Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 12

Show that if \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and \(f \geq g\) a.c., then $$ \int_{E} f d \mu \geq \int_{E} g d \mu $$

Short Answer

Expert verified
Integrals preserve order. That is, if \(f \geq g\) almost everywhere, then the integral of \(f\) over \(E\), \(\int_{E} f d \mu\), is greater than or equal to the integral of \(g\) over \(E\), \(\int_{E} g d \mu\).

Step by step solution

01

Establish the condition

We are given \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and that \(f \geq g\) almost everywhere (a.c.) in \(E\). This means that the set \({x \in E: f(x) < g(x)}\) has measure zero. Since \(f\) and \(g\) are Lebesgue integrable, we can subtract them without issue.
02

Formulate the measure of difference inequality

We define the difference \(h\) by \(h = f - g\). Because \(f \geq g\) almost everywhere, \(h(x) \geq 0\) for almost all \(x\) in \(E\). By the non-negativity property of Lebesgue integrals, which states that for any non-negative function \(h\), \(int_E h d \mu \geq 0\). In this case, the inequality becomes, \(int_E f d \mu - int_E g d \mu = int_E (f - g) d \mu = int_E h d \mu \geq 0\).
03

Rearrange inequality to show condition

Re-arrange the inequality obtained in the previous step to obtain \(int_E f d \mu \geq int_E g d \mu\). This shows that if \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and \(f \geq g\) almost everywhere, then the integral of \(f\) over \(E\) will be greater than or equal to the integral of \(g\) over \(E\).

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.

Start your free trial

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

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

Show that a subset \(E\) of \(\mathbb{R}\) is measurable if for all \(\epsilon>0\) there exists an open set \(U \supset E\) such that \(\mu^{*}(U-E)<\epsilon\)

\(\operatorname{lf}(X, M)\) and \((Y, N)\) are measurable spaces, show that the projection maps \(\mathrm{pr}_{1}: X \times\) \(Y \rightarrow X\) and \(\mathrm{pr}_{2}: X \times Y \rightarrow Y\) defined by \(\mathrm{pr}_{1}(x, y)=x\) and \(\mathrm{pr}_{2}(x, y)=y\) are measurable functions,

The inner measure \(\mu_{*}(E)\) of a set \(E\) is defined as the least upper bound of the measures of all measurable subsets of \(E\). Show that \(\mu_{*}(E) \leq \mu^{*}(E)\). For any open set \(U \supset E\), show that $$ \mu(U)=\mu_{*}(U \cap E)+\mu^{*}(U-E) $$ and that \(E\) is measurable with finite measure if and only if \(\mu_{*}(E)=\mu^{*}(E)<\infty\).

Show that the union of a sequence of sets of measure zero is a set of Lebesgue measure zero,

Find a step function \(s(x)\) that approximates \(f(x)=x^{2}\) uniformly to within \(\varepsilon>0\) on \([0,1]\), in the sense that \(|f(x)-s(x)|<\varepsilon\) cierywhere in \([0,1]\).
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
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