Product Groups and Product Haar Measures

Why Product Groups Have Product Haar Measure

When two locally compact groups are combined, their product group has a natural topology and coordinatewise group operation. The product of their Haar measures is not merely a convenient way to compute integrals: it is itself invariant under left translation on the product group. This gives a consistent measure for studying functions that depend on both coordinates, and makes iterated integration available when its hypotheses hold.

Let \(G\) and \(H\) be locally compact Hausdorff groups, with left Haar measures \(\mu\) and \(\nu\), respectively. A left Haar measure is a nonzero Radon measure that is finite on compact sets and invariant under left translation. The product group \(G\times H\) has coordinatewise multiplication, \((a,b)(x,y)=(ax,by)\), and the product topology. Haar measures need not be finite or sigma-finite on arbitrary locally compact groups, so the general construction is stated using Radon measures.

For sigma-finite measures, the usual product measure theorem also gives the product measure on the product sigma-algebra, with the rectangle formula \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\) for measurable \(A\subseteq G\) and \(B\subseteq H\). In particular, this applies when \(G,H\) are second-countable locally compact groups: they are sigma-compact, and Haar measure is sigma-finite. In that setting the Borel sigma-algebra of the product is \(\mathcal{B}_G\otimes\mathcal{B}_H\). For nonnegative measurable functions, Tonelli’s theorem permits the two orders of integration and allows the value \(+\infty\).

The Product Haar Measure Theorem

Proof. By the Radon product measure theorem, \(\mu\times\nu\) is a Radon measure on \(G\times H\). It remains to show that it is nonzero and left invariant. Since \(\mu\) and \(\nu\) are nonzero Radon measures, there exist nonnegative continuous compactly supported functions \(u\) on \(G\) and \(v\) on \(H\) with \(\int_G u\,d\mu>0\) and \(\int_H v\,d\nu>0\). For \(F(x,y)=u(x)v(y)\), the product integral formula gives \(\int_{G\times H}F\,d(\mu\times\nu)=(\int_Gu\,d\mu)(\int_Hv\,d\nu)>0\), so \(\mu\times\nu\) is nonzero.

Fix \((a,b)\in G\times H\) and a real-valued \(F\in C_c(G\times H)\), where \(C_c\) denotes the continuous compactly supported functions. The translate \(F_{a,b}(x,y)=F(ax,by)\) is also continuous and compactly supported. The product integral formula applies to both functions. For each fixed \(x\), left invariance of \(\nu\) gives \(\int_H F(ax,by)\,d\nu(y)=\int_H F(ax,y)\,d\nu(y)\). The function \(x\mapsto\int_H F(ax,y)\,d\nu(y)\) is integrable with respect to \(\mu\): its support lies in the compact projection onto \(G\) of the support of \(F_{a,b}\), and its absolute value is bounded by a finite constant on that compact set. Left invariance of \(\mu\) therefore gives \(\int_G\int_H F(ax,by)\,d\nu(y)\,d\mu(x)=\int_G\int_H F(x,y)\,d\nu(y)\,d\mu(x)\). Thus the translate and original measure have equal integrals against every function in \(C_c(G\times H)\). Uniqueness of Radon measures from their integrals against \(C_c\) functions implies that left translation preserves \(\mu\times\nu\). The product measure is nonzero, Radon, and left invariant, so it is a left Haar measure. \(\square\)

The theorem identifies a Haar measure on the product group; it does not impose a normalization. If \(\mu\) is replaced by \(c\mu\) and \(\nu\) by \(d\nu\), where \(c,d>0\), their product is replaced by \(cd(\mu\times\nu)\). When the factors are compact and their Haar measures are normalized to have total mass one, the product measure also has total mass one. For noncompact groups a Haar measure generally has infinite total mass and cannot be normalized to be a probability measure.

Worked Product Integrals

What Product Integration Does—and Does Not—Say

Tonelli’s theorem applies to nonnegative measurable functions on a product of sigma-finite measure spaces and equates the product integral with either iterated integral, allowing \(+\infty\). Fubini’s theorem applies to integrable functions (in particular, when the integral of the absolute value is finite) and gives finite iterated integrals equal to the product integral. Neither theorem says that the integral of a general function is the product of two one-variable integrals. That factorization is justified when the integrand itself separates as \(f(x,y)=g(x)h(y)\) and the relevant integrals satisfy the applicable hypotheses.

Common mistake. On \([0,1]^2\) with Lebesgue product measure, someone might replace \(1+xy\) by the separated product \((1+x)(1+y)\), and then report \((\int_0^1(1+x)\,dx)^2=\tfrac94\). This changes the function: \((1+x)(1+y)=1+x+y+xy\), not \(1+xy\). The correct integral is \(\int_0^1\int_0^1(1+xy)\,dy\,dx=\int_0^1(1+\tfrac{x}{2})\,dx=1+\tfrac14=\tfrac54\). As a check, integrating first in \(x\) gives \(\int_0^1(1+\tfrac{y}{2})\,dy=\tfrac54\) as well. The product-integral theorem permits iterated integration; it does not permit an unsupported factorization.

The product construction is useful whenever a group action or a function naturally has separate coordinates: lattice sums become double sums, Euclidean volume integrals can be computed by sections, and mixed continuous-discrete models fit the same framework. The hypotheses remain important. In the displayed examples, the measures are sigma-finite and the integrands are nonnegative, so Tonelli applies directly. For signed or complex-valued integrands, one must instead verify integrability before invoking Fubini.