Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Assume Fatou's lemma and deduce the monotone convergence theorem from it.

Short Answer

Expert verified
Using Fatou's Lemma and the non-decreasing property of integrals, we deduce the monotone convergence theorem.

Step by step solution

01

Understand Fatou's Lemma

Fatou's Lemma states that for a sequence of non-negative measurable functions \( f_n \), we have \( \int \liminf_{n \to \infty} f_n \, d\mu \leq \liminf_{n \to \infty} \int f_n \, d\mu \). This lemma is a crucial tool in measure theory and will form the basis to prove the monotone convergence theorem.
02

Review the Monotone Convergence Theorem

The Monotone Convergence Theorem states that if \( f_n \) is a sequence of non-negative measurable functions such that \( f_n \uparrow f \) pointwise, then \( \int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu \). Our goal is to derive this theorem using Fatou's Lemma.
03

Apply the Pointwise Limit Condition

Since \( f_n \uparrow f \), we have \( \liminf_{n \to \infty} f_n(x) = f(x) \) for each \( x \). Therefore, according to Fatou's Lemma, \( \int \liminf_{n \to \infty} f_n(x) \, d\mu = \int f \, d\mu \leq \liminf_{n \to \infty} \int f_n \, d\mu \).
04

Use the Monotonicity of the Sequence

Given that \( f_n \uparrow f \), for each \( n \), \( \int f_n \, d\mu \leq \int f_{n+1} \, d\mu \) and these integrals are non-decreasing, which implies that \( \liminf_{n \to \infty} \int f_n \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu \).
05

Combine Results to Conclude

Combining steps 3 and 4, we have \( \int f \, d\mu \leq \lim_{n \to \infty} \int f_n \, d\mu \). But since \( f_n \uparrow f \), we also have \( \lim_{n \to \infty} \int f_n(x) \, d\mu \leq \int f \, d\mu \). Hence, we conclude \( \int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu \), thus proving the monotone convergence theorem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fatou's lemma
Fatou's Lemma is an important element of measure theory. It helps us perform calculations on a sequence of functions even when direct computation seems daunting. The lemma is typically applied to non-negative measurable functions. It states that if we have a sequence of non-negative measurable functions \( f_n \), then:- \( \int \liminf_{n \to \infty} f_n \, d\mu \leq \liminf_{n \to \infty} \int f_n \, d\mu \).This essentially tells us that the limit inferior of the integrals is at least as large as the integral of the limit inferior, providing a vital inequality that aids in proving convergence theorems. Fatou's Lemma is an especially useful tool when dealing with sequences where we know the behavior of individual elements but need to deduce the behavior of the sequence as a whole. It's crucial to note that Fatou's Lemma does not guarantee equality, only an inequality.
This inequality becomes foundational when we attempt to derive other theorems, such as the monotone convergence theorem.
Measure Theory
Measure theory allows us to generalize the notion of length, area, and volume from simple geometrical objects to much more complex objects. In essence, measure theory provides the mathematical foundation needed to "measure" things mathematically. It is a part of mathematical analysis that finds applications across different fields such as probability theory.- It deals with \(\sigma\)-algebras, which help in defining measurable spaces.
- It helps in defining functions that can be integrated over these spaces.The main aim of measure theory is to go beyond basic calculus concepts, to extend these ideas to a wider class of functions and processes. One critical ideal of measure theory is the Lebesgue integral, which, as opposed to the more traditional Riemann integral, allows for greater flexibility in integration.
This flexibility is key when dealing with complex functions in advanced calculus problems. Understanding measure theory is fundamental when diving into the study of functional analysis, probability, and dynamical systems.
Non-negative measurable functions
Non-negative measurable functions are those functions which take values greater than or equal to zero, and are measurable with respect to some \(\sigma\)-algebra. These functions often appear in analysis because they have very convenient properties that simplify integration and limit operations.- A measurable function is crucial because it allows us to use measures to integrate with respect to real functions, especially within the contexts of probability and complex analysis.- Such functions are bounded from below, meaning they cannot take on negative values.This non-negativity is particularly beneficial when dealing with integration because it avoids the complexities that arise with cancelation of positive and negative areas. In measure theory, focusing on non-negative functions simplifies many arguments and proofs.
This makes apparent why such functions are involved in foundational theorems like Fatou's lemma and the monotone convergence theorem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(X=A \cup B\) where \(A, B \in \mathcal{M}\), a function \(f\) on \(X\) is measurable iff \(f\) is measurable on \(A\) and on \(B\).

Suppose that \(f\) is a function on \(\mathbb{R} \times \mathbb{R}^{k}\) such that \(f(x, \cdot)\) is Borel measurable for each \(x \in \mathbb{R}\) and \(f(\cdot, y)\) is continuous for each \(y \in \mathbb{R}^{k}\). For \(n \in \mathbb{N}\), define \(f_{n}\) as follows. For \(i \in \mathbb{Z}\) let \(a_{i}=i / n\), and for \(a_{i} \leq x \leq a_{i+1}\) let $$ f_{n}(x, y)=\frac{f\left(a_{i+1}, y\right)\left(x-a_{i}\right)-f\left(a_{i}, y\right)\left(x-a_{i+1}\right)}{a_{i+1}-a_{i}} $$ Then \(f_{n}\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\) and \(f_{n} \rightarrow f\) pointwise; hence \(f\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\). Conclude by induction that every function on \(\mathbb{R}^{n}\) that is continuous in each variable separately is Borel measurable.

Let \(\mu\) be counting measure on \(\mathbb{N}\). Interpret Fatou's lemma and the monotone and dominated convergence theorems as statements about infinite series.

Let \((X, \mathcal{M}, \mu)\) and \((Y, \mathcal{N}, \nu)\) be arbitrary measure spaces (not necessarily \(\sigma\) finite). a. If \(f: X \rightarrow \mathbb{C}\) is \(\mathcal{M}\)-measurable, \(g: Y \rightarrow \mathbb{C}\) is N-measurable, and \(h(x, y)=\) \(f(x) g(y)\), then \(h\) is \(\mathcal{M} \otimes \mathcal{N}\)-measurable. b. If \(f \in L^{1}(\mu)\) and \(g \in L^{1}(\nu)\), then \(h \in L^{1}(\mu \times \nu)\) and \(\int h d(\mu \times \nu)=\) \(\left[\int f d \mu\right]\left[\int g d \nu\right]\).

Suppose that \(f_{n}\) and \(f\) are measurable complex-valued functions and \(\phi: \mathbb{C} \rightarrow C\). a. If \(\phi\) is continuous and \(f_{n} \rightarrow f\) a.e., then \(\phi \circ f_{n} \rightarrow \phi \circ f\) a.e. b. If \(\phi\) is uniformly continuous and \(f_{n} \rightarrow f\) uniformly, almost uniformly, or in measure, then \(\phi \circ f_{n} \rightarrow \phi \circ f\) uniformly, almost uniformly, or in measure, respectively. c. There are counterexamples when the continuity assumptions on \(\phi\) are not satisfied.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free