L1 Space

Introduction

Let $(X,{\mathcal {M}},\mu )$ be a measure space. From our study of integration, we know that if $f,g$ are integrable functions, the following functions are also integrable:

1. $f+g$
2. $cf$, for $c\in \mathbb {R}$

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 $\mathbb {R}$, 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 $f=0$ almost everywhere, then $\int f=\int 0=0$. In other words, there are non zero functions which has a zero integral. This motivates our definition of $L^{1}(\mu )$ to be the set of integrable functions up to equivalence to sets of measure zero.

$L^{1}(\mu )$ Space

In this section, we will construct $L^{1}(\mu )$.

Definition

Let $L^{1}$ denote the set of integrable functions on $X$, ie $\int |f|<\infty$. Define an equivalence relation: $f\sim g$ if $f=g$ a.e. Then $L^{1}(\mu )=L^{1}/\sim$.

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

Proposition: Let $f,g\in L^{1}$, then the following are equivalent:

1. $\int _{E}f=\int _{E}g,$ for all $E\in {\mathcal {M}}$
2. $\int |f-g|=0$
3. $f=g$ a.e.

Proof

(to be filled in)

With the proposition, we define our norm on $L^{1}(\mu )$ to be $\lVert f\rVert =\int |f-g|$. This is indeed a norm since:

1. $\int |f+g|\leq \int |f|+\int |g|$
2. $\int |cf|=c\int |c||f|,c\in \mathbb {R}$
3. $\int |f|=0\iff f=0$ a.e

   Convergence in $L^{1}(\mu )$

   With our notion of norm defined, we can have the notion of a metric, that is $d(f,g)=\lVert f-g\rVert =\int |f-g|$.

   References