Problem 23 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

3-23. Let \(C \subset A \times B\) be a set of content \(0 .\) Let \(A^{\prime} \subset A\) be the set of all \(x \in A\) such that \(\\{y \in B:(x, y) \in C\\}\) is not of content \(0 .\) Show that \(A^{\prime}\) is a set of measure 0. Hint \(: \chi C\) is integrable and \(\int_{A \times B \chi C}=\int_{A} \mathfrak{U}=\int_{A} \mathcal{L}\), so \(\int_{A} \mathfrak{U}-\mathcal{L}=0\)

Step-by-Step Solution

Answer

The set \( A' \) has measure 0 because the integral of the characteristic function over \(A \times B\) being 0 implies that the function itself is 0 almost everywhere.

Step 1: Understand the Sets and Their Properties

Given a set \(C \subset A \times B\) of content 0, understand that this set is a subset of the Cartesian product of sets \(A\) and \(B\). The content of the set \(C\) being 0 implies that its measure (in terms of Lebesgue measure) is 0. Let \(A' \subset A\) be the set of all \(x \in A\) such that \[ \{ y \in B : (x, y) \in C\} \] does not have content 0.

Step 2: Define the Characteristic Function

Let \(\chi_C\) be the characteristic function of the set \(C\). This means \(\chi_C(x, y) = 1 \) if \((x, y) \in C\) and \(\chi_C(x, y) = 0\) otherwise. Recall the property that \(\chi_C\) is integrable, meaning \(\int_{A \times B} \chi_C d\mu = 0\) where \(\mu\) is the measure on \(A \times B\).

Step 3: Set Up the Integral Equation

Given that \(\chi_C\) is integrable, we can write \[ \int_{A \times B} \chi_C \, d(\mu_A \times \mu_B) = \int_{A} \left( \int_{B} \chi_C(x, y) \, d\mu_B(y) \right) d\mu_A(x). \] This integral setup utilizes Fubini's theorem, which allows us to switch the order of integration.

Step 4: Evaluate the Inner Integral

Let \( f(x) = \int_{B} \chi_C(x, y) \, d\mu_B(y) \). Because \[ \int_{A \times B} \chi_C d(\mu_A \times \mu_B) = 0, \] it follows that \( \int_{A} f(x) d\mu_A(x) = 0 \). Note that \( f(x) = 0 \) for almost every \( x \in A \), since the integral of a non-negative function being 0 implies that the function itself must be 0 almost everywhere.

Step 5: Identify the Set of Measure 0

Now, consider the set \( A' \) defined as \[ A' = \{x \in A : \mu_B(\{y \in B : (x, y) \in C\}) eq 0\} . \] The function \( f(x) = \mu_B(\{y \in B : (x, y) \in C\}) \) is non-zero only for \( x \in A' \), but since the integral \( \int_{A} f(x) d\mu_A(x) = 0 \), the set \( A' \) must have \( \mu_A(A') = 0 \), meaning it has measure 0.

Key Concepts

Lebesgue Measure

The Lebesgue measure is a way to assign a consistent 'size' or 'measure' to subsets of a given space. In Measure Theory, this is the most common way to measure the idea of size, enabling a better understanding of the properties of different sets.
The Lebesgue measure extends the idea of length and area to more complex sets. For a set in the Cartesian product space, having a Lebesgue measure of zero means that the set is negligible or 'small' in a very precise sense.
When the exercise says the set \(C \subset A \times B\) has content 0, it means the Lebesgue measure of \(C\) is zero. This concept is crucial for understanding the integration and functions defined over these sets.

Characteristic Function

The characteristic function, denoted as \(\chi_C\), is a simple yet powerful concept in Measure Theory. It is defined for a set \(C\) and takes values:

• \(\chi_C(x, y) = 1\) if \((x, y) \in C\)
• \(\chi_C(x, y) = 0\) otherwise

This function helps in simplifying integrals, especially when working with complex sets. In the given solution, \(\chi_C\) is used to transform the problem into an integration problem. It makes it easier to apply Fubini's theorem and calculate the required integrals.
Integrability of \(\chi_C\) is derived from the fact that the measure of \(C\) is zero, thus making the integral of \(\chi_C\) over \(A \times B\) equal to zero.

Fubini's Theorem

Fubini's theorem is an essential tool in Measure Theory. It allows for switching the order of integration in a double integral. The theorem states that if you have a function that is integrable on the product space, you can integrate it iteratively.
In mathematical terms: \[ \int_{A \times B} \, f \, d(\mu_A \times \mu_B) = \int_A \, \left( \int_B \, f(x, y) \, d\mu_B(y) \right) d\mu_A(x) \]
In our solution, Fubini's theorem is used to split the integral into two parts: an inner integral over \(B\) and an outer integral over \(A\). By doing so, we can focus on the inner function \( f(x) = \int_B \, \chi_C(x, y) \, d\mu_B(y) \). This separates the variables and simplifies the process of finding the measure.

Integrable Function

An integrable function is one that has a finite integral. This is crucial in ensuring that the setup used in our solution works out correctly. When we say \(\chi_C\) is integrable, we mean that \[ \int_{A \times B} \, \chi_C \, d(\mu_A \times \mu_B) = 0 \]
In our problem, this property is used to conclude that the integral of \(\chi_C\) over any measurable set will also be zero. This final step lets us deduce that \(A'\), the set of points in \(A\) where \(\chi_C\) is not zero, must have measure zero.
By showing that the integral of \(f(x)\) over \(A\) is zero, we prove that \(f(x)\) must be zero almost everywhere, solidifying our conclusion through the properties of integrable functions.