/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Let \(f_{n}(x)=\min (f(x), n)\) ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 4

Let \(f_{n}(x)=\min (f(x), n)\) where \(f \geqslant 0\) is measurable. Show that \(\int f_{n} \mathrm{~d} x \uparrow \int f \mathrm{~d} x\).

Short Answer

Expert verified
Using the Monotone Convergence Theorem, \( \int f_n \, \mathrm{d}x \) converges to \( \int f \, \mathrm{d}x \) as \( n \) increases.

Step by step solution

01

Understand the Definitions

The function \( f_n(x) = \min(f(x), n) \), where \( n \) is a real number, essentially "caps" the values of \( f(x) \) to \( n \). This means if \( f(x) > n \), then \( f_n(x) = n \), otherwise \( f_n(x) = f(x) \). Given that \( f \geq 0 \) and is measurable, \( f_n \) is also measurable.
02

Analyze Monotonicity

Observe that for each \( n \), \( f_n(x) \leq f_{n+1}(x) \leq f(x) \) due to the capping function. Therefore, \( \{ f_n(x) \} \) is a non-decreasing sequence of functions that is bounded above by \( f(x) \).
03

Apply Monotone Convergence Theorem

According to the Monotone Convergence Theorem (MCT), if \( f_n(x) \) increases to \( f(x) \), the integrals \( \int f_n \, \mathrm{d}x \) converge to \( \int f \, \mathrm{d}x \) as \( n \to \infty \). Thus, \( \int f_n \, \mathrm{d}x \uparrow \int f \, \mathrm{d}x \).
04

Confirm Boundedness

Since \( f_n(x) \leq f(x) \), and \( f(x) \) is integrable, it implies that \( \int f_n(x) \, \mathrm{d}x \) is bounded above by \( \int f(x) \, \mathrm{d}x \). Thus, we have \( \int f_n \, \mathrm{d}x \to \int f \, \mathrm{d}x \) as it is a non-decreasing and bounded sequence.
05

Conclude the Proof

Since we have shown that \( f_n \to f \) pointwise, \( 0 \leq f_n \leq f \), and \( f_n \) is non-decreasing, the MCT assures us that \( \int f_n \, \mathrm{d}x \uparrow \int f \, \mathrm{d}x \). We conclude that the integration of \( f_n \) approaches the integration of \( f \) as \( n \) increases.

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Key Concepts

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

Measurable Functions
A measurable function is a type of function that interacts well with the measure of a set, making it possible to apply integrals in the context of measure theory.
For a function to be measurable, the inverse image of every open set should be a measurable set within the domain.
  • This makes it easier to calculate integrals when working with functions on complex domains.
  • Essentially, measurable functions help bridge the gap between raw data points and the smooth world of calculus.
In our exercise, we have a function \( f \) that is both non-negative and measurable, guaranteeing that the function \( f_n(x) = \min(f(x), n) \) retains the property of measurability.
Non-decreasing Sequence
A non-decreasing sequence is formed when each term is equal to or greater than the preceding term. In the context of functions, this relates to a sequence of functions where each function is pointwise less than or equal to the next.
For example, if you have a sequence of functions \( \{ f_n(x) \} \), it is non-decreasing if:
  • For all \( n \), \( f_n(x) \leq f_{n+1}(x) \) for all values of \( x \).
  • This ensures that as \( n \) increases, \( f_n(x) \) moves upward towards a limiting function.
In our case, \( \{ f_n(x) \} \) is a non-decreasing sequence as \( f_n(x) = \min(f(x), n) \), which will either increase or remain the same as \( n \) increases.
Bounded Functions
Bounded functions are essential in analysis since they prevent values from going beyond a certain limit, making calculations more manageable.
  • A function \( g(x) \) is said to be bounded above if there exists a number \( M \) such that for all \( x \), \( g(x) \leq M \).
  • In our exercise, the sequence of functions \( \{ f_n(x) \} \) is bounded above by \( f(x) \), meaning no \( f_n(x) \) exceeds \( f(x) \).
This property is pivotal when using integrals, as having an upper bound ensures that integrals exist and are finite, thus aiding in convergence analysis.
Integration
Integration is a mathematical process that determines the accumulation of quantities, which can be represented as areas under curves. It serves as a powerful tool for dealing with functions defined by limits.
  • In this scenario, the Monotone Convergence Theorem plays a crucial role in linking the integration of a sequence of non-decreasing functions to the integral of their limit.
  • The theorem ensures that if \( f_n(x) \to f(x) \) pointwise, and each \( f_n(x) \) is measurable and non-decreasing, then \( \int f_n \mathrm{~d}x \uparrow \int f \mathrm{~d}x \).
This means the integral of \( f_n \) approaches the integral of \( f \), reinforcing the outcome demonstrated in the original exercise.

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Most popular questions from this chapter

Show that if \(\alpha>0\), then $$ \lim _{n \rightarrow \infty} \int_{0}^{n}(1-x / n)^{n} x^{\alpha-1} \mathrm{~d} x=\int_{0}^{\infty} e^{-x} x^{\alpha-1} \mathrm{~d} x $$

Show that \(\lim _{b \rightarrow 1-0} \int_{0}^{b} \sum_{n=1}^{\infty} \frac{x^{n-1}}{\sqrt{n}} \mathrm{~d} x=\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}} .\)

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