Question: I am trying to follow the proof the following theorem from page 21 of Deitmar et al's Principles of Harmonic Analysis
Let $G$ be a locally compact group, $K \subset G$ a compact subgroup and $H \subset G$ a closed subgroup such that $G = HK$. Then one can arrange the Haar measures on $G,H,K$ in a way that for every $f \in L^1(G)$ one has $$\int_G f(x) \, dx= \int_H \int_K f(hk) \, dk \, dh.$$
The group $H \times K$ acts on $G$ by $(h,k).g = h g k^{-1}$ thus we identify $G \cong (H \times K)/(H \cap K)$ where $H \cap K$ is embedded diagonally into $H \times K$. As a subgroup of $H \times K$, $H \cap K$ is compact in $H\times K$ thus $\Delta_{H \cap K} \equiv 1$ and in fact $\Delta_{H \times K}\vert_{H \cap K} \equiv 1 \equiv \Delta_{H \cap K}$. Thus, there exists a nonzero invariant Radon measure on $G \cong (H \times K)/(H\cap K)$ so that $$\int_{H \times K} f( (h,k)) \, d( (h,k)) = \int_{G \cong (H \times K)/(H\cap K)} \int_{H \cap K} f(xy) \, dy \, dx$$ However, I am not sure how to proceed from here to obtain the above formula.
Could I check if there is any gap in the following argument:
Following the proof in the text, given any Haar measure $\mu_G$ on $G$ is also $H \times K$ -invariant, that is if $E \subset G$ is $\mu_G$-measurable then for any $(h,k) \in H \times K$, we have $$\mu_G((h,k).E) = \mu_G(h E k^{-1}) = \mu_G(E k^{-1}) =\Delta_G(k^{-1}) \mu_G(E) = \mu_G(E)$$ where we first use the left invariance of $\mu_G$ and $\Delta_G$ is trivial on the compact subgroup $K$.
By uniqueness of Haar measure, there exists $c >0$ so that $c \mu_{(H\times K)/(H \cap K)} = c \mu_G$ and for any $f \in L^1(G) = L^1((H\times K)/(H \cap K))$, we can view $f$ as a function of $H \times K$ we have \begin{align*} \int_{H \times K} f(h,k) \, d\mu_{H \times K}(h,k) = \int_{(H\times K)/(H \cap K)} \int_{H \cap K}f(\alpha \beta) \,d \beta \, d \alpha. \end{align*} For the right hand side of the above expression, we choose $\mu_{H \cap K}$ to be so $\mu_{H \cap K} = 1$ and note that if $\alpha_1, \alpha_2 \in H \times K$ is so that $\alpha_1(H\cap K) = \alpha_2(H\cap K)$ then $$\int_{H \cap K} f(\alpha_1 \beta) \, d \beta = \int_{H \cap K}f(\alpha_2 \beta) \,d \beta = f(\alpha_1) \mu_{H \cap K}(H \cap K) = f(\alpha_2) \mu_{H \cap K}(H \cap K) $$ implying that $f(\alpha_1) = f(\alpha_2)$. So the right hand side of the above expression can be written as \begin{align*} \int_{(H\times K)/(H \cap K)} \int_{H \cap K}f(\alpha \beta) \,d \beta \, d \alpha &= \int_{(H\times K)/(H \cap K)} f(\alpha) \mu_{H \cap K}(H \cap K) \, d \alpha\\ &= \int_{(H\times K)/(H \cap K)} f(\alpha)\, d \alpha\\ &= c \int_G f(x) \, dx. \end{align*}
For the left hand side of the above expression, assume $\mu_{H \times K}$ is the unique Haar measure is so that $\mu_{H \times K} = c \mu_{H} \mu_{K}$ and $$\int_{H \times K} f(h,k) \, d\mu_{H \times K}(h,k) = c\int_H \int_K f(hk) \, dk \, dh.$$ Assume $\int_G f(x) \, dx \neq 0$ we have $ \int_H \int_K f(hk) \, dk \, dh = \int_G f(x) \, dx.$