How Right Translation Changes Haar Measure
Haar measure is invariant under translation on one side of a group, but a noncommutative group need not have a measure invariant on both sides. The factor by which right translation changes a left Haar measure is encoded by a positive function on the group. That function is the modular function; weighting left Haar measure by it gives a right Haar measure.
Throughout, \(G\) is a locally compact Hausdorff group, \(\mathcal{B}_G\) is its Borel sigma-algebra, and \(\mu\) is a nonzero left Haar measure on \((G,\mathcal{B}_G)\). Thus \(\mu\) is a regular Borel measure, finite on compact sets, and \(\mu(gE)=\mu(E)\) for \(g\in G\) and \(E\in\mathcal{B}_G\). No assumption that \(\mu\) is finite or sigma-finite is made. The Haar existence and uniqueness theorem, established earlier, says in particular that any two left Haar measures on \(G\) are positive scalar multiples of one another.
The reciprocal convention is important: some sources define the modular function as the factor \(c(x)\) itself. Here \(\Delta\) is the inverse of that factor. The convention will be fixed by the displayed scaling identity and used consistently below.
Homomorphism, Continuity, and Right Haar Measure
Proof. First, \(c(x)\) is positive and finite: \(\mu_x\) is a nonzero left Haar measure, so uniqueness makes it a positive finite scalar multiple of \(\mu\). For \(x,y\in G\), the scaling identity applied first to \(y\) and then to \(x\) gives, for every Borel \(E\),
Consequently \(c(xy)=c(x)c(y)\), and taking reciprocals gives \(\Delta(xy)=\Delta(x)\Delta(y)\). Also \(c(e)=1\), since \(Ee=E\), so \(\Delta(e)=1\). Thus \(\Delta\) is a homomorphism, including at the identity.
To prove continuity, choose a nonnegative continuous function \(f\) with compact support and with \(0<\int_G f\,d\mu<\infty\). Such a function exists for a nonzero Radon measure on a locally compact Hausdorff space. For fixed \(x\), the measure of a set under the map \(y\mapsto yx\) is governed by \(\mu(Ax^{-1})=c(x^{-1})\mu(A)=\Delta(x)\mu(A)\). It follows first for indicators, then for nonnegative simple functions, and finally for nonnegative Borel functions by monotone convergence that $$ \int_G f(yx)\,d\mu(y)=\Delta(x)\int_G f(y)\,d\mu(y). $$ The limit passage here is the Monotone Convergence Theorem applied to increasing simple approximations of \(f\) and their compositions with the translation map.
The left side is continuous in \(x\). Indeed, fix \(x_0\) and take a compact neighborhood \(V\) of \(x_0\). Whenever \(x\in V\), the integrand \(f(yx)\) can be nonzero only for \(y\) in the compact set \((\operatorname{supp}f)V^{-1}\). On this compact set, continuity of \((y,x)\mapsto f(yx)\) gives uniform convergence in \(y\) as \(x\to x_0\). The compact set has finite \(\mu\)-measure, so the integrals converge. Dividing the displayed identity by the fixed positive number \(\int_G f\,d\mu\) proves continuity of \(\Delta\).
Since \(\Delta\) is positive and continuous, it is bounded on each compact set. Hence the weighted measure \(\mu_R(E)=\int_E\Delta\,d\mu\) is locally finite and Radon. For a Borel set \(E\) and \(a\in G\), translation of the measure gives $$ \int_{Ea}\Delta(z)\,d\mu(z) =\Delta(a)^{-1}\int_E\Delta(ya)\,d\mu(y). $$ This identity follows for indicators from \(\mu(Aa)=\Delta(a)^{-1}\mu(A)\), then for nonnegative simple functions and by monotone convergence for nonnegative measurable functions. Since \(\Delta(ya)=\Delta(y)\Delta(a)\), the right side is \(\int_E\Delta(y)\,d\mu(y)\). Thus \(\mu_R(Ea)=\mu_R(E)\), proving right invariance. The measure is nonzero because its density \(\Delta\) is everywhere positive. Finally, uniqueness for right Haar measures follows from the same Haar uniqueness theorem with left and right translations interchanged, so any right Haar measure is a positive scalar multiple of \(\mu_R\). \(\square\)
Worked Example: An Abelian Group Has Trivial Modular Function
Let \(G=(\mathbb{R},+)\) with its Borel sigma-algebra and Lebesgue measure \(m\), a left Haar measure. For \(t\in\mathbb{R}\), right translation sends a Borel set \(E\) to \(E+t\). Lebesgue translation invariance gives \(m(E+t)=m(E)\), so the scaling identity yields \(\Delta(t)=1\) for every \(t\). The associated right Haar measure is therefore \(d\mu_R=dm\).
