Remark

Tag 09D8, permanent. Cite it as [Bourbaki, Tag 09D8].

Integration, chapter VII, § 1. Construction of a Haar measure

— If $\varphi$ is an isomorphism of $G$ onto a locally compact group $G'$ , then $\Delta_{G'} \circ \varphi = \Delta_G$ . In particular: 1) Since $x \mapsto x^{-1}$ is an isomorphism of $G$ onto the opposite group $G^0$ , one has $\Delta_{G^0} = \Delta_G^{-1}$ . 2) If $\varphi$ is an automorphism of $G$ , then $\Delta_G \circ \varphi = \Delta_G$ .

Let $s \in G$ . Then:

\delta(s)(\Delta_G^{-1} \cdot \mu) = (\delta(s)\Delta_G^{-1}) \cdot (\delta(s)\mu) = (\Delta_G(s)^{-1}\Delta_G^{-1}) \cdot (\Delta_G(s)\mu) = \Delta_G^{-1} \cdot \mu,

therefore $\Delta_G^{-1} \cdot \mu = \mu'$ is a right Haar measure. From this, one deduces that $\gamma(s)\mu' = (\gamma(s)\Delta_G^{-1}) \cdot \mu = \Delta_G(s)(\Delta_G^{-1} \cdot \mu) = \Delta_G(s)\mu'$ , therefore, for every right Haar measure $\nu$ , we have $\gamma(s)\nu = \Delta_G(s)\nu$ . Since $\dot{\mu}$ is a right Haar measure, $\dot{\mu} = a \Delta_G^{-1} \cdot \mu$ with a constant $a > 0$ ; it follows that

\mu = a(\Delta_G^{-1} \cdot \mu)^{\vee} = a \Delta_G \cdot \dot{\mu} = a^2 \mu,

therefore $a = 1$ and finally $\dot{\mu} = \Delta_G^{-1} \cdot \mu$ . One sees similarly that $\dot{\nu} = \Delta_G \cdot \nu$ .

We therefore have the following results:

Formulary. — Let G be a locally compact group, $\Delta$ its modulus, $\mu$ a left Haar measure, and $\nu$ a right Haar measure.

1) One has

\gamma(s)\mu = \mu \quad \delta(s)\mu = \Delta(s)\mu \quad \check{\mu} = \Delta^{-1} \cdot \mu .

If $f$ is $\mu$ -integrable on G, then the left and right translates of $f$ are $\mu$ -integrable, and

\int f(sx)\, d\mu(x) = \int f(x)\, d\mu(x)

(21)

\int f(xs)\, d\mu(x) = \Delta(s)^{-1} \int f(x)\, d\mu(x) .

Moreover, $\check{f}$ is integrable for $\Delta^{-1} \cdot \mu$ and

\int f(x^{-1})\Delta(x)^{-1}\, d\mu(x) = \int f(x)\, d\mu(x) .

(22)

If A is a $\mu$ -integrable subset of G, then sA and As are $\mu$ -integrable and

\mu(sA) = \mu(A) \qquad \mu(As) = \Delta(s)\mu(A) .

(23)

2) One has

\delta(s)\nu = \nu \quad \gamma(s)\nu = \Delta(s)\nu \quad \check{\nu} = \Delta \cdot \nu .

If $f$ is $\nu$ -integrable on G, then the left and right translates of $f$ are $\nu$ -integrable and

\int f(xs)\, d\nu(x) = \int f(x)\, d\nu(x)

(25)

\int f(xs)\, d\nu(x) = \Delta(s) \int f(x)\, d\nu(x) .

Moreover, $\check{f}$ is integrable for $\Delta \cdot \nu$ and

\int f(x^{-1})\Delta(x)\, d\nu(x) = \int f(x)\, d\nu(x) .

(26)

If A is a $\nu$ -integrable subset of G, then sA and As are $\nu$ -integrable and

\nu(As) = \nu(A) \qquad \nu(sA) = \Delta(s)^{-1} \nu(A) .

(27)

3) $\nu$ is proportional to $\Delta^{-1} \cdot \mu$ , and $\mu$ is proportional to $\Delta \cdot \nu$ .

4) Suppose G is unimodular. Let $\mu$ be a Haar measure on G. Then:

\gamma(s)\mu = \delta(s)\mu = \dot{\mu} = \mu .

If $f$ is $\mu$ -integrable on G, then the left and right translates of $f$ are $\mu$ -integrable, as is $\check{f}$ , and

\int f(sx)\,d\mu(x) = \int f(xs)\,d\mu(x) = \int f(x^{-1})\,d\mu(x) = \int f(x)\,d\mu(x) .

If A is a $\mu$ -integrable subset of G, then sA, As and $A^{-1}$ are $\mu$ -integrable and

\mu(sA) = \mu(As) = \mu(A^{-1}) = \mu(A) .

The analogous properties hold for the essential integral.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.