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 $T_H(f) : G\to \mathbf{C}$ by

\[

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

\[

|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 $G\to G/H$, then $\hat{T}_H(f)$ is supported on $q(K)$, so $\hat{T}_H:C_c(G)\to C_c(G/H)$. Finally, define $\lambda: C_c(G)\to \mathbf{C}$ by

\[

\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

\[

\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.

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