Consider $(\mathbb{R},\mathcal{B}^1,\mu)$ where $$\mu(A):=\mathcal{H}^0(A\cap\mathbb{Q}),\quad A\in\mathcal{B}^1,$$ where $\mathcal{H}^0$ is the counting measure. According to my course this measure space is $\sigma$-finite.
Reviewing this statement, I remembered the example of Vitali and considered $\mathbb{R}/\mathbb{Q}$ and some set of representatives $R$. Now it is $\mathbb{R}=\bigcup_{q\in\mathbb{Q}} q+R$ and $(\mu(q+R)\leq 2,\ q\in\mathbb{Q})$, if only $q+R$ was in $\mathcal{B}^1$ for every $q\in\mathbb{Q}$. But this isn't the case, right?
So, is this measure space really $\sigma$-finite, or do we need to consider $\mathcal{P}(\mathbb{R})$ as the $\sigma$-algebra?
The countable family of Borel sets $$ \{s\},\qquad \text{where }s \in \mathbb Q \\ \mathbb R \setminus \mathbb Q $$ covers the space; and all of these have finite measure. The sets $\{s\}$ have measure $1$, and the set $\mathbb R \setminus \mathbb Q$ has