The following was stated in my course on automorphic forms and I think there is at least missing an assumption.
Let $G$ be a locally compact Hausdorff group and $H\subseteq G$ compact. Then we can fix a Haar measure $\mu$ on $G$ by requiring $\mu(H)=1$. Here a Haar measure is locally finite, inner regular and left invariant measure on the Borel $\sigma$-algebra generated by the open sets of $G$.
Setting $G=\Bbb{R}$ and $H=\{-1,0,1\}$ which is a compact subgroup of $G$ that has Lebesgue measure $0$, the above is clearly not possible. I have already ruled out the possible problem that $H$ has infinite measure by using the local finiteness of the measure.
So am I missing something or should be added $H\subseteq G$ compact that has nonzero Haar measure (for any non trivial ($\not \equiv 0$) Haar measure)?