The Banach Algebra of Borel Measures on Euclidean Space

This blog post is intended to deliver a quick explanation of the algebra of Borel measures on $\mathbb{R}^n$. It will be broken into pieces. All complex-valued complex Borel measures $M(\mathbb{R}^n)$ clearly form a vector space over $\mathbb{C}$. The main goal of this post is to show that this is a Banach space and also a Banach algebra.

In fact, the $\mathbb{R}^n$ case can be generalised into any locally compact abelian group (see any abstract harmonic analysis books), this is because what really matters here is being locally compact and abelian. But at this moment we stick to Euclidean spaces. Note since $\mathbb{R}^n$ is $\sigma$-compact, all Borel measures are regular.

To read this post you need to be familiar with some basic properties of Banach algebra, complex Borel measures, and the most important, Fubini’s theorem.

Banach space

The norm on $M(\mathbb{R}^n)$ is the total variation:

$$\lVert \mu \rVert = |\mu|(\mathbb{R}^n) = \sup \sum_{i=1}^{\infty}|\mu(E_i)|$$

the supremum being taken over all partitions $(E_i)$ of $\mathbb{R}^n$. The supremum on the right-hand side is finite because $\mu$ is assumed to be complex. This norm makes $M(\mathbb{R}^n)$ normed but we are interested in proving this space to be Banach.

Note each measure in $M(\mathbb{R})$ gives rise to a bounded complex functional

$$\begin{aligned} \Phi_\mu:C_0(\mathbb{R}^n) &\to \mathbb{C} \\ f &\to \int_{\mathbb{R}^n}fd\mu. \end{aligned}$$

Note we have $\vert \Phi_\mu(f)\vert = |\int f d\mu| \le \int |f|d|\mu| <\infty$. Indeed the norm of $\Phi_\mu$ is $\lVert \mu \rVert$.

Conversely, every bounded linear functional $\Phi$ gives rise to a regular Borel measure $\mu$ such that $\Phi(f)=\int fd\mu$ and $\lVert \Phi \rVert = \lVert \mu \rVert$, which is ensured by Riesz representation theorem. This is to say

$$C_0(\mathbb{R}^n)^\ast \cong M(\mathbb{R}^n)$$

in the sense of vector space isomorphism and homeomorphism (in fact, isometry). But it is well known that the dual space of a normed vector space is a Banach space, hence $M(\mathbb{R}^n)$ is Banach as is expected.

Convolution and Banach algebra

A vector space $V$ over a field $\mathbb{F}$ is called an algebra if there is an $\mathbb{F}$-bilinear form

$$B:V \times V \to V$$

. It is a Banach algebra if $V$ itself is Banach and the bilinear form is associative, i.e. $B(x,B(y,z)) = B(B(x,y),z)$ and

$$\lVert B(x,y) \rVert \le \lVert x \rVert \lVert y \rVert.$$

We show that $M(\mathbb{R}^n)$ is Banach by taking $B(\lambda,\mu)=\lambda \ast \mu$.

Convolution of two Borel measures a Borel measure

The convolution of measures is defined in the style of convolution of functions, in a natural sense. For any Borel set $E \subset \mathbb{R}^n$, we can consider the set restricted by addition:

$$E_2 = \{(x,y):x+y \in E\} \subset \mathbb{R}^{2n}.$$

Then we define the convolution of $\mu,\lambda \in M(\mathbb{R}^{2n})$ by product measure

$$(\mu \ast \lambda)(E) = (\mu \times \lambda)(E_2).$$

It looks natural but we need many routine verifications.

First, we need to show that $E_2$ is Borel. In fact, we have

$$\chi_{E_2}(x,y) = \chi_E(x+y).$$

Since $E$ is Borel, we see $\chi_E$ is Borel. Meanwhile $\varphi(x,y)=x+y$ is continuous hence Borel. Therefore $\chi_{E_2}$ is Borel as well. It follows that $E_2$ is a Borel set.