More generally, in any abelian locally compact group, \(Ex=xE\), and left invariance gives \(\mu(Ex)=\mu(E)\). The defining identity then forces \(\Delta(x)=1\). Check: on the interval \(E=[2,5]\subset\mathbb{R}\), translation by \(t=3\) gives \(E+t=[5,8]\); both intervals have Lebesgue measure \(3\), matching \(\Delta(3)^{-1}m(E)=1\cdot3=3\).
Worked Example: Counting Measure on a Discrete Group
Let \(G=S_3\), the six-element permutation group with the discrete topology, and let \(\mu\) be counting measure on its Borel sets (all subsets). Left and right multiplication by any fixed permutation are bijections of \(S_3\). Therefore they preserve the number of elements in every set, so \(\mu(Ea)=\mu(E)\) and \(\Delta(a)=1\) for every \(a\in S_3\). This applies even though \(S_3\) is nonabelian.
Check: take \(E=\{e,(12)\}\) and \(a=(123)\). Right multiplication gives \(Ea=\{(123),(12)(123)\}\). The two products are distinct because cancellation would otherwise imply \(e=(12)\); thus \(Ea\) has exactly two elements, as does \(E\). The defining identity gives \(2=\Delta(a)^{-1}\cdot2\), hence \(\Delta(a)=1\).
Worked Example: The Affine Group of the Line
Consider \(G=\{(a,b):a>0,\ b\in\mathbb{R}\}\), representing the affine map \(u\mapsto au+b\). Its group law is \((a,b)(r,s)=(ar,b+as)\). On this locally compact group, take the measure with density \(d\mu(a,b)=da\,db/a^2\) on the open half-plane \(a>0\). To verify left invariance, left multiplication by \((a_0,b_0)\) maps \((a,b)\) to \((a_0a,b_0+a_0b)\). Its Jacobian determinant is \(a_0^2\), while the new first coordinate squared is \(a_0^2a^2\). Thus the transformed density is \(a_0^2\,da\,db/(a_0^2a^2)=da\,db/a^2\), as required.
Now right multiplication by \((r,s)\), with \(r>0\), maps \((a,b)\) to \((ar,b+as)\). Its Jacobian determinant is \(r\). The image density is therefore $$ \frac{r\,da\,db}{(ar)^2} =\frac{1}{r}\frac{da\,db}{a^2}. $$ Consequently \(\mu(E(r,s))=r^{-1}\mu(E)\) for every Borel set \(E\), and the convention in this tutorial gives \(\Delta(r,s)=r\). The right Haar measure is \(d\mu_R(a,b)=\Delta(a,b)d\mu(a,b)=da\,db/a\).
Check: under the same right translation, the density \(da\,db/a\) transforms to \(r\,da\,db/(ar)=da\,db/a\). It is right invariant, in agreement with the theorem. Since \(\Delta((a,b)(r,s))=ar=\Delta(a,b)\Delta(r,s)\), this computation also directly verifies the homomorphism property in this group.
A Convention Pitfall and Why the Function Matters
A frequent error is to report the reciprocal scaling factor as the modular function while retaining the convention \(\mu(Ea)=\Delta(a)^{-1}\mu(E)\). In the affine example, the Jacobian calculation gives \(\mu(E(r,s))=r^{-1}\mu(E)\). A wrong calculation would identify that factor as \(\Delta(r,s)=r^{-1}\). Substitution into the defining identity would then predict \(\mu(E(r,s))=(r^{-1})^{-1}\mu(E)=r\mu(E)\), contradicting the Jacobian calculation unless \(r=1\). The correction is \(\Delta(r,s)=r\), whose reciprocal is exactly the measured scaling factor.
The modular function quantifies the obstruction to having one Haar measure invariant on both sides. If \(\Delta\equiv1\), the defining identity says that the left Haar measure is also right invariant; groups with this property are called unimodular. Abelian and discrete groups in the examples are unimodular, while the affine group has \(\Delta(a,b)=a\), which is not identically \(1\). The weighted right Haar measure records the same discrepancy in a way that is useful when integrating on the group. The precise inversion formula for Haar integrals is a separate matter.
Check Your Understanding
For each question, keep the convention \(\mu(Ea)=\Delta(a)^{-1}\mu(E)\) in view.
- Why is the measure \(E\mapsto\mu(Ex)\) a left Haar measure, and which uniqueness result makes its scaling factor independent of \(E\)?
- Use the scaling identity twice to show that \(\Delta(xy)=\Delta(x)\Delta(y)\).
- For \(G=(\mathbb{R},+)\), compute \(\Delta(t)\) from translation invariance and identify the resulting right Haar measure.
- For the affine group, if right translation scales \(da\,db/a^2\) by \(r^{-1}\), what is \(\Delta(r,s)\), and what is the density of the associated right Haar measure?
- Explain why counting measure on a discrete group is right invariant even when the group is nonabelian.