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 Rn. It will be broken into pieces. All complex-valued complex Borel measures M(Rn) clearly form a vector space over C. The main goal of this post is to show that this is a Banach space and also a Banach algebra.

In fact the Rn 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 Rn is σ-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(Rn) is the total variation: ‖μ‖=|μ|(Rn)=sup∞∑i=1|μ(Ei)| the supremum being taken over all partitions (Ei) of Rn. The supremum on the right hand side is finite because μ is assumed to be complex. This norm makes M(Rn) normed but we are interested in proving this space to be Banach.

Note each measure in M(R) gives rise to a bounded complex functional Φμ:C0(Rn)→Cf→∫Rnfdμ. Note we have |Φμ(f)|=|∫fdμ|≤∫|f|d|μ|<∞. Indeed the norm of Φμ is ‖μ‖.

Conversely, every bounded linear functional Φ gives rise to a regular Borel measure μ such that Φ(f)=∫fdμ and ‖Φ‖=‖μ‖, which is ensured by Riesz representation theorem. This is to say C0(Rn)∗≅M(Rn) 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(Rn) is Banach as is expected.

Convolution and Banach algebra

A vector space V over a field F is called an algebra if there is an F-bilinear form B:V×V→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 ‖B(x,y)‖≤‖x‖‖y‖. We show that M(Rn) is Banach by taking B(λ,μ)=λ∗μ.

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⊂Rn, we can consider the set restricted by addition: E2={(x,y):x+y∈E}⊂R2n. Then we define the convolution of μ,λ∈M(R2n) by product measure (μ∗λ)(E)=(μ×λ)(E2). It looks natural but we need many routine verification.

First we need to show that E2 is Borel. In fact we have χE2(x,y)=χE(x+y). Since E is Borel, we see χE is Borel. Meanwhile φ(x,y)=x+y is continuous hence Borel. Therefore χE2 is Borel as well. It follows that E2 is a Borel set.

Next, is μ∗λ an element of M(Rn)? For any Borel set E, the value of μ∗λ(E) is defined in C, so we only need to verify that the definition of measure is satisfied. It shall be shown that (μ∗λ)(∞⋃k=1Ek)=∞∑k=1(μ∗λ)(Ek) where Ek are pairwise disjoint. Since the "measure" of E is connected to E2, we first show that if E and F are disjoint, then so are E2 and F2. Indeed, if (x,y)∈E2∩F2, then we have x+y∈E∩F, and the set cannot be empty. Hence pairwise disjoint is preserved. Putting E=⋃∞k=1Ek, we also need to show that E2=⋃∞k=1Ek2. If x+y∈E, then it lies in one of Ek, hence (x,y)∈E2⟹(x,y)∈Ek2 for some k. It follows that E2⊂⋃∞k=1Ek2. Conversely, for (x,y)∈⋃∞k=1Ek2, we must have some k such that x+y∈Ek⊂E, hence (x,y)∈E2, which is to say that ⋃∞k=1Ek2⊂E2. Therefore (μ∗λ)(E)=(μ×λ)(E2)=(μ×λ)(∞⋃k=1Ek2)=∞∑k=1(μ×λ)(Ek2)=∞∑k=1(μ∗λ)(Ek) as is desired.

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

For any f∈C0(Rn), we have a linear functional Φ:f↦∬f(x+y)dμ(x)dλ(y)=∫fd(μ∗λ) By Riesz representation theorem, there exists a unique measure ν such that Φ(f)=∫fdν, it follows that ν=μ∗λ is uniquely determined. However we have ∬f(x+y)dμ(x)dλ(y)=∬f(x+y)dλ(x)dμ(y)=∫fd(λ∗μ) It follows that λ∗μ=ν=μ∗λ. This convolution is commutative. Note for complex measures we always have |μ|(Rn)<∞ so Fubini's theorem is always valid.

Next we show that ∗ is associative. It can be carried out by Riesz's theorem. Put ν1=λ∗(μ∗γ) and ν2=(λ∗μ)∗γ. It follows that ∫fdν1=∬f(x+y)dλ(x)d(μ∗γ)(y)=∭f(x+y+z)dλ(x)dμ(y)dγ(z)=∬f(x+y+z)dγ(z)dλ(x)dμ(y)=∬f(x+y)dγ(x)d(λ∗μ)(y)=∫fd(γ∗(λ∗μ))=∫fdν2. Hence ν1=ν2, which delivers the associativity of the convolution. To show that ∗ makes M(Rn) a Banach space, we need to show the distribution law. This follows from the definition of product measure because μ∗(λ1+λ2)(E)=∫(λ1+λ2)(Ex2)dμ(x)=∫λ1(Ex2)dμ(x)+∫λ2(Ex2)dμ(x) which is to say μ∗λ1+μ∗λ2=μ∗(λ1+λ2). Therefore M(Rn) is a complex algebra. It remains to show that M(Rn) is a Banach algebra. Let E1,E2,⋯ be a partition of Rn, we see ∞∑k=1|μ∗λ(Ek)|=∞∑k=1|(μ×λ)(Ek2)|=∞∑k=1|∬χEk2dμdλ|≤∞∑k=1∬χEk2d|μ|d|λ|≤|μ|(Rn)⋅|λ|(Rn)≤‖μ‖⋅‖λ‖. Hence ‖μ∗λ‖≤‖μ‖‖λ‖.

