6.4 Haar measure

Suppose $G$ is a locally compact Hausdorff topological group and $B$ is the Borel $\sigma$-algebra on $G$. Question: does there exist a measure function $\mu :B\to [0,\infty ]$ that is left-translation invariant, that is, $\mu (gE)=\mu (E)$ for $g\in G,E\in B$?

Theorem

Suppose $G$ is a locally compact Hausdorff topological group and $B$ is the Borel $\sigma$-algebra on $G$. There is a unique (up to scalar multiplication) non-trivial measure $\mu :B\to [0,\infty ]$ satisfying

• $\mu$ is left-translation invariant
• $\mu$ is finite on every compact subset of $G$
• $\mu$ is outer regular on Borel sets, that is, $\mu (E)=\inf \{\mu (U):U\supseteq E\}$, $E\in B$.
• $\mu$ is inner regular on open sets, that is, $\mu (E)=\sup \{\mu (F):F\subseteq E,F\text{ is open}\}$.

such a measure is called a Haar measure on $G$.

Rmk. i) If $\mu$ is left-invariant, then ${\mu }^{\prime }(E):=\mu (E^{-1})$ is right-invariant.

ii) A Haar measure $\mu$ satisfies $\mu (U)>0$ for any non-empty open subset.

Here is an example. To verify it, only need ${\mathrm{d}}^{\times }(ga)={\mathrm{d}}^{\times }(a)$ and $\mathrm{d}(g_1{x}_1+x_1{a}_2)=\mathrm{d}(g_1{x}_1)=g_1\mathrm{d}(x_1)$.