L1 Space

Very much not finished

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

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

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

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

With the proposition, we define our norm on $L^{1}(\mu )$ to be $\lVert f\rVert =\int |f|$. 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 norm defined, we can the metric to be $d(f,g)=\lVert f-g\rVert =\int |f-g|$. With a metric, one can talk about convergence in $L^{1}(\mu )$. This gives us a fourth mode of convergence for a sequence of functions. It is useful to compare these mode of convergence: Uniform Convergence $\implies$ Pointwise Convergence $\implies$ Pointwise a.e. Convergence

However, convergence in $L^{1}(\mu )$ does not imply pointwise a.e. convergence and vice versa. To see that, we look at the following examples: