Three Changes of Variables on a Group
On \(\mathbb{R}^n\), translations and reflection preserve Lebesgue measure. On a general locally compact group, left translation preserves a left Haar measure, but right translation and inversion need not. Their integral formulas involve the modular function. Keeping track of its convention is essential: using the reciprocal convention without changing the formulas reverses the factor.
Throughout, let \(G\) be a second-countable locally compact Hausdorff group, and let \(\mu\) be a nonzero left Haar measure on its Borel sigma-algebra \(\mathcal{B}_G\). Thus \(\mu\) is a regular Borel measure, finite on compact sets, and left invariant. Under the second-countability assumption it is sigma-finite. It need not be finite. The formulas below hold for nonnegative Borel functions \(f:G\to[0,\infty]\), with infinite integrals allowed, and for complex-valued Borel functions integrable with respect to \(\mu\). Equalities are equalities of integrals, not assertions about functions holding almost everywhere.
The modular-function properties and the relation between left and right Haar measures are the standard results on right Haar measure. We derive the three integral substitutions from these measure identities. In particular, the convention above makes the factor for right translation \(\Delta(y)^{-1}\), while the inversion integrand has factor \(\Delta(x)^{-1}\).
The Change-of-Variables Formulas
Proof. Fix \(y\in G\). For left translation, the preimage of a Borel set \(A\) under \(x\mapsto yx\) is \(y^{-1}A\). Left invariance gives \(\mu(y^{-1}A)=\mu(A)\). Therefore the measure obtained by transporting \(\mu\) under this map agrees with \(\mu\) on Borel sets. Integrating nonnegative simple functions first, and then using their increasing approximation to any nonnegative Borel function and the Monotone Convergence Theorem, gives the left-translation formula.
For the map \(x\mapsto xy\), the preimage of \(A\) is \(Ay^{-1}\). By the modular-function identity and the homomorphism property, $$ \mu(Ay^{-1})=\Delta(y^{-1})\mu(A)=\Delta(y)^{-1}\mu(A). $$ Thus the transported measure is \(\Delta(y)^{-1}\mu\). The same simple-function argument followed by the Monotone Convergence Theorem gives the right-translation formula.
For inversion \(x\mapsto x^{-1}\), the preimage of \(A\) is \(A^{-1}\), so the transported measure is \(\mu_R\). The stated relation between right and left Haar measure gives \(\mu_R(A)=\int_A\Delta(x)^{-1}\,d\mu(x)\). Integrating simple functions and then applying the Monotone Convergence Theorem proves the inversion formula. Finally, for complex integrable \(f\), apply the nonnegative formulas to \(|f|\) to see that each transformed function is integrable. Applying the formulas to the positive and negative parts of the real and imaginary parts proves the complex-valued identities. \(\square\)
The formulas apply to compositions as well. For example, first applying the left-translation identity and then the right-translation identity gives \(\int_G f(yxb)\,d\mu(x)=\Delta(b)^{-1}\int_G f\,d\mu\) for nonnegative Borel \(f\). The left factor contributes no modular term; the right factor does.
Worked Example: Translation on the Real Line
Take \(G=(\mathbb{R},+)\) with Lebesgue measure \(m\). It is a left Haar measure and is invariant under translations, so \(\Delta(t)=1\) for every \(t\). Let \(f=\chi_{[0,2]}\) and translate by \(3\). The strictness of the interval endpoints does not affect the integral because singletons have Lebesgue measure zero.
The translated function \(f(x+3)\) equals \(1\) exactly when \(x+3\in[0,2]\), or \(x\in[-3,-1]\). Hence $$ \int_{\mathbb{R}} f(x+3)\,dm(x)=m([-3,-1])=2. $$ The formula for left translation gives \(\int_{\mathbb{R}}f\,dm=m([0,2])=2\). For inversion, \(f(-x)=1\) exactly on \([-2,0]\), also of measure \(2\); the inversion formula gives \(\int f(x)\Delta(x)^{-1}\,dm(x)=2\cdot1=2\). Check: direct interval lengths and all three formulas give \(2\).
Worked Example: Left and Right Translation in the Affine Group
Consider \(G=\{(a,b):a>0,\ b\in\mathbb{R}\}\), with multiplication \((a,b)(c,d)=(ac,b+ad)\). A left Haar measure is \(d\mu(a,b)=da\,db/a^2\), and its modular function in our convention is \(\Delta(a,b)=1/a\). Let \(E=[1,2]\times[0,1]\) and \(y=(2,0)\). The rectangle is Borel; including its boundary does not alter its measure.
First compute \(\mu(E)\): $$ \mu(E)=\int_1^2\int_0^1\frac{1}{a^2}\,db\,da =\int_1^2\frac{1}{a^2}\,da =\left[-\frac1a\right]_1^2=\frac12. $$ Left multiplication by \(y\) maps \((a,b)\) to \((2a,2b)\), so \(yE=[2,4]\times[0,2]\). Its measure is $$ \mu(yE)=\int_2^4\int_0^2\frac{1}{a^2}\,db\,da =2\left[-\frac1a\right]_2^4 =2\left(\frac12-\frac14\right)=\frac12. $$ This agrees with left invariance.
Right multiplication by \(y\) maps \((a,b)\) to \((2a,b)\), so \(Ey=[2,4]\times[0,1]\). Therefore $$ \mu(Ey)=\int_2^4\int_0^1\frac{1}{a^2}\,db\,da =\frac12-\frac14=\frac14. $$ The set form of the modular identity predicts \(\Delta(y)\mu(E)=\frac12\cdot\frac12=\frac14\). Equivalently, for \(f=\chi_E\), \(\int f(xy)\,d\mu(x)=\mu(Ey^{-1})=\Delta(y)^{-1}\mu(E)=2\cdot\frac12=1\). Here \(Ey^{-1}=[1/2,1]\times[0,1]\), whose measure is \(\int_{1/2}^1a^{-2}\,da=2-1=1\). Check: the direct preimage calculation and the right-translation integral formula both give \(1\).
Worked Example: Inversion in the Affine Group
Use the same group and measure, and again let \(E=[1,2]\times[0,1]\). The inverse of \((a,b)\) is \((1/a,-b/a)\). Consequently, $$ E^{-1}=\{(u,v):1/2\le u\le1,\ -u\le v\le0\}. $$ For each fixed \(u\in[1/2,1]\), the \(v\)-section has length \(u\). Thus $$ \mu(E^{-1})=\int_{1/2}^1\int_{-u}^0\frac{1}{u^2}\,dv\,du =\int_{1/2}^1\frac{u}{u^2}\,du =\int_{1/2}^1\frac{1}{u}\,du =\ln 2. $$ The inversion formula, applied to \(f=\chi_E\), gives the same value: $$ \int_E\Delta(a,b)^{-1}\,d\mu(a,b) =\int_1^2\int_0^1\frac{a}{a^2}\,db\,da =\int_1^2\frac1a\,da =\ln 2. $$ Check: the direct measure of \(E^{-1}\) and the modularly weighted integral over \(E\) both equal \(\ln 2\).
Discrete Groups and a Common Pitfall
For a discrete group \(G\), counting measure is both left and right invariant: multiplying by a fixed group element is a bijection of \(G\). Thus \(\Delta\equiv1\), and all three formulas reduce to reindexing a sum. This remains true for noncommutative discrete groups; commutativity is not required.
Worked Example: Reindexing on a Finite Noncommutative Group
Let \(G=S_3\), the permutation group on three symbols, with counting measure, and let \(E=\{e,(12)\}\). For \(f=\chi_E\), \(\int_G f\,d\mu=2\). Choose \(y=(123)\). The sum \(\sum_{x\in G}f(yx)\) has value \(1\) precisely when \(yx\in E\), that is, when \(x\in y^{-1}E=\{y^{-1},y^{-1}(12)\}\). These are two distinct elements, so the sum is \(2\). Likewise \(\sum_{x\in G}f(xy)\) counts \(x\in Ey^{-1}\), again two elements. Finally, \(\sum_{x\in G}f(x^{-1})\) counts \(x\in E^{-1}=E\), also two elements. Check: each direct count is \(2\), as the formulas with \(\Delta\equiv1\) require.
Common mistake: omitting the modular factor. In the affine-group example, one might incorrectly write \(\mu(Ey)=\mu(E)\) because left translation preserves Haar measure. But for \(y=(2,0)\), the calculations give \(\mu(E)=1/2\) and \(\mu(Ey)=1/4\), not \(1/2\). The correction is to use the identity for the side of translation actually being performed: \(\mu(Ey)=\Delta(y)\mu(E)=(1/2)(1/2)=1/4\). Equivalently, in the integral formula for \(f(xy)\), use \(\Delta(y)^{-1}=2\), not \(1\).
A locally compact group is called unimodular when \(\Delta\equiv1\). In that case inversion and both translations preserve the left Haar integral. The affine group above is not unimodular, as \(\Delta(2,0)=1/2\). Its example shows why the distinction between left and right translation matters, and why inversion cannot generally be treated as measure-preserving.
Check Your Understanding
In each question, use the convention \(\mu(Ey)=\Delta(y)\mu(E)\).
- Write the right-translation integral formula for a nonnegative Borel function \(f\), and identify its modular factor.
- If \(\Delta(y)=3\), what factor appears in \(\int_G f(xy)\,d\mu(x)\)? What factor appears in the set identity for \(\mu(Ey)\)?
- For the affine group with \(\Delta(a,b)=1/a\), calculate the factor in the inversion formula at \((a,b)=(4,1)\).
- Explain why counting measure on a discrete group has modular function \(1\), even if the group is noncommutative.
- State the additional condition on \(f\) under which the theorem's formulas apply to complex-valued functions.