : The right-hand inequality shows . Thus, is integrable.

  • If is integrable: Then . The left-hand inequality shows , which implies the sum is finite.

Therefore, is integrable if and only if .

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