Next, is $\mu \ast \lambda$ an element of $M(\mathbb{R}^n)$? For any Borel set $E$, the value of $\mu \ast \lambda(E)$ is defined in $\mathbb{C}$, so we only need to verify that the definition of measure is satisfied. It shall be shown that

$$(\mu \ast \lambda)\left(\bigcup_{k=1}^{\infty}E^k\right)=\sum_{k=1}^{\infty}(\mu \ast \lambda)(E^k)$$

where $E^k$ are pairwise disjoint. Since the “measure” of $E$ is connected to $E_2$, we first show that if $E$ and $F$ are disjoint, then so are $E_2$ and $F_2$. Indeed, if $(x,y) \in E_2 \cap F_2$, then we have $x+y \in E \cap F$, and the set cannot be empty. Hence pairwise disjoint is preserved. Putting $E= \bigcup_{k=1}^{\infty}E^k$, we also need to show that $E_2 = \bigcup_{k=1}^{\infty}E_2^k$. If $x+y \in E$, then it lies in one of $E^k$, hence $(x,y) \in E_2 \implies (x,y) \in E_2^k$ for some $k$. It follows that $E_2 \subset \bigcup_{k=1}^{\infty}E_2^k$. Conversely, for $(x,y) \in \bigcup_{k=1}^{\infty}E_2^k$, we must have some $k$ such that $x+y \in E^k \subset E$, hence $(x,y) \in E_2$, which is to say that $\bigcup_{k=1}^{\infty}E_2^k \subset E_2$. Therefore

$$(\mu \ast \lambda)(E) = (\mu \times \lambda)(E_2) = (\mu \times \lambda)\left( \bigcup_{k=1}^{\infty}E_2^k\right) = \sum_{k=1}^{\infty}(\mu \times \lambda)(E_2^k) = \sum_{k=1}^{\infty}(\mu \ast \lambda)(E^k)$$

as is desired.

Properties of convolution resulting in a Banach algebra over the complex field

For any $f \in C_0(\mathbb{R}^n)$, we have a linear functional

$$\Phi:f \mapsto \iint f(x+y)d\mu(x)d\lambda(y) = \int fd(\mu \ast \lambda)$$

By Riesz representation theorem, there exists a unique measure $\nu$ such that $\Phi(f)=\int fd\nu$, it follows that $\nu = \mu \ast \lambda$ is uniquely determined. However, we have

$$\iint f(x+y)d\mu(x)d\lambda(y) = \iint f(x+y)d\lambda(x)d\mu(y)=\int fd(\lambda \ast \mu)$$

It follows that $\lambda \ast \mu = \nu = \mu \ast \lambda$. This convolution is commutative. Note for complex measures we always have $|\mu|(\mathbb{R}^n)<\infty$ so Fubini’s theorem is always valid.

Next we show that $\ast$ is associative. It can be carried out by Riesz’s theorem. Put $\nu_1 = \lambda \ast (\mu \ast \gamma)$ and $\nu_2 = (\lambda \ast \mu) \ast \gamma$. It follows that

$$\begin{aligned} \int fd\nu_1 &= \iint f(x+y)d\lambda(x)d(\mu \ast \gamma)(y) \\ &= \iiint f(x+y+z)d\lambda(x)d\mu(y)d\gamma(z) \\ &= \iint f(x+y+z)d\gamma(z)d\lambda(x)d\mu(y) \\ &= \iint f(x+y)d\gamma(x)d(\lambda \ast \mu)(y) \\ &= \int fd(\gamma \ast (\lambda \ast \mu)) \\ &= \int fd\nu_2. \end{aligned}$$

Hence $\nu_1 = \nu_2$, which delivers the associativity of the convolution. To show that $\ast$ makes $M(\mathbb{R}^n)$ a Banach space, we need to show the distribution law. This follows from the definition of product measure because

$$\mu \ast (\lambda_1 + \lambda_2)(E) = \int (\lambda_1 + \lambda_2)(E_{2}^{x})d\mu(x) = \int \lambda_1(E_2^x)d\mu(x) + \int \lambda_2(E_2^x)d\mu(x)$$

which is to say $\mu \ast \lambda_1 + \mu \ast \lambda_2 = \mu \ast (\lambda_1 + \lambda_2)$. Therefore $M(\mathbb{R}^n)$ is a complex algebra. It remains to show that $M(\mathbb{R}^n)$ is a Banach algebra. Let $E^1, E^2, \cdots$ be a partition of $\mathbb{R}^n$, we see

$$\begin{aligned} \sum_{k=1}^{\infty}|\mu \ast \lambda(E^k)| &= \sum_{k=1}^{\infty}|(\mu \times \lambda)(E^k_2)| \\ &= \sum_{k=1}^{\infty} \left|\iint \chi_{E_2^k}d\mu d\lambda\right| \\ &\leq \sum_{k=1}^{\infty} \iint \chi_{E_2^k}d|\mu|d|\lambda| \\ &\leq |\mu|(\mathbb{R}^n) \cdot |\lambda|(\mathbb{R}^n) \\ &\leq \lVert \mu \rVert \cdot \lVert \lambda \rVert. \end{aligned}$$

Hence $\lVert \mu \ast \lambda\rVert \le \lVert \mu \rVert \lVert \lambda \rVert$.

