Question

Let \(\mu\) be a positive measure. A collection of functions \(\left\\{f_{\alpha}\right\\}_{\alpha \in A} \subset L^{1}(\mu)\) is called uniformly integrable if for every \(\epsilon>0\) there exists \(\delta>0\) such that \(\left|\int_{E} f_{\alpha} d \mu\right|<\epsilon\) for all \(\alpha \in A\) whenever \(\mu(E)<\delta .\) a. Any finite subset of \(L^{1}(\mu)\) is uniformly integrable. b. If \(\left\\{f_{n}\right\\}\) is a sequence in \(L^{1}(\mu)\) that converges in the \(L^{1}\) metric to \(f \in L^{1}(\mu)\). then \(\left\\{f_{n}\right\\}\) is uniformly integrable.

Short answer

a. True; b. True

Step-by-step solution

Understanding Uniform Integrability

A collection of functions \(\{f_{\alpha}\}_{\alpha \in A} \subset L^{1}(\mu)\) is called uniformly integrable if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that \(|\int_{E} f_{\alpha} \, d\mu| < \epsilon\) for all \(\alpha \in A\) whenever \(\mu(E) < \delta\). This means the functions do not "concentrate" their mass on sets of arbitrarily small measure.

Proving (a) - Finite Subsets of \(L^1(\mu)\)

Consider a finite collection of functions \(\{f_1, f_2, \ldots, f_n\}\) in \(L^1(\mu)\). Let \(M = \max\{\|f_1\|_{1}, \|f_2\|_{1}, \ldots, \|f_n\|_{1}\}\). Given \(\epsilon > 0\), choose \(\delta = \frac{\epsilon}{M}\). Then for any measurable set \(E\) such that \(\mu(E) < \delta\), and any \(f_i\), the integral \( |\int_E f_i \, d\mu| \leq \|f_i\|_1 \cdot \mu(E) < M \cdot \frac{\epsilon}{M} = \epsilon\). Thus, all functions in the finite set are uniformly integrable.

Proving (b) - Convergence in \(L^1\)

Consider a sequence \(\{f_n\}\) in \(L^1(\mu)\) converging to \(f\) in \(L^1\). This means \(\|f_n - f\|_1 \to 0\). For small \(\epsilon > 0\), find \(N\) such that for all \(n > N\), \(\|f_n - f\|_1 < \frac{\epsilon}{2}\). The function \(f\) itself is integrable, choose \(\delta\) such that \( |\int_E f \, d\mu| < \frac{\epsilon}{2}\) whenever \(\mu(E) < \delta\). For \(n > N\), \( |\int_E f_n \, d\mu| \leq |\int_E f \, d\mu| + |\int_E (f_n - f) \, d\mu| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). For finite \(n \leq N\), uniform integrability follows as in step 2. Thus, \(\{f_n\}\) is uniformly integrable.

Key concepts

Positive Measure

A positive measure is a foundational concept in measure theory, which is a branch of mathematical analysis dealing with the quantification of "size" or "volume" of sets. In essence, when we say that \( \mu \) is a positive measure, it implies that \( \mu(E) \geq 0 \) for any measurable set \( E \). This property ensures that measures give a non-negative value to the size of any set, which is crucial in defining integrals, particularly in spaces like \( L^1(\mu) \).

A measure is a function that assigns a real number to a set, making it possible to integrate over these sets with respect to the measure. The importance of maintaining positivity cannot be overstated because it aligns with our intuitive understanding of area, volume, and probability.

• Ensures consistent interpretations of size, meaning no set gets a negative measure.
• Provides a basis for many integral and differentiation concepts that follow.

In terms of practical application, positive measures are used in probability to assign weights to outcomes and in geometry to measure dimensions of shapes.

Convergence in L1

Convergence in \( L^1 \) or \( L^1(\mu) \) plays a significant role in understanding the behavior of functions within this space. The \( L^1 \) space consists of all integrable functions with respect to a measure \( \mu \), and an important type of convergence here is \( L^1 \)-convergence.

Consider a sequence of functions \( \{f_n\} \) in this space. If \( f_n \) converges to a function \( f \) in \( L^1 \), this means that the integral of their difference over the whole space goes to zero as \( n \) approaches infinity. Mathematically, it is expressed as \( \|f_n - f\|_1 \to 0 \, \text{as}\, n \to \infty \).

• The key idea is that the total "distance" between \( f_n \) and \( f \) gets arbitrarily small.
• This form of convergence implies that \( f \) is the limit in terms of the \( L^1 \)-norm, which measures the average size of their difference.

The significance of \( L^1 \)-convergence lies in its utility for proving more complex theorems and working with function sequences in analysis. It's a stronger notion compared to pointwise or uniform convergence, ensuring stability and robustness in applications like probability and statistics.

Finite Subsets of L1

Finite subsets of \( L^1(\mu) \), where \( \mu \) is a positive measure, are automatically uniformly integrable. To understand why, recall that uniform integrability requires the integral over small sets to be controllably small, no matter which function is used within the finite set. This is achievable because there is a bound on the integrals of all functions in the finite collection.

Suppose you have a finite set of functions \( \{f_1, f_2, \ldots, f_n\} \) in \( L^1(\mu) \). Here's why they are uniformly integrable:

• Each function's integral is bounded, since they belong to the finite set.
• By taking the maximum of their \( L^1 \)-norms, you can choose a small enough \( \delta \) for any \( \epsilon \) desired, ensuring controlled integrability over small measures \( \mu(E) \).

This concept simplifies many theoretical and practical problems, as handling a finite set involves direct, concrete steps. When confronting an infinite collection, additional criteria might be necessary to ensure uniform integrability. Thus, finite subsets often serve as a stepping stone in broader analytical work.