Lebesgue spaces

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The Lebesgue spaces are function spaces defined by regularity of certain integrals. They play an important role in functional analysis and partial differential equations. We will use them to better understand the integration of real-valued functions, in contrast with the integration of functions taking values in $[0,\mathrm{\infty }]$ that we looked at in the last section. In this section we define the Lebesgue spaces and cover some of their basic properties, most notably that they are complete.

Integrating measurable functions

Given a measurable function $f:X\to \mathbb{R}$ we define $$f^{+}(x)=\{\begin{array}{ll}f(x)& f(x)\ge 0\\ 0& f(x)<0\end{array}$$ and $$f^{-}(x)=\{\begin{array}{ll}0& f(x)\ge 0\\ -f(x)& f(x)<0\end{array}$$ for all $x\in X$. Both $f^{+}$ and $f^{-}$ are measurable functions from $X$ to $[0,\mathrm{\infty })$ so their integrals are defined.

Definition

The integral of a measurable function $f:X\to \mathbb{R}$ is $$\int f\mathsf{d}\mu =\int f^{+}\mathsf{d}\mu -\int f^{-}\mathsf{d}\mu$$ whenever this makes sense i.e. when at least one of $f^{+}$ or $f^{-}$ does not have an integral with respect to $\mu$ of $\mathrm{\infty }$.

Note that the integral of $|f|$ is always defined, and is equal to $$\int f^{+}\mathsf{d}\mu +\int f^{-}\mathsf{d}\mu$$ but may be infinite.

It is fairly straight-forward using linearity of the integral for functions taking values in $[0,\mathrm{\infty }]$ to check the the integral is linear. The main result we want to prove about the integral of real-valued functions is the premier convergence theorem associated with the Lebesgue integral.

Theorem (Dominated convergence theorem)

Fix a measure space $(X,\mathcal{B},\mu )$ and a sequence $f_1,f_2,\dots$ of measurable functions from $X$ to $\mathbb{R}$ that converges pointwise. If there is a function $b:X\to [0,\mathrm{\infty }]$ with

1. ${\displaystyle \int b\mathsf{d}\mu <\mathrm{\infty }}$
2. $|f_n|\le b$ for all $n\in \mathbb{N}$

then $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu =\int \underset{n\to \mathrm{\infty }}{lim}{f}_n\mathsf{d}\mu$$ holds.

Proof:

Let $h$ be the pointwise limit of the sequence $n\mapsto f_n$. The sequences $$n\mapsto g+f_{n}n\mapsto g-f_n$$ take non-negative vales. Apply Fatou's lemma. ▮

A norm on functions

Fix $d\in \mathbb{N}$. We are familiar with the norm $$\Vert v\Vert =\sqrt{v_1^2+v_2^2+\cdots +v_d^2}$$ on ${\mathbb{R}}^d$. The quantity $\Vert v\Vert$ represents the Euclidean length of a vector. Interpreting $v\in {\mathbb{R}}^d$ as a function from $\{1,\dots ,d\}$ to $\mathbb{R}$ and writing $\mu$ for the counting measure on $\{1,\dots ,d\}$ we can rewrite this formula as $$\Vert v\Vert ={(\int |v{|}^2\mathsf{d}\mu )}^{1/2}$$ where $|v{|}^2$ is the function on $\{1,\dots ,d\}$ sending $i$ to $|v(i){|}^2$. We also have, more generally, the quantity $$\Vert v{\Vert }_p=(|v_1{|}^p+\cdots +|v_d{|}^p{)}^{1/p}$$ which is a norm for every $p\ge 1$.

Written this way, a generalization is immediately suggested. Fix a measure space $(X,\mathcal{B},\mu )$. Given a measurable function $f:X\to \mathbb{R}$ one could attempt to define a norm via $$\Vert f{\Vert }_p={(\int |f{|}^p\mathsf{d}\mu )}^{1/p}$$ but this only plausible if $|f{|}^p$ is integrable. Thus, in general we cannot use the above expression to define a norm on the space of all functions from $X$ to $\mathbb{R}$ that are measurable. To get around this problem we introduce the set $${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )=\{f:X\to \mathbb{R}\text{ measurable}:\int |f{|}^p\mathsf{d}\mu <\mathrm{\infty }\}$$ of measurable functions $f$ for which $|f{|}^p$ is integrable. We now have two questions to answer.

1. Is ${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$ a vector space?
2. Is ${\displaystyle f\mapsto {(\int |f{|}^p\mathsf{d}\mu )}^{1/p}}$ a norm on ${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$?

The answer to the first question is yes, although proving it will take some time.

Some inequalities

We run through the proofs of some fundamental inequalities.

Theorem (Young's inequality)

We have $$xy\le {\displaystyle \frac{x^p}{p}}+{\displaystyle \frac{y^q}{q}}$$ whenever $x,y\ge 0$ and $p,q>0$ with $\frac{1}{p}+\frac{1}{q}=1$.

Proof: This is a consequence of the convexity of the logarithm. Assume $x>0$ and $y>0$ as otherwise the inequality is true by direct verification. Then $$\log (\frac{1}{p}{x}^p+\frac{1}{q}{x}^q)\ge \frac{1}{p}\log (x^p)+\frac{1}{q}\log (y^q)=\log (xy)$$ because $\frac{1}{p}+\frac{1}{q}=1$.▮

Theorem (Hölder's inequality)

If $f,g$ belong to ${\mathcal{L}}^{\mathsf{1}}(X,\mathcal{B},\mu )$ then $$\int |fg|\mathsf{d}\mu \le {(\int |f{|}^p\mathsf{d}\mu )}^{1/p}{(\int |g{|}^q\mathsf{d}\mu )}^{1/q}$$ whenever $p,q>0$ and $\frac{1}{p}+\frac{1}{q}=1$.