To conclude, $M(\mathbb{R}^n)$ is a commutative Banach algebra. Even better, this space has a unit which is customarily called the Dirac measure. Let $\delta$ be the measure determined by the evaluation functional $\Lambda:f \mapsto f(0)$. It follows that

$$\begin{aligned} \int f d(\delta \ast \mu) &= \iint f(x+y)d\delta(x)d\mu(y) \\ &= \int f(y)d\mu(y) \end{aligned}$$

Hence $\delta \ast \mu = \mu$ for all $\mu \in M(\mathbb{R}^n)$. Besides, $\delta$ has norm $1$ because it attains value $1$ at any Borel subset $E \subset \mathbb{R}^n$ containing the origin and value $0$ at any other Borel sets.

The subalgebra of discrete measures and subspace of (absolutely) continuous measures

A measure $\mu$ is said to be discrete if there is a countable set $E$ such that $\mu(A)=\mu(A \cap E)$ for all measurable sets $A$ (in general we say $\mu$ is concentrated on $E$). $\mu$ is said to be continuous if $\mu(A)=0$ whenever $A$ only contains a single point. We write $\mu \ll \lambda$, $\mu$ is absolutely continuous with respect to $\lambda$, if $\lambda(A)=0 \implies \mu(A)=0$.

We now play some games between continuous and discrete measures. First, we study the subspace of discrete measures $M_d(\mathbb{R}^n)$. For sum, things are quite straightforward. Suppose $\mu$ is concentrated on $A$ and $\lambda$ is concentrated on $B$, then

$$\begin{aligned} \mu(E) + \lambda(E) &= \mu(E \cap A) + \lambda(E \cap B) \\ &= \mu(E \cap (A \cap (A \cup B))) + \lambda(E \cap (B \cap (A \cup B))) \\ &= \mu(E \cap (A \cup B))+ \lambda(E \cap (A \cup B)). \end{aligned}$$

Hence $\mu+\lambda$ is concentrated on $A \cup B$.

For convolution, things are a little trickier. Suppose $\mu = \sum_{i=1}^{\infty}a_i\delta_{x_i}$, $\lambda=\sum_{i=1}^{\infty}b_i\delta_{y_i}$, where the $x_i$ and $y_i$ are distinct points, $\delta_x$ is the Dirac measure concentrated on $\{x\}$ (hence $\delta=\delta_0$), i.e. $\mu$ is concentrated on $A=\{x_i\}_{i=1}^{\infty}$ and $\lambda$ is concentrated on $\{y_i\}_{i=1}^{\infty}$, we see

$$\begin{aligned} (\mu \ast \lambda)(E) &= \iint \chi_E(x+y)d\mu(x)d\lambda(y) \\ &= \int \sum_{i=1}^{\infty}a_i\chi_E(x_i+y)d\lambda(y) \\ &= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_ib_j\chi_E(x_i+y_j) \\ &= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_ib_j\chi_{E \cap (A+B)}(x_i+y_j) \\ &= (\mu \ast \lambda)(E \cap (A+B)). \end{aligned}$$

Therefore $M_d(\mathbb{R}^n)$ forms a subalgebra of $M(\mathbb{R}^n)$.

Next, we focus on the subspace of continuous measures $M_c(\mathbb{R}^n)$. To begin with, we first consider the following identity:

$$\begin{aligned} (\mu \ast \lambda)(E) &= \iint \chi_E(x+y)d\mu(x)d\lambda(y) \\ &= \iint \chi_{E-y}(x)d\mu(x)d\lambda(y) \\ &= \int \mu(E-y)d\lambda(y). \end{aligned}$$

Suppose $\mu$ is continuous and $E$ is a singleton, then $E-y$ is still a singleton and hence $\mu(E-y)=0$ for all $y$, hence $(\mu \ast \lambda)(E)=0$, i.e. $\mu \ast \lambda$ is still continuous. Therefore the subspace of continuous measures actually forms an ideal.

