I'm working on an exercise from Cohn's book Measure Theory, and I think I have a solution but I'm not sure. My issue rests on whether or not a measure defined to be infinity on every nonempty set can be a Haar measure or not - I imagine the answer is no, but I can't see why from the definition.
Let $(G, +)$ be the topological group of rational numbers with addition, with the subspace topology induced from $\mathbb{R}$. Show that there does not exist a nonzero translation invariant regular Borel measure on $G$ .
Let $\mu$ be a nonzero translation invariant regular Borel measure measure on $G$, and label $G = \{q_1, q_2, q_3, \ldots\}$. Then by countable additivity and translation invariance of $\mu$, $$\mu(G) = \mu\left(\bigcup\limits_{i=1}^{\infty}\{q_i\}\right) = \sum\limits_{i=1}^{\infty}\mu(\{q_i\}) = \sum\limits_{i=1}^{\infty}\mu(\{0\}). $$ We can immediately discount the possibility that $\mu(\{0\}) = 0$, as then $\mu$ would be identically zero. So let $\mu(\{0\}) > 0$.
Any open set in $G$ is a countable union of sets of the form $(a, b) \cap \mathbb{Q}$, where $a,b \in \mathbb{R}$, and $a < b$. There are infinitely many rationals in any real open interval, so hence the measure of every open set in $G$ is infinity.
Regularity of $\mu$ tells us that $$\mu(\{0\}) = \text{inf}\{\mu(U) : \{0\} \subseteq U, U \text{ is open}\} = \infty. $$ So the measure of any nonempty subset of $G$ is infinite. At this point I feel the proof should be done, but like I said I can't see why the infinity measure isn't a nonzero regular translation invariant Borel measure.