This blog post is intended to deliver a quick explanation of thealgebra of Borel measures on \(\mathbb{R}^n\). It will be broken intopieces. All complex-valued complex Borel measures \(M(\mathbb{R}^n)\) clearly form a vectorspace over \(\mathbb{C}\). The maingoal of this post is to show that this is a Banach space and also aBanach algebra.

In fact, the \(\mathbb{R}^n\) casecan be generalised into any locally compact abelian group (see anyabstract harmonic analysis books), this is because what really mattershere is being locally compact and abelian. But at this moment we stickto Euclidean spaces. Note since \(\mathbb{R}^n\) is \(\sigma\)-compact, all Borel measures areregular.

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

Banach space

The norm on \(M(\mathbb{R}^n)\) isthe 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 theright-hand side is finite because \(\mu\) is assumed to be complex. This normmakes \(M(\mathbb{R}^n)\) normed but weare interested in proving this space to be Banach.

Note each measure in \(M(\mathbb{R})\) gives rise to a boundedcomplex 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 Borelmeasure \(\mu\) 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 (infact, isometry). But it is well known that the dual space of a normedvector space is a Banach space, hence \(M(\mathbb{R}^n)\) is Banach as isexpected.

Convolution and Banachalgebra

A vector space \(V\) over a field\(\mathbb{F}\) is called an algebra ifthere is an \(\mathbb{F}\)-bilinearform \[B:V \times V \to V\]

. 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 \(B(\lambda,\mu)=\lambda \ast \mu\).

Convolutionof two Borel measures a Borel measure

The convolution of measures is defined in the style of convolution offunctions, in a natural sense. For any Borel set \(E \subset \mathbb{R}^n\), we can considerthe 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 henceBorel. Therefore \(\chi_{E_2}\) isBorel as well. It follows that \(E_2\)is a Borel set.

Next, is \(\mu \ast \lambda\) anelement of \(M(\mathbb{R}^n)\)? For anyBorel set \(E\), the value of \(\mu \ast \lambda(E)\) is defined in \(\mathbb{C}\), so we only need to verifythat 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 pairwisedisjoint. Since the "measure" of \(E\)is connected to \(E_2\), we first showthat 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 setcannot be empty. Hence pairwise disjoint is preserved. Putting \(E= \bigcup_{k=1}^{\infty}E^k\), we alsoneed 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.

Propertiesof 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 uniquelydetermined. However, we have \[\iint f(x+y)d\mu(x)d\lambda(y) = \iint f(x+y)d\lambda(x)d\mu(y)=\intfd(\lambda \ast \mu)\] It follows that \(\lambda \ast \mu =\nu = \mu \ast \lambda\). This convolution is commutative. Notefor complex measures we always have \(|\mu|(\mathbb{R}^n)<\infty\) so Fubini'stheorem is always valid.

Next we show that \(\ast\) isassociative. 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 needto show the distribution law. This follows from the definition ofproduct 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 acomplex 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, \(M(\mathbb{R}^n)\) isa commutative Banach algebra. Even better, this space has a unit whichis customarily called the Dirac measure. Let \(\delta\) be the measure determined by theevaluation functional \(\Lambda:f \mapstof(0)\). 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 \(\mu \inM(\mathbb{R}^n)\). Besides, \(\delta\) has norm \(1\) because it attains value \(1\) at any Borel subset \(E \subset \mathbb{R}^n\) containing theorigin and value \(0\) at any otherBorel sets.

Thesubalgebra of discrete measures and subspace of (absolutely) continuousmeasures

A measure \(\mu\) is said to bediscrete if there is a countable set \(E\) such that \(\mu(A)=\mu(A \cap E)\) for all measurablesets \(A\) (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 continuouswith 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 arequite 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 \(\mu+\lambda\) isconcentrated 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 measureconcentrated 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, wefirst 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 continuousand \(E\) 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 ofcontinuous 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 absolutelycontinuous measures \(M_{ac}(\mathbb{R}^n)\) also forms anideal.

Finally, we consider the Radon-Nikodym derivatives (which exists(surjective) and is unique almost everywhere (injective)) of absolutelycontinuous measures. If \[\mu(E) = \int_E fdm, \quad \lambda(E) = \int_E gd\mu,\] then the coincide \(\mu \ast\lambda\) coincide with \(f \astg\) in the following sense: \[\begin{aligned}(\mu \ast \lambda)(E) &= \int_{\mathbb{R^n}} \mu(E-t)d\lambda(t) \\ &=\int_{\mathbb{R^n}}\left(\int_{E}f(x+t)dm(x) \right)g(t)dm(t) \\ &= \int_{\mathbb{R}^n}\int_Ef(x+t)g(t)dm(x)d(t) \\ &= \int_E (f \ast g)dm\end{aligned}\] In other words, we have \(d(\mu \ast\lambda) = (f \ast g)dm\). Through this, we established analgebraic 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'vebeen a Banach algebra, but the unit is missing. However one can embed itinto \(M(\mathbb{R}^n)\) as a subspaceof the subalgebra \(M_{L^1}(\mathbb{R}^n)\) which contains allcomplex 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 aunique decomposition \[\mu = \mu_a + \mu_s\] where \(\mu_a \ll m\) and\(\mu_s \perp m\). With this being saidwe 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 Gelfandtransform on \(L^1(\mathbb{R}^n,m)\)can be identified as the Fourier transform. Hence to study the Gelfandtransform on \(M(\mathbb{R}^n)\) itsuffices to work on \(M_s(\mathbb{R}^n)\). This shows therelation between \(L^1\) and \(C_0\).

The Group of invertibleelements

\(G\) be the group of invertibleelements 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 offinite 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\(\mu_\alpha \in M\). PerformingFourier transform on both sides gives \[\int e^{-ixt}d\delta_\alpha = e^{-i\alpha t} = \inte^{-ixt}d\exp(\mu_\alpha)(x)=e^{\hat{\mu}_\alpha(t)}\] Hence \[-i\alpha t = \hat{\mu}_\alpha(t)+2k\pi{i}\] Since \(\mu_\alpha\) isbounded, 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 onlyDirac measure. If not, however, then \(\lambdaG_1\) contains no Dirac measures. Hence we have obtained aninjective 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, \(G/G_1\) isuncountable.