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Suppose \(\left\\{f_{n}\right\\} \subset L^{+}, f_{n} \rightarrow f\) pointwise, and \(\int f=\lim \int f_{n}<\infty\). Then \(\int_{E} f=\lim \int_{E} f_{n}\) for all \(E \in \mathcal{M}\). However, this need not be true if \(\int f=\lim \int f_{n}=\) \(\infty\).

Short Answer

Expert verified
If \( \int f < \infty \), then \( \int_E f = \lim \int_E f_n \) for all measurable sets \( E \).

Step by step solution

01

Understanding the Setup

We are given a sequence of functions \( f_n \) that belong to \( L^+ \), which means they are measurable and non-negative. The sequence converges pointwise to a function \( f \). We also have the condition that \( \int f = \lim \int f_n < \infty \). Our goal is to prove that \( \int_E f = \lim \int_E f_n \) for all measurable sets \( E \).
02

Applying Fatou's Lemma

Fatou's Lemma tells us that for any non-negative measurable functions \( g_n \), \( \int \liminf g_n \leq \liminf \int g_n \). Applying this to our sequence \( f_n \), we get \( \int \liminf f_n \leq \liminf \int f_n = \int f \). Since \( f_n \rightarrow f \) pointwise, \( \liminf f_n = f \), so this simplifies to \( \int f \leq \int f \), which holds true.
03

Using the Dominated Convergence Theorem

The condition \( \int f = \lim \int f_n < \infty \) along with \( f_n \rightarrow f \) pointwise allows us to use the Dominated Convergence Theorem. This theorem states that if \( g_n \rightarrow g \) pointwise and there exists an integrable function \( h \) such that \( |g_n| \leq h \) for all \( n \), then \( \int g_n \rightarrow \int g \). In our case, since \( f \) is finite and \( f_n \leq f \), the theorem applies, giving us \( \int_E f = \lim \int_E f_n \) for any measurable set \( E \).
04

Counterexample for Infinite Integrals

To see why the previous result might not hold when \( \int f = \lim \int f_n = \infty \), consider a sequence of functions \( f_n(x) = n \) on an interval of length \( 1/n \) and zero elsewhere. Each integral \( \int f_n = 1 \), so their limit, and \( \int f = \infty \), preventing us from using the same reasoning.

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Key Concepts

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

Fatou's Lemma
Fatou's Lemma is a vital concept in measure theory, especially when dealing with sequences of non-negative measurable functions. It provides useful inequality relations that help in understanding limiting behaviors of sequences and their integrals. Fatou's Lemma states that if you have a sequence of non-negative measurable functions, \( \{g_n\} \), then the integral of the limit inferior of \( g_n \) is less than or equal to the limit inferior of the integrals of \( g_n \). This can be expressed as: \[ \int \liminf g_n \leq \liminf \int g_n. \] The lemma is particularly useful in situations where convergence is not uniform, allowing us to draw meaningful conclusions about the integrals. In practical terms, it tells us that even if we only know information about the sequence's limit inferior, there is still a relationship to the sequence's integrals.
This concept often comes into play when we cannot directly apply the Dominated Convergence Theorem due to lack of uniform or dominated convergence. In the context of our exercise, we applied Fatou's Lemma to deduce that \( \int \liminf f_n \leq \liminf \int f_n \), which helped us move forward in understanding the behavior of \( f_n \) as it converges to \( f \).
Pointwise Convergence
Pointwise convergence is a type of convergence where each point in the domain of a sequence of functions converges individually to a function. More formally, a sequence of functions \( \{f_n\} \) is said to converge pointwise to a function \( f \) if, for every \( x \) in the domain, as \( n \) approaches infinity, \( f_n(x) \to f(x) \).
  • This means, no matter which point \( x \) you pick from the domain, the value of \( f_n(x) \) will get closer and closer to \( f(x) \) as \( n \) increases.

In our exercise, we are dealing with functions \( f_n \) that converge pointwise to \( f \). While pointwise convergence might not imply the convergence of integrals directly—hence, why we cannot always apply the Dominated Convergence Theorem—it is a fundamental part of the setup. Notably, it's the convergence condition required to consider using tools like Fatou's Lemma and the Dominated Convergence Theorem.
Pointwise convergence is often contrasted with uniform convergence, which requires that the functions converge to \( f \) uniformly across the entire domain, ensuring a stronger form of convergence.
Measurable Functions
Measurable functions are a cornerstone in the field of integration and measure theory. A function \( f \) is said to be measurable if the preimage of every open set under \( f \) is a measurable set. In simpler terms, this means that for any value, you pick up to which \( f \) is mapped, we can measure the size of the input space mapping to that value.
  • This concept is crucial because measurable functions enable us to apply the theory of Lebesgue integration.
  • The functions that we can integrate over a given domain belong to this category.

In the exercise, it's given that the sequence \( \{f_n\} \) consists of measurable functions. This condition guarantees that integrals of these functions can be properly evaluated. Additionally, when applying theorems like Fatou's and the Dominated Convergence Theorem, having measurable functions in your sequence is prerequisite for their application. Overall, understanding this concept helps us understand why and how we can manipulate such sequences in exercises that involve integration.

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Most popular questions from this chapter

Suppose \((X, \mathcal{M}, \mu)\) is a \(\sigma\)-finite measure space and \(f \in L^{+}(X) .\) Let $$ G_{f}=\\{(x, y) \in X \times[0, \infty]: y \leq f(x)\\} $$ Then \(G_{f}\) is \(\mathrm{M} \times \mathrm{B}_{\mathrm{R}}\)-measurable and \(\mu \times m\left(G_{f}\right)=\int f d \mu ;\) the same is also true if the inequality \(y \leq f(x)\) in the definition of \(G_{f}\) is replaced by \(y

Let \(\mu\) be counting measure on \(\mathrm{N}\). Then \(f_{n} \rightarrow f\) in measure iff \(f_{n} \rightarrow f\) uniformly.

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 \(B\).

Define \(G: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) by \(G\left(r, \phi_{1}, \ldots, \phi_{n-2}, \theta\right)=\left(x_{1}, \ldots, x_{n}\right)\) where \(x_{1}=r \cos \phi_{1}, \quad x_{2}=r \sin \phi_{1} \cos \phi_{2}, \quad x_{3}=r \sin \phi_{1} \sin \phi_{2} \cos \phi_{3}, \ldots\), \(x_{n-1}=r \sin \phi_{1} \cdots \sin \phi_{n-2} \cos \theta, \quad x_{n}=r \sin \phi_{1} \cdots \sin \phi_{n-2} \sin \theta .\) a. \(G\) maps \(\mathbb{R}^{n}\) onto \(\mathbb{R}^{n}\), and \(\left|G\left(r, \phi_{1}, \ldots, \phi_{n-2,} \theta\right)\right|=|r|\). b. \(\operatorname{det} D_{\left(r, \phi_{1}, \ldots, \phi_{n-2}, \theta\right)} G=r^{n-1} \sin ^{n-2} \phi_{1} \sin ^{n-3} \phi_{2} \cdots \sin \phi_{n-2}\). c. Let \(\Omega=(0, \infty) \times(0, \pi)^{n-2} \times(0,2 \pi)\). Then \(G \mid \Omega\) is a diffeomorphism and \(m\left(\mathbb{R}^{n} \backslash G(\Omega)\right)=0\). d. Let \(F\left(\phi_{1}, \ldots, \phi_{n-2}, \theta\right)=G\left(1, \phi_{1}, \ldots, \phi_{n-2}, \theta\right)\) and \(\Omega^{\prime}=(0, \pi)^{n-2} \times\) \((0,2 \pi)\). Then \(\left(F \mid \Omega^{\prime}\right)^{-1}\) defines a coordinate system on \(S^{n-1}\) except on a \(\sigma\)-null set, and the measure \(\sigma\) is given in these coordinates by $$ d \sigma\left(\phi_{1}, \ldots \phi_{n-2}, \theta\right)=\sin ^{n-2} \phi_{1} \sin ^{n-3} \phi_{2} \cdots \sin \phi_{n-2} d \phi_{1} \cdots d \phi_{n-2} d \theta $$

Let \(f(x)=x^{-1} \sin x\). a. Show that \(\int_{0}^{\infty}|f(x)| d x=\infty\). b. Show that \(\lim _{b \rightarrow \infty} \int_{0}^{b} f(x) d x=\frac{1}{2} \pi\) by integrating \(e^{-x y} \sin x\) with respect to \(x\) and \(y\). (In view of part (a), some care is needed in passing to the limit as \(b \rightarrow \infty .)\)

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