Translation as an Action on Sets, Functions, and Measures
On a group, translation depends on the order of multiplication. This matters even before any measure is chosen: the left translate of a set is not generally its right translate, and the corresponding formula for translating a measure must use a preimage. Once these conventions are fixed, invariance can be expressed in two equivalent ways: as an equality of measures of translated sets, or as an identity for integrals of translated functions.
Throughout, let \(G\) be a locally compact Hausdorff topological group, with identity \(e\) and Borel sigma-algebra \(\mathcal{B}_G\). Let \(\mu\) be a Radon measure on \((G,\mathcal{B}_G)\). We do not assume that \(\mu\) is finite, sigma-finite, or complete. Since group multiplication and inversion are continuous, the map \(L_g(x)=gx\) is a homeomorphism for each \(g\in G\), so it carries Borel sets to Borel sets. Functions in the main integral statements below are Borel measurable and take values in \([0,\infty]\); the integrable real- or complex-valued case will also be specified.
The inverse in the measure definition is forced by preimages: \(L_g^{-1}(E)=g^{-1}E\). It also makes the measure-translation formula agree with the usual pushforward integration formula. For every nonnegative Borel function \(f\),
To justify this identity, first take \(f=\chi_E\). Its left side is \((g\cdot\mu)(E)=\mu(g^{-1}E)\), and its right side is \(\int\chi_E(gx)\,d\mu(x)=\mu(\{x:gx\in E\})=\mu(g^{-1}E)\). Additivity gives the identity for nonnegative simple functions. Every nonnegative Borel function is the increasing pointwise limit of nonnegative Borel simple functions; applying the Monotone Convergence Theorem on both sides proves the formula, including when either side is infinite. If \(f\) is integrable and real- or complex-valued, apply the nonnegative formula to its positive and negative parts, or to the positive and negative parts of its real and imaginary parts. In particular, the formula also shows that \(f(g\,\cdot)\) is integrable with respect to \(\mu\) whenever \(f\) is integrable with respect to \(g\cdot\mu\).
Translations compose according to the group law. Indeed, \(g(hE)=(gh)E\), and for functions \(\tau_g(\tau_h f)(x)=f(h^{-1}g^{-1}x)=f((gh)^{-1}x)=\tau_{gh}f(x)\). Likewise, \(g\cdot(h\cdot\mu)=(gh)\cdot\mu\), as follows by evaluating both measures on \(E\): each gives \(\mu(h^{-1}g^{-1}E)=\mu((gh)^{-1}E)\). These formulas depend on left multiplication and do not require the group to be commutative.
Set Invariance and Integral Invariance
- \(\mu(gE)=\mu(E)\) for every \(g\in G\) and \(E\in\mathcal{B}_G\).
- For every \(g\in G\) and every nonnegative Borel function \(f:G\to[0,\infty]\), \(\int_G f(g^{-1}x)\,d\mu(x)=\int_G f(x)\,d\mu(x)\).
Proof. Suppose first that \(\mu(gE)=\mu(E)\) for every Borel set \(E\) and every \(g\). This also gives \(\mu(g^{-1}E)=\mu(E)\), by applying the assumed equality with \(g^{-1}\) in place of \(g\). For a Borel set \(E\), the set \(\{x:g^{-1}x\in E\}\) is \(gE\). Thus \(\int\chi_E(g^{-1}x)\,d\mu(x)=\mu(gE)=\mu(E)=\int\chi_E\,d\mu\). By finite additivity of the integral, the same equality holds for every nonnegative simple function. For a general nonnegative Borel \(f\), choose nonnegative Borel simple functions \(\varphi_n\) increasing pointwise to \(f\). Then \(\varphi_n(g^{-1}\,\cdot)\) increases pointwise to \(f(g^{-1}\,\cdot)\). The Monotone Convergence Theorem applied to both sequences gives \(\int f(g^{-1}x)\,d\mu(x)=\int f(x)\,d\mu(x)\), allowing the value \(+\infty\).
Now suppose the integral identity holds for all nonnegative Borel functions. For any \(E\in\mathcal{B}_G\), use \(f=\chi_E\). Since \(\chi_E(g^{-1}x)=\chi_{gE}(x)\), the identity gives \(\mu(gE)=\mu(E)\). This proves both directions. Finally, if \(f\) is integrable and real-valued, apply the nonnegative identity to \(f^+\) and \(f^-\); both have finite integrals, so their difference gives the result for \(f\). For an integrable complex-valued function, apply the real-valued conclusion to its real and imaginary parts. The same argument applied to \(|f|\) verifies integrability after translation. \(\square\)
Equivalently, left invariance says \(g\cdot\mu=\mu\) for every \(g\), because \((g\cdot\mu)(E)=\mu(g^{-1}E)\). Under invariance, the pushforward formula therefore yields \(\int f(gx)\,d\mu(x)=\int f(x)\,d\mu(x)\) for nonnegative Borel \(f\), and for integrable real- or complex-valued \(f\). Replacing \(g\) by \(g^{-1}\) gives the equivalent formula with \(f(g^{-1}x)\).
Worked Example: Translating a Simple Function on the Real Line
Take \(G=(\mathbb{R},+)\), with its usual topology, and let \(m\) be Lebesgue measure on the Lebesgue sigma-algebra \(\mathcal{L}\). This measure is complete and sigma-finite, but not finite on \(\mathbb{R}\). Its translation invariance follows directly from the construction of Lebesgue measure: translating every interval in a countable cover preserves its length, so the outer measure satisfies \(m^*(E+a)=m^*(E)\). The Carathéodory measurability criterion then shows that translating a Lebesgue measurable set preserves measurability; applying the outer-measure equality gives \(m(E+a)=m(E)\).
Let \(f=2\chi_{[0,1)}+\chi_{[1,3)}\) and translate by \(a=2\). The two half-open intervals are disjoint, with measures \(1\) and \(2\), so \(\int_{\mathbb{R}}f\,dm=2\cdot1+1\cdot2=4\). Here \((\tau_2f)(x)=f(x-2)=2\chi_{[2,3)}(x)+\chi_{[3,5)}(x)\). The translated intervals again have lengths \(1\) and \(2\), and hence \(\int_{\mathbb{R}}\tau_2f\,dm=2\cdot1+1\cdot2=4\). Check: the integral identity gives \(\int\tau_2f\,dm=\int f\,dm\), and the direct calculations give \(4=4\). The half-open endpoint convention makes the pieces disjoint; changing endpoint membership would not change these Lebesgue integrals.
Worked Example: A Dirac Measure Is Not Usually Left Invariant
Let \(G\) be any locally compact Hausdorff group with an element \(a\ne e\), and take the Radon measure \(\mu=\delta_e\), the unit point mass at \(e\). This is a finite measure. For \(E=\{e\}\), we have \(\mu(E)=1\), whereas \(aE=\{a\}\) and \(\mu(aE)=\delta_e(\{a\})=0\). Thus \(\delta_e\) is not left invariant.
The function identity fails in the same calculation. Set \(f=\chi_{\{e\}}\). Then \(\int f\,d\delta_e=f(e)=1\). But \((\tau_a f)(x)=\chi_{\{e\}}(a^{-1}x)\), which equals \(1\) exactly when \(x=a\). Since \(a\ne e\), \(\int\tau_af\,d\delta_e=(\tau_af)(e)=0\). Check: the set and function computations both give the same failure, \(0\ne1\), as the equivalence theorem requires.
Worked Example: A Weighted Measure That Fails Translation Invariance
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), define \(\mu(E)=\int_E\chi_{[0,1]}\,dm\). This is a finite Borel Radon measure, with total mass \(1\). It is not left invariant. For \(E=[0,1]\) and \(a=1\), \(\mu(E)=m([0,1])=1\). However, \(E+a=[1,2]\), and \([1,2]\cap[0,1]=\{1\}\), a Lebesgue null set. Therefore \(\mu(E+a)=m([1,2]\cap[0,1])=0\).
The integral test gives the same result. For \(f=\chi_{[0,1]}\), \(\int f\,d\mu=\int_{[0,1]}\chi_{[0,1]}\,dm=1\). Its translate is \((\tau_1f)(x)=\chi_{[1,2]}(x)\), so \(\int\tau_1f\,d\mu=m([1,2]\cap[0,1])=0\). Check: both calculations give \(1\) before translation and \(0\) after translation. The single shared endpoint contributes zero because Lebesgue measure assigns measure zero to a singleton.
A Common Convention Error
A frequent mistake is to define the translated measure by \(\mu(gE)\) instead of \(\mu(g^{-1}E)\). That expression is a set evaluation, but it does not give the pushforward under \(L_g\): the pushforward evaluates a set using its preimage. For the Dirac measure above, the correct translated measure satisfies \((a\cdot\delta_e)(\{a\})=\delta_e(a^{-1}\{a\})=\delta_e(\{e\})=1\). Using \(\delta_e(a\{a\})=\delta_e(\{a^2\})\) instead would generally give \(0\) and would not describe the translated point mass at \(a\). The corresponding function convention is consistent: \(\tau_gf(x)=f(g^{-1}x)\) moves a function supported at \(e\) to one supported at \(g\).
Left invariance is a condition on the measure, not a claim that every Radon measure on a group has it. The examples show why both formulations are useful: set identities make the geometry of translated regions visible, while integral identities are suited to changing the argument of a function. The equivalence lets one verify either formulation and use the other without an extra hypothesis.
Check Your Understanding
Use the definitions and theorem to justify each answer, taking care with the inverse in the translated measure.
- For \(g\in G\), express \((g\cdot\mu)(E)\) using the preimage of \(E\) under \(L_g\).
- Show directly that \(\chi_E(g^{-1}x)=\chi_{gE}(x)\).
- If \(\mu\) is left invariant and \(f\) is nonnegative Borel measurable, which integral identity follows for \(f(gx)\)?
- For the measure \(\mu(E)=\int_E\chi_{[0,1]}\,dm\), compute \(\mu([2,3])\) and explain why this does not contradict finiteness of \(\mu\).
- For \(a\ne e\), evaluate \((a\cdot\delta_e)(\{a\})\) and \(\delta_e(\{a\})\). What does the comparison show?