Next, suppose $\mu \ll m$ and $m(E)=0$. We see

$$(\mu \ast \lambda)(E) = \int \mu(E-y)d\lambda(y) = 0$$

because $m(E)=0$ implies $m(E-y)=0$ for all $y$. Hence the subspace of absolutely continuous measures $M_{ac}(\mathbb{R}^n)$ also forms an ideal.

Finally, we consider the Radon-Nikodym derivatives (which exists (surjective) and is unique almost everywhere (injective)) of absolutely continuous measures. If

$$\mu(E) = \int_E fdm, \quad \lambda(E) = \int_E gd\mu,$$

then the coincide $\mu \ast \lambda$ coincide with $f \ast g$ in the following sense:

$$\begin{aligned} (\mu \ast \lambda)(E) &= \int_{\mathbb{R^n}} \mu(E-t)d\lambda(t) \\ &= \int_{\mathbb{R^n}}\left(\int_{E}f(x+t)dm(x) \right)g(t)dm(t) \\ &= \int_{\mathbb{R}^n}\int_E f(x+t)g(t)dm(x)d(t) \\ &= \int_E (f \ast g)dm \end{aligned}$$

In other words, we have $d(\mu \ast \lambda) = (f \ast g)dm$. Through this, we established an algebraic isomorphism $M_{ac}(\mathbb{R}^n) \cong L^1(\mathbb{R}^n,m)$.

The relation of $M(\mathbb{R}^n)$ and $L^1(\mathbb{R}^n,m)$

$L^1(\mathbb{R}^n,m)$ could’ve been a Banach algebra, but the unit is missing. However one can embed it into $M(\mathbb{R}^n)$ as a subspace of the subalgebra $M_{L^1}(\mathbb{R}^n)$ which contains all complex Borel measures $\mu$ satisfying

$$d\mu = fdm + \lambda d\delta, \quad \lambda \in \mathbb{C}.$$

Conversely, by the Lebesgue decomposition theorem, to every $\mu \in M(\mathbb{R}^n)$, we have a unique decomposition

$$\mu = \mu_a + \mu_s$$

where $\mu_a \ll m$ and $\mu_s \perp m$. With this being said we have a direct sum

$$M(\mathbb{R}^n) = L^1(\mathbb{R}^n,m) \oplus M_s(\mathbb{R}^n)$$

where $M_s(\mathbb{R}^n)$ is the subspace of complex measures singular to $m$. Informally speaking, the Gelfand transform on $L^1(\mathbb{R}^n,m)$ can be identified as the Fourier transform. Hence to study the Gelfand transform on $M(\mathbb{R}^n)$ it suffices to work on $M_s(\mathbb{R}^n)$. This shows the relation between $L^1$ and $C_0$.

The Group of invertible elements

\Let $G$ be the group of invertible elements of $M=M(\mathbb{R})$, and $G_1$ be the component of $G$ that contains $\delta$. $G_1$ is an open normal subgroup of $G$. Since $M$ is commutative, $G_1=\exp(M)$, and $G/G_1$ contains no nontrivial element of finite order. We will show that $G/G_1$ is actually uncountable. Pick $\alpha \in \mathbb{R}$, assume $\delta_\alpha \in G_1$, then $\delta_\alpha = \exp(\mu_\alpha)$ for some $\mu_\alpha \in M$. Performing Fourier transform on both sides gives

$$\int e^{-ixt}d\delta_\alpha = e^{-i\alpha t} = \int e^{-ixt}d\exp(\mu_\alpha)(x)=e^{\hat{\mu}_\alpha(t)}$$

Hence

$$-i\alpha t = \hat{\mu}_\alpha(t)+2k\pi{i}$$

Since $\mu_\alpha$ is bounded, so is $\hat{\mu}_\alpha(t)$. Hence $\alpha=0$. This is to say $\delta_\alpha \in G_1 \implies \alpha=0$. Next, consider any $\lambda{G_1} \in G/G_1$. If $\lambda=\delta_\alpha$ for some real $\alpha$, then $\delta_\alpha \in \lambda G_1$ is the only Dirac measure. If not, however, then $\lambda G_1$ contains no Dirac measures. Hence we have obtained an injective but not surjective map

$$\begin{aligned} \Lambda:\mathbb{R} &\to G/G_1, \\ \alpha &\mapsto \delta_\alpha G_1. \end{aligned}$$

This is to say, $G/G_1$ is uncountable.