To conclude, M(Rn) is a commutative Banach algebra. Even better, this space has a unit which is customarily called the Dirac measure. Let δ be the measure determined by the evaluation functional Λ:f↦f(0). It follows that ∫fd(δ∗μ)=∬f(x+y)dδ(x)dμ(y)=∫f(y)dμ(y) Hence δ∗μ=μ for all μ∈M(Rn). Besides, δ has norm 1 because it attains value 1 at any Borel subset E⊂Rn containing the origin and value 0 at any other Borel sets.

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

A measure μ is said to be discrete if there is a countable set E such that μ(A)=μ(A∩E) for all measurable sets A (in general we say μ is concentrated on E). μ is said to be continuous if μ(A)=0 whenever A only contains a single point. We write μ≪λ, μ is absolutely continuous with respect to λ, if λ(A)=0⟹μ(A)=0.

We now play some games between continuous and discrete measures. First we study the subspace of discrete measures Md(Rn). For sums things are quite straightforward. Suppose μ is concentrated on A and λ is concentrated on B, then μ(E)+λ(E)=μ(E∩A)+λ(E∩B)=μ(E∩(A∩(A∪B)))+λ(E∩(B∩(A∪B)))=μ(E∩(A∪B))+λ(E∩(A∪B)). Hence μ+λ is concentrated on A∪B.

For convolution things are a little trickier. Suppose μ=∑∞i=1aiδxi, λ=∑∞i=1biδyi, where the xi and yi are distinct points, δx is the Dirac measure concentrated on {x} (hence δ=δ0), i.e. μ is concentrated on A={xi}∞i=1 and λ is concentrated on {yi}∞i=1, we see (μ∗λ)(E)=∬χE(x+y)dμ(x)dλ(y)=∫∞∑i=1aiχE(xi+y)dλ(y)=∞∑j=1∞∑i=1aibjχE(xi+yj)=∞∑j=1∞∑i=1aibjχE∩(A+B)(xi+yj)=(μ∗λ)(E∩(A+B)). Therefore Md(Rn) forms a subalgebra of M(Rn).

Next we focus on the subspace of continuous measures Mc(Rn). To begin with we first consider the following identity: (μ∗λ)(E)=∬χE(x+y)dμ(x)dλ(y)=∬χE−y(x)dμ(x)dλ(y)=∫μ(E−y)dλ(y). Suppose μ is continuous and E is a singleton, then E−y is still a singleton and hence μ(E−y)=0 for all y, hence (μ∗λ)(E)=0, i.e. μ∗λ is still continuous. Therefore the subspace of continuous measures actually forms an ideal.

Next suppose μ≪m and m(E)=0. We see (μ∗λ)(E)=∫μ(E−y)dλ(y)=0 because m(E)=0 implies m(E−y)=0 for all y. Hence the subspace of absolutely continuous measures Mac(Rn) 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 μ(E)=∫Efdm,λ(E)=∫Egdμ, then the coincide μ∗λ coincide with f∗g in the following sense: (μ∗λ)(E)=∫Rnμ(E−t)dλ(t)=∫Rn(∫Ef(x+t)dm(x))g(t)dm(t)=∫Rn∫Ef(x+t)g(t)dm(x)d(t)=∫E(f∗g)dm In other words we have d(μ∗λ)=(f∗g)dm. Through this we established an algebraic isomorphism Mac(Rn)≅L1(Rn,m).

The relation of M(Rn) and L1(Rn,m)

L1(Rn,m) could've been a Banach algebra, but the unit is missing. However one can embed it into M(Rn) as a subspace of the subalgebra ML1(Rn) which contains all complex Borel measures μ satisfying dμ=fdm+λdδ,λ∈C. Conversely, by Lebesgue decomposition theorem, to every μ∈M(Rn), we have a unique decomposition μ=μa+μs where μa≪m and μs⊥m. With this being said we have a direct sum M(Rn)=L1(Rn,m)⊕Ms(Rn) where Ms(Rn) is the subspace of complex measures singular to m. Informally speaking, the Gelfand transform on L1(Rn,m) can be identified as the Fourier transform. Hence to study the Gelfand transform on M(Rn) it suffices to work on Ms(Rn). This shows the relation between L1 and C0.

The Group of invertible elements

G be the group of invertible elements of M=M(R), and G1 be the component of G that contains δ. G1 is an open normal subgroup of G. Since M is commutative, G1=exp(M), and G/G1 contains no nontrivial element of finite order. We will show that G/G1 is actually uncountable. Pick α∈R, assume δα∈G1, then δα=exp(μα) for some μα∈M. Performing Fourier transform on both sides gives ∫e−ixtdδα=e−iαt=∫e−ixtdexp(μα)(x)=eˆμα(t) Hence −iαt=ˆμα(t)+2kπi Since μα is bounded, so is ˆμα(t). Hence α=0. This is to say δα∈G1⟹α=0. Next consider any λG1∈G/G1. If λ=δα for some real α, then δα∈λG1 is the only Dirac measure. If not however, then λG1 contains no Dirac measures. Hence we have obtained an injective but not surjective map Λ:R→G/G1,α↦δαG1. This is to say, G/G1 is uncountable.