Haar-measure on homogenous spaces

Question

Consider a locally compact group $G$ and a compact, closed subgroup $H$. It is well known that we have a Haar measure $\mu$ on $G$ and can then construct a left-invariant measure on $G/H$, which, as far as chapter 2 of Follands "A course in Abstract Harmonic" is concerned, satisfies $$ \int_G f\,d\nu=\int_{G/H}\int_{H} f(g\xi)\,d\nu(\xi)\,d\nu(gH).$$ This is theorem 2.49 in the forementioned book. Now I wonder what would happen, if, just like it is the case for $\mathbb{R}^k$ and the stabilizer of the addition, $H$ was a nullset. Then above equality cannot hold, since the inner integral would always be $0$. How do I solve this problem? Do I even need to solve it or is this example just way too pathological? Can $H$ be a nullset whilst being more than a singleton set?

Answer 1

If I understand right, you refer to Theorem 2.49 in Folland, but you wrote it wrong. Note that the result says that $$\int_G f\,d\nu=\int_{G/H}\int_{H} f(g\xi)\, {\bf d \xi}\,d\nu(gH).$$

The mistake you are making is that the inside integral $\int_{H} f(g\xi)\, {\bf d \xi}$ is calculated with respect to the Haar measure on $H$, not on $G$.

For example, if $G= \mathbb R$ and $H= \mathbb Z$ the formula becomes $$\int_\mathbb{R} f(x)\,d x=\int_{\mathbb R/ \mathbb Z} \left( \sum_{n \in \mathbb Z} f(n+y) \right) d \theta_{\mathbb R/ \mathbb Z}(y+\mathbb Z)$$

This formula holds for compactly supported continuous functions on $\mathbb R$. It also holds for $L^1(\mathbb R)$ with the small issue that the inside integral (sum in this case) only exists for almost all $y+\mathbb Z \in \mathbb R/\mathbb Z$.

P.S. Very likely you assumed implicitly that the Haar measure on $H$ is the restriction to $H$ of the Haar measure on $G$. While this holds if $H$ is open in $G$, it cannot hold in general exactly for the reason you mentioned: $H$ can have zero measure in $G$. Just compare the Haar measure on $\mathbb Z$ and $\mathbb R$.