Proof:

If $|f{|}^p$ or $|g{|}^q$ has infinite integral then there is nothing to prove. If either $|f{|}^p$ or $|g{|}^q$ has zero integral there is nothing to prove because then necessarily either $f=0$ almost-surely or $g=0$ almost surely respectively. We may therefore assume $\Vert f{\Vert }_p$ and $\Vert g{\Vert }_q$ are positive and finite. We can then calculate $$\begin{array}{rl}\int {\displaystyle \frac{|f(x)|}{\Vert f{\Vert }_p}}{\displaystyle \frac{|g(x)|}{\Vert g{\Vert }_q}}\mathsf{d}\mu (x)& \le \int {\displaystyle \frac{1}{p}}{\displaystyle \frac{|f(x){|}^p}{(\Vert f{\Vert }_p{)}^p}}+{\displaystyle \frac{1}{q}}{\displaystyle \frac{|g(x){|}^q}{(\Vert g{\Vert }_q{)}^q}}\mathsf{d}\mu (x)\\ & ={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1\end{array}$$ by applying Young's inequality for every $x$.▮

Theorem (Minkowski's inequality)

For every $p\ge 1$ we have $$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$$ whenever $f,g\in {\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$.

Proof:

If $p=1$ the results follows from the triangle inequality. Otherwise set $$q={\displaystyle \frac{p}{p-1}}$$ and note that $$|f+g{|}^p\le |f|\cdot |f+g{|}^{p-1}+|g|\cdot |f+g{|}^{p-1}$$ which gives $$\begin{array}{rl}& {(\Vert f+g{\Vert }_p)}^p\\ \le & \Vert f{\Vert }_p{(\int |f+g{|}^{q(p-1)}\mathsf{d}\mu )}^{1/q}+\Vert g{\Vert }_p{(\int |f+g{|}^{q(p-1)}\mathsf{d}\mu )}^{1/q}\\ =& (\Vert f{\Vert }_p+\Vert g{\Vert }_p){(\Vert f+g{\Vert }_p)}^{p/q}\end{array}$$ after applying Hölder's inequality. If $|f+g|$ is zero almost everywhere then there is again nothing to prove, so we may assume otherwise and divide through by $(\Vert f+g{\Vert }_p{)}^{p/q}$ to get $$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$$ because $pq-p=q$. ▮

For the moment, the main reason for going through these inequalities is to arrive at Minkowski's inequality, which is the crucial ingredient in verifying that ${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$ is a vector space and that $\Vert \cdot {\Vert }_p$ is a seminorm on ${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$. It is not generally a norm because there will in general be non-empty sets of measure zero. For instance, the functions $1_{\mathbb{Q}}$ and $1_{\{i\}}$ both have an integral of zero with respect to Lebesgue measure and therefore $\Vert 1_{\mathbb{Q}}-1_{\{\pi \}}{\Vert }_1=0$.

There is a general procedure for taking a seminorm on a vector space and creating a normed vector space. The set $${\mathcal{Z}}^{\mathsf{p}}(X,\mathcal{B},\mu )=\{f\in {\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu ):\Vert f{\Vert }_p=0\}$$ is a subspace of ${\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$ and $\Vert \cdot {\Vert }_p$ descends to a norm on the quotient $${\mathsf{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )={\mathcal{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )/{\mathcal{Z}}^{\mathsf{p}}(X,\mathcal{B},\mu )$$ whenever $p\ge 1$.

This is the definition of the ${\mathsf{L}}^{\mathsf{p}}$ or Lebesgue of $(X,\mathcal{B},\mu )$. We finish this section by verifying that Lebesgue spaces are always complete.

Theorem

For every measure space $(X,\mathcal{B},\mu )$ and every $p\ge 1$ the Lebesgue space ${\mathsf{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$ is complete as a normed vector space.

Proof:

Fix a Cauchy sequence $n\mapsto f_n$ in ${\mathsf{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$. We first choose indices $N(1)<N(2)<\cdots$ such that $$n,m\ge N(j)\Rightarrow \Vert f_n-f_m{\Vert }_p\le {\displaystyle \frac{1}{2^j}}$$ for all $j\in \mathbb{N}$. Next define $g_1=f_{N(1)}$ and $$g_{n+1}=f_{N(n+1)}-f_{N(n)}$$ for all $n\in \mathbb{N}$. Since $n\mapsto f_n$ is Cauchy and $$g_1+\cdots +g_n=f_{N(n)}$$ it suffices to prove that the series $n\mapsto g_n$ converges with respect to $\Vert \cdot {\Vert }_p$.

Define $$G_n=\sum _{i=1}^n|g_i|H=\sum _{i=1}^{\mathrm{\infty }}|g_i|$$ for all $n\in \mathbb{N}$ and note that $\Vert G_n{\Vert }_p\le \Vert f_1{\Vert }_p+1$. We can apply the monotone convergence theorem to get $${(\Vert H{\Vert }_p)}^p=\underset{n\in \mathrm{\infty }}{lim}{(\Vert G_n{\Vert }_p)}^p\le (\Vert f_1{\Vert }_p+1{)}^p$$ from which we conclude that $H$ belongs to ${\mathsf{L}}^{\mathsf{p}}(X,\mathcal{B},\mu )$. In particular, the series $n\mapsto G_n$ converges on a set of full measure. We conclude that the series $$n\mapsto \sum _{i=1}^n{g}_i$$ converges on a set of full measure and we can conclude by applying the dominated convergence theorem. ▮