What Translation Averaging Reveals
A left Haar measure is invariant under left translations, but a translated function can still encode useful information about that invariance. Averaging a compactly supported continuous function over all left translates produces a constant function. Comparing the constants for two measures is one way to understand the scale ambiguity in Haar measure: invariance determines the measure up to a positive multiplicative constant, while a normalization chooses that constant.
Throughout, \(G\) is a locally compact Hausdorff topological group with identity \(e\). A left Haar measure on \(G\) means a nonzero regular Borel measure that is finite on compact sets and satisfies \(\mu(aE)=\mu(E)\) for every Borel set \(E\subseteq G\) and every \(a\in G\). The measures considered here need not be finite on \(G\); they are locally finite, and no assumption that \(G\) is sigma-finite is made. The functions used in the averaging argument are real-valued, continuous, and compactly supported. Integrals may also be applied to nonnegative such functions.
The Translation-Averaging Identity
For a left Haar measure \(\eta\) and a function \(f\in C_c(G)\), define its left-translation average by \(A_\eta f(x)=\int_G f(y^{-1}x)\,d\eta(y)\). This integral is finite for each \(x\). Indeed, if \(K\) is the compact support of \(f\), a nonzero integrand requires \(y^{-1}x\in K\), or \(y\in xK^{-1}\), a compact set. The function being integrated is continuous and bounded, and \(\eta\) is finite on that compact set.
Proof. Fix \(a,x\in G\). In the integral defining \(A_\eta f(ax)\), make the substitution \(y=az\). Left invariance means that integration is unchanged under this substitution: for any integrable function \(q\), \(\int q(y)\,d\eta(y)=\int q(az)\,d\eta(z)\). Then \(y^{-1}ax=(az)^{-1}ax=z^{-1}x\), so \(A_\eta f(ax)=\int f(z^{-1}x)\,d\eta(z)=A_\eta f(x)\). Since \(a\) can be any group element, the value is independent of \(x\). Evaluating at \(x=e\) gives \(A_\eta f(x)=A_\eta f(e)=\int f(y^{-1})\,d\eta(y)\), as claimed. \(\square\)
The inversion in this formula matters. Left invariance alone does not say that \(y\mapsto y^{-1}\) preserves \(\eta\). The constant is therefore an integral of \(f\) against the image of \(\eta\) under inversion, not necessarily \(\int f\,d\eta\). No change-of-variables formula for inversion is needed here.
The comparison of two left Haar measures is a statement about their integrals on all test functions, not just about one average. Haar uniqueness says that a single scale \(c>0\) works for every test function. The averaging identity is consistent with that scale: if \(\mu=c\nu\), then for every \(x\) and \(f\in C_c(G)\), \(A_\mu f(x)=cA_\nu f(x)\). This follows directly from linearity of the integral. The examples below show how a normalization or a known test-function integral determines \(c\).
Worked Examples
Worked Example: Scaling Lebesgue Measure on the Real Line
Let \(G=(\mathbb{R},+)\). Let \(\nu=m\) be Lebesgue measure and let \(\mu=7m\). Both are left Haar measures, because addition by any \(a\in\mathbb{R}\) preserves Lebesgue measure. For \(f=\chi_{[0,1]}\), which is Borel measurable and integrable for both measures, compute the scale and the translation average. (The indicator is used for the integral comparison; the averaging proposition itself is stated for continuous compactly supported functions.)
The interval has Lebesgue measure \(m([0,1])=1\). Therefore \(\nu([0,1])=1\) and \(\mu([0,1])=7m([0,1])=7\), so the ratio is \(\mu([0,1])/\nu([0,1])=7\). Thus \(\mu=7\nu\), including on every Borel set.
For a continuous compactly supported function \(f\), the additive-group average is \(A_\nu f(x)=\int_{\mathbb R}f(x-y)\,dy\). With \(z=x-y\), Lebesgue translation invariance gives \(A_\nu f(x)=\int_{\mathbb R}f(z)\,dz\), independent of \(x\); also \(A_\mu f(x)=7A_\nu f(x)\). Check: take \(f(x)=\max(1-|x|,0)\). Its integral is the area of a triangle with base \(2\) and height \(1\), namely \(1\). Hence \(A_\nu f(x)=1\) and \(A_\mu f(x)=7\) for every \(x\), agreeing with the scale computed from \([0,1]\).
Worked Example: Counting Measure on a Discrete Group
Let \(G=\mathbb{Z}\) with the discrete topology. Its compact subsets are finite, and counting measure \(\nu\) is a left Haar measure. Let \(\mu=3\nu\). Use \(f=\chi_{\{0,2\}}\) to compute the ratio of the integrals, and then calculate \(A_\nu f(k)\) directly for any integer \(k\).
The set \(\{0,2\}\) contains exactly two points, so \(\int f\,d\nu=\nu(\{0,2\})=2\). Since \(\mu=3\nu\), \(\int f\,d\mu=3\cdot2=6\), and the ratio is \(6/2=3\).
For this additive group, inversion is \(y\mapsto-y\), and the average is a sum: \(A_\nu f(k)=\sum_{y\in\mathbb Z}f(k-y)\). The summand is \(1\) exactly when \(k-y=0\) or \(k-y=2\), that is, when \(y=k\) or \(y=k-2\). These are distinct integers, so the sum has exactly two terms equal to \(1\) and all other terms equal to \(0\); hence \(A_\nu f(k)=2\). Consequently \(A_\mu f(k)=3\cdot2=6\). Check: the average values have ratio \(6/2=3\), exactly the ratio of the test-function integrals.
Worked Example: Normalizing Haar Measure on a Compact Group
Let \(G\) be a compact group, and let \(\nu\) be a left Haar measure with total mass \(\nu(G)=12\). Let \(\mu\) be the normalized Haar probability measure, so \(\mu(G)=1\). Find the scale \(c\) in \(\mu=c\nu\), and compute the two integrals of the constant function \(f=1\).
Since \(G\) is compact, the constant function \(1\) is continuous and compactly supported. Applying \(\mu=c\nu\) to \(G\) gives \(1=c\cdot12\), hence \(c=1/12\). Directly, \(\int_G1\,d\mu=\mu(G)=1\) and \(\int_G1\,d\nu=\nu(G)=12\). Thus \(\int_G1\,d\mu=(1/12)\int_G1\,d\nu\), as required.
For any \(f\in C(G)\), the constant function is integrable because the measures are finite. The uniqueness theorem gives \(\int_G f\,d\mu=(1/12)\int_G f\,d\nu\). Check: substituting \(f=1\) gives \((1/12)\cdot12=1\), which is the prescribed probability normalization.
Using the Scale and Avoiding Common Errors
When two Haar measures are given by explicit formulas, a convenient way to identify their proportionality constant is to choose a Borel set \(E\) with \(0<\nu(E)<\infty\), compute both measures of \(E\), and use \(c=\mu(E)/\nu(E)\). The set need not be compact, but finiteness and positivity of the denominator are essential for this quotient. Once the measures are known to be Haar measures, uniqueness ensures that this one ratio applies to every Borel set. For compact groups, total mass is an especially convenient choice; for discrete groups, a singleton often suffices.
A common mistake is to infer from left invariance that the inversion map preserves the measure. In the additive example, inversion is reflection, which does preserve Lebesgue measure; that special fact does not justify the same step for an arbitrary locally compact group. The averaging proposition avoids the mistake: it evaluates the constant at the identity as \(\int f(y^{-1})\,d\eta(y)\), without replacing this by \(\int f(y)\,d\eta(y)\). A change-of-variables formula for inversion requires additional information and is treated separately in the later material on inversion and Haar integrals.
A second error is to regard the scale as determined by left invariance alone. If \(\nu\) is left Haar and \(c>0\), then \(c\nu\) is also left Haar: \((c\nu)(aE)=c\nu(aE)=c\nu(E)=(c\nu)(E)\). Thus uniqueness is uniqueness up to a positive factor, not uniqueness of a numerically fixed measure. A normalization, such as total mass one on a compact group, fixes the factor.
Check Your Understanding
For each question, identify the group, the measures, and the normalization being used before computing.
- For \(G=(\mathbb{R},+)\), let \(\mu=5m\) and \(\nu=2m\). What is the constant \(c\) such that \(\mu=c\nu\)? Verify it by integrating \(\chi_{[0,1]}\).
- For a discrete group with counting measure \(\nu\), express \(A_\nu f(x)\) as a sum and explain why only finitely many terms are nonzero when \(f\in C_c(G)\).
- In the translation-averaging proposition, why is \(A_\eta f(x)\) finite for each fixed \(x\)? Identify a compact set containing the possible \(y\)-values where the integrand is nonzero.
- On a compact group, a Haar measure \(\nu\) has total mass \(4\). What scale gives a probability measure, and what does the scale identity give for the integral of the constant function \(1\)?
- Why does the averaging identity evaluate its constant using \(f(y^{-1})\), rather than automatically using \(f(y)\)?