# Inducing a Haar measure from a quotient

Suppose that ${G}$ is a locally compact group and that ${H}$ is a closed subgroup with an ${H}$-left-invariant regular Borel measure ${\mu_H}$ such that ${G/H}$ possesses a ${G}$-left-invariant regular Borel measure ${\mu_{G/H}}$. For instance, ${G = \mathbf{R}}$, ${H=\mathbf{Z}}$, and ${G/H= S^1}$. The following is how you then induce a Haar measure on ${G}$. (Technically, it’s easier to construct Haar measure on compact groups, so this extends that construction slightly.)

For ${f\in C_c(G)}$, define ${T_H(f) : G\rightarrow \mathbf{C}}$ by $\displaystyle T_H(f)(x) = \int_H f(xh) d\mu_H.$

Let ${K=\text{supp}f}$. Then ${f(xh)}$, as a function of ${h}$ is supported on ${x^{-1} K}$, so the above integral is finite. Moreover, if ${x,y\in U}$ and ${U}$ is compact, then $\displaystyle |T_H(f)(x) - T_H(f)(y)| = \left| \int_H (f(xh)-f(yh))d\mu_H\right| \leq \mu_H(H\cap U^{-1} K) \sup_h |f(xh)-f(yh)|.$

Because a continuous function on a compact set is uniformly continuous (in the sense that there exists a neighbourhood ${V}$ of ${e}$ such that ${gh^{-1} \in V}$ implies ${|f(g)-f(h)| <\epsilon}$), ${T_H(f)}$ is continous. Since ${T_H(f)}$ is ${H}$-right-invariant, ${T_H(f)}$ descends to a continuous function ${\hat{T}_H(f)}$ defined on ${G/H}$. Moreover, if ${q}$ is the quotient map ${G\rightarrow G/H}$, then ${\hat{T}_H(f)}$ is supported on ${q(K)}$, so ${\hat{T}_H:C_c(G)\rightarrow C_c(G/H)}$. Finally, define ${\lambda: C_c(G)\rightarrow \mathbf{C}}$ by $\displaystyle \lambda(f) = \int_{G/H} \hat{T}_H(f) d\mu_{G/H}.$

This linear functional ${\lambda}$ is positive (in the sense that ${f\geq0}$ implies ${\lambda(f)\geq0}$), so the Riesz representation theorem guarantees the existence of a regular Borel measure ${\mu_G}$ on ${G}$ such that $\displaystyle \lambda(f) = \int_G f d\mu_G$

for all ${f\in C_c(G)}$. It is now a simple matter to check that ${\mu_G}$ is ${G}$-left-invariant.