Representing measures

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Fix a compact metric space $X$. We equip $X$ with the Borel σ-algebra $\mathsf{Bor}(X)$. This is the σ-algebra generated by the open subsets of $X$. Continuous functions from $X$ to $\mathbb{R}$ are always $(\mathsf{Bor}(X),\mathsf{Bor}(\mathbb{R}))$ measurable.

Fix a measure $\mu$ on $(X,\mathsf{Bor}(X))$. As every continuous function from $X$ to $\mathbb{R}$ is bounded, if $\mu (X)<\mathrm{\infty }$ then every continuous function belongs to ${\mathsf{L}}^{\mathsf{1}}(X,\mu )$. We can therefore think of $\mu$ as a mapping from $\mathsf{C}(X)$ to $\mathbb{R}$.

We can interpret the above in terms of functional analysis. The space $\mathsf{C}(X)$ of continuous functions from $X$ to $\mathbb{R}$ is a normed vector space if we take $$||f||=sup\{|f(x)|:x\in X\}$$ and is in fact complete with respect to this norm. Given a measure $\mu$ on $(X,\mathsf{Bor}(X))$ with $\mu (X)<\mathrm{\infty }$ we can define $${\varphi }_{\mu }:\mathsf{C}(X)\to \mathbb{R}$$ by $${\varphi }_{\mu }(f)=\int f\mathsf{d}\mu$$ for all $f\in \mathsf{C}(X)$.

Proposition

Fix a compact metric space $X$ and a measure $\mu$ on $(X,\mathsf{Bor}(X))$ with $\mu (X)<\mathrm{\infty }$. The map ${\varphi }_{\mu }:\mathsf{C}(X)\to \mathbb{R}$ defined above is linear and bounded.

Proof:

We have to prove that ${\varphi }_{\mu }$ is linear and that there is $R>0$ such that $$|{\varphi }_{\mu }(f)|\le R||f||$$ for all $f\in \mathsf{C}(X)$. But linearity is immediate, because we have already proved that integration is linear on ${\mathsf{L}}^{\mathsf{1}}(X,\mu )$. Boundedness follows from $$|{\varphi }_{\mu }(f)|=|\int f\mathsf{d}\mu |\le \int |f|\mathsf{d}\mu \le \int ||f||\mathsf{d}\mu =||f||\mu (X)$$ if one takes $R=\mu (X)$.▮

Linear functionals

Definition

A linear functional on $\mathsf{C}(X)$ is any map $\mathsf{C}(X)\to \mathbb{R}$ that is bounded and linear. Write $\mathsf{C}(X{)}^{\ast }$ for the set of linear functionals on $\mathsf{C}(X)$.

It is straightforward to check that $\mathsf{C}(X{)}^{\ast }$ is itself a vector space if we define $$(a\varphi )(f)=a\cdot \varphi (f)(\varphi +\psi )(f)=\varphi (f)+\psi (f)$$ for all $f\in \mathsf{C}(X)$ and all $a\in \mathbb{R}$ and all $\varphi ,\psi \in \mathsf{C}(X{)}^{\ast }$. It is in fact a normed vector space if we define $$||\varphi ||=inf\{R>0:|\varphi (f)|\le R||f||\text{ for all }f\in \mathsf{C}(X)\}$$ for all $\varphi \in \mathsf{C}(X{)}^{\ast }$.

The previous proposition shows that Borel measures are sources of linear functionals. It is a major result that in fact every Borel measure arises from a special type of linear function. The special property is precisely positivity.

Definition

A linear map $\psi$ from $\mathsf{C}(X)$ to $\mathbb{R}$ is non-negative if $\psi (f)\ge 0$ whenever $f\ge 0$.

Our final major theorem about measures is that every positive linear functional on $\mathsf{C}(X)$ comes from a measure. The result is known as the Riesz representation theorem, as the version for $X$ a closed interval in $\mathbb{R}$ was first proved by Riesz. More general versions were proved by Markov and Kakutani.

Theorem

Let $X$ be a compact metric space. For every non-negative linear functional $\psi$ on ${\mathsf{C}}_{\mathsf{c}}(X)$ there is a measure $\mu$ on $(X,\mathsf{Bor}(X))$ such that $$\psi (f)=\int f\mathsf{d}\mu$$ for all $f\in {\mathsf{C}}_{\mathsf{c}}(X)$.

Proof:

We will not go through this proof. The idea is to approximate the indicator functions of open or closed subsets of $X$ by continuous functions, and to define an outer measure on $X$ via such approximations.▮

The non-compact case

We can relax the assumption that we are in a compact metric space if we change the space of functions with which we work. If $X$ is not compact then some continuous functions may have infinite integral and therefore it would not make same to think of the value of the integral as the output of a linear functional. To get around this one works with functions of comapct support.

Definition

Fix a topological space $X$. The support of a function $f:X\to \mathbb{R}$ is the closure of $\{x\in X:f(x)\ne 0\}$. A function $f:X\to \mathbb{R}$ has compact support if its support is compact as a subset of $X$. Write ${\mathsf{C}}_{\mathsf{c}}(X)$ for the set of continuous functions from $X$ to $\mathbb{R}$ that have compact support.

In this case also the uniform norm $||\cdot |{|}_{\mathsf{u}}$ makes ${\mathsf{C}}_{\mathsf{c}}(X)$ into a normed vector space but when $X$ is not compact it will likely not be complete. The closure of ${\mathsf{C}}_{\mathsf{c}}(X)$ with respect to the uniform norm is, when $X$ is not too wild, the space ${\mathsf{C}}_{\mathsf{0}}(X)$ of continuous functions from $X$ to $\mathbb{R}$ that have the property $$\underset{x\to \mathrm{\infty }}{lim}f(x)=0$$ where one assesses $x\to \mathrm{\infty }$ in terms of leaving compact subsets of $X$. Thus the definition of the above limit is that, for every $ϵ>0$ there is a compact set $K\subset X$ with the property that $|f(x)|<ϵ$ for all $x\in X\setminus K$.

By "not too wild" we mean that $X$ is Hausdorff and that for every point $x\in X$ there is an open set $x\in U\subset X$ whose closure is compact. Such topological spaces are known as LCH or locally compact Hausdorff spaces. This is the context for the following general version of the Riesz representation theorem, due to Kakutani. Its proof is beyond the scope of the course.

Theorem

Fix a locally compact Hausdorff space $X$. For every positive linear functional $\psi$ on ${\mathsf{C}}_{\mathsf{0}}(X)$ there is a Borel measure $\mu$ on $X$ such that $$\int f\mathsf{d}\mu =\psi (f)$$ for all $f\in {\mathsf{C}}_{\mathsf{c}}(X)$.