L1 Space: Difference between revisions

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<b> Proposition:</b> Let <math>f,g\in L^1</math>, then the following are equivalent:
<b> Proposition:</b> Let <math>f,g\in L^1</math>, then the following are equivalent:
<ol>
<ol>
<li> <math>\int_E f=\int_E g,</math> for all <math>E\in\mathcal{M}</math>
<li> <math>\int_E f=\int_E g,</math> for all <math>E\in\mathcal{M}</math>

Revision as of 08:46, 15 December 2020

Introduction

Let be a measure space. From our study of integration, we know that if are integrable functions, the following functions are also integrable:

  1. , for

This shows that the set of integrable functions on any measurable space is a vector space. Furthermore, integration is a linear functional on this vector space, ie a linear function sending elements in our vector space to , one would like to use integration to define a norm on our vector space. However, if one were to check the axioms for a norm, one finds integration fails to be a norm by taking almost everywhere, then . In other words, there are non zero functions which has a zero integral. This motivates our definition of to be the set of integrable functions up to equivalence to sets of measure zero.

Space

In this section, we will construct .

Definition

Let denote the set of integrable functions on , ie . Define an equivalence relation: if a.e. Then .

To make sense of the definition, we need the following proposition:

Proposition: Let , then the following are equivalent:

  1. for all
  2. a.e.

Proof

References