# L1 Space

## Introduction

Let ${\displaystyle (X,{\mathcal {M}},\mu )}$ be a measure space. From our study of integration[1], we know that if ${\displaystyle f,g}$ are integrable functions, the following functions are also integrable:

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

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

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

In this section, we will construct ${\displaystyle L^{1}(\mu )}$. These are sometimes called Lebesgue spaces.

### Definition

Let ${\displaystyle L^{1}}$ denote the set of integrable functions on ${\displaystyle X}$, ie ${\displaystyle \int |f|<\infty }$. Define an equivalence relation: ${\displaystyle f\sim g}$ if ${\displaystyle f=g}$ a.e. Then ${\displaystyle L^{1}(\mu )=L^{1}/\sim }$. In some abuse of notation, we often refer to an element ${\displaystyle f\in L^{1}(\mu )}$ as a function, even though it really denotes the equivalence class of all functions in ${\displaystyle L^{1}}$ which are a.e. equivalent to ${\displaystyle f}$.

To see that ${\displaystyle \sim }$ is indeed an equivalence relation, reflexivity and symmetry are immediate. Transitivity when ${\displaystyle f\sim g}$ and ${\displaystyle g\sim h}$ follows by considering the null set where ${\displaystyle f}$ and ${\displaystyle g}$ differ and similarly for ${\displaystyle g}$ and ${\displaystyle h}$. Then see that the set where ${\displaystyle f}$ and ${\displaystyle h}$ differ is a subset of the union of the previous two null sets and hence is also a null set, so ${\displaystyle f\sim h}$.

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

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

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

### Proof[1]

${\displaystyle 2\implies 1:{\bigg \vert }\int _{E}f-\int _{E}g{\bigg \vert }\leq 1_{E}|f-g|\leq \int |f-g|=0}$

${\displaystyle 3\implies 2:}$ Since ${\displaystyle f=g}$ a.e., ${\displaystyle |f-g|=0}$ a.e. Take a simple function, ${\displaystyle \phi }$, such that ${\displaystyle 0\leq \phi \leq |f-g|}$, such ${\displaystyle \phi }$ must be ${\displaystyle 0}$ a.e. Therefore, ${\displaystyle \int |f-g|=\sup \left\{\int \phi :0\leq \phi \leq |f-g|,\phi {\text{ simple}}\right\}=0}$

${\displaystyle 1\implies 3:}$ Suppose the set ${\displaystyle \{x\in X:f(x)\neq g(x)\}}$ does not have measure zero. Then either ${\displaystyle E_{+}=\{x\in X:(f-g)_{+}(x)\neq 0\}}$ or ${\displaystyle E_{-}=\{x\in X:(f-g)_{-}(x)\neq 0\}}$ has nonzero measure, where ${\displaystyle (f-g)_{+}}$ denotes ${\displaystyle \max\{f-g,0\}}$ and ${\displaystyle (f-g)_{-}}$ denotes ${\displaystyle \max\{-(f-g),0\}}$. WLOG, assume ${\displaystyle E_{+}}$ has nonzero measure. Define the following sets ${\displaystyle E_{+,n}=\{x\in X:(f-g)_{+}>1/n\}}$, then from continuity from below, ${\displaystyle \mu (E_{+})=\mu \left(\cup _{i}^{\infty }E_{+,i}\right)=\lim _{i\to \infty }\mu (E_{+,i})>0}$. This shows that there exists some ${\displaystyle i\in \mathbb {N} }$ such that ${\displaystyle \mu (E_{+,i})>0}$, which implies that ${\displaystyle \int _{E}(f-g)\geq \int _{E}(f-g)_{+}\geq \int _{E_{+,i}}(f-g)_{+}\geq \mu (E_{+,i})/i>0}$, contradicting 1.

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

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

## Completeness of ${\displaystyle L^{1}}$ space

A space ${\displaystyle V}$ with a metric ${\displaystyle d}$ is said to be complete if for every Cauchy sequence ${\displaystyle {x_{k}}}$ in ${\displaystyle V}$ (that is, ${\displaystyle d(x_{l},x_{k})\to 0}$ as ${\displaystyle k,l\to \infty }$) there exist ${\displaystyle x\in V}$ such that ${\displaystyle x_{k}\to x}$ in the sense that

${\displaystyle d(x_{k},x)\to 0}$ as ${\displaystyle k\to \infty }$

### Riesz-Fischer Theorem

The vector space ${\displaystyle L^{1}}$ is complete in its metric induced by the ${\displaystyle p}$-norm. [2]

#### Proof

See Stein and Shakarchi

## References

1. Folland, Gerald B. (1999). Real Analysis: Modern Techniques and Their Applications, John Wiley and Sons, ISBN 0471317160, Second edition.
2. Elias M. Stein and Rami Shakarchi(2005), Real Analysis: measure theory, integration, & hilbert spaces, first edition