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If \(\left\\{f_{n}\right\\} \subset L^{+}, f_{n}\) decreases pointwise to \(f\), and \(\int f_{1}<\infty\), then \(\int f=\lim \int f_{n}\).

Short Answer

Expert verified
By the Monotone Convergence Theorem, \(\int f = \lim \int f_n\).

Step by step solution

01

Understand the Context

We are given a sequence of nonnegative measurable functions \(\{f_n\}\) that decrease pointwise to a function \(f\). The integral of the first function in the sequence, \(\int f_1\), is finite, i.e., \(\int f_1 < \infty\). We need to show that the integral of \(f\) is the limit of the integrals of \(f_n\).
02

Apply the Monotone Convergence Theorem

Since \(\{f_n\}\) is a sequence of nonnegative measurable functions that decreases pointwise to \(f\), and \(\int f_1 < \infty\), we can apply the Monotone Convergence Theorem, which states that for a decreasing sequence of measurable functions \(f_n\) converging to \(f\), \(\int f = \lim_{n \to \infty} \int f_n \). This is exactly what we want to prove.

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Key Concepts

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

Nonnegative Measurable Functions
Nonnegative measurable functions are a crucial concept when dealing with integrals and convergence theorems. These functions are defined within a measure space and take on values that are either nonnegative or equal to zero.
This property ensures that when we integrate these functions, the result is also nonnegative, making it easier to handle and interpret results. A function is said to be measurable if it aligns with the structure of a given measure space, usually meaning it can be broken down into simpler parts where we can effectively measure the size or probability of various subsets.
By being measurable, functions can integrate well with Lebesgue integrals and other convergence theorems to achieve accurate and meaningful calculations. Moreover, in the context of our exercise, we're dealing with a sequence of such functions, \({f_n}\), ensuring that the tools from measure theory are applicable throughout the problem.
Pointwise Convergence
Pointwise convergence refers to the behavior of a sequence of functions converging to a limit function at each point in its domain. With pointwise convergence, for every point \(x\) in the domain, the values of the sequence of functions \(f_n(x)\) approach the value of the function \(f(x)\) as \(n\) tends to infinity.
In our exercise, the sequence of nonnegative measurable functions \(\{f_n\}\) decreases pointwise to \(f\). This means for every point \(x\), \(f_n(x)\) is greater than or equal to \(f(x)\) and gets closer as \(n\) increases.
This type of convergence is beneficial because it provides a simple framework to illustrate how a sequence gradually tightens around the target function point-by-point.However, note that pointwise convergence alone doesn’t ensure the integrability of \(f\). But coupled with the monotonicity (the sequence being decreasing), we can apply the Monotone Convergence Theorem effectively.
Lebesgue Integral
The Lebesgue integral is a powerful tool in mathematical analysis, particularly useful for integrating more complex and broader classes of functions than the Riemann integral. It allows us to deal with functions whose traditional "area under the curve" might be difficult to define using earlier methods.
The strength of the Lebesgue integral lies in its construction. Instead of summing slices in the domain (as Riemann does with rectangles), Lebesgue sums over values of the function range. This provides a robust way to manage functions with discontinuities and unique convergence attributes.
The exercise involves using the Lebesgue integral to affirm the Monotone Convergence Theorem.The theorem leverages the Lebesgue integral for nonnegative measurable functions by stating that if we have a sequence of such functions \(f_n\) decreasing pointwise to \(f\), then \(\int f = \lim \int f_n\). This conclusion offers a solid step towards understanding how convergence behaves under different integrals and solidifying why Lebesgue's method is favored in modern analysis.

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Most popular questions from this chapter

For which real values of \(a\) and \(b\) is \(|x|^{a}|\log | x||^{b}\) integrable over \(\left\\{x \in \mathbb{R}^{n}\right.\) : \(\left.|x|<\frac{1}{2}\right\\}\) ? Over \(\left\\{x \in \mathbb{R}^{n}:|x|>2\right\\}\) ?

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 \(f(x)=x^{-1} \sin x\). 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 \(b \rightarrow \infty .)\)

If \(f: X \rightarrow \mathbb{R}\) and \(f^{-1}((r, \infty]) \in \mathcal{M}\) for each \(r \in \mathbb{Q}\), then \(f\) is 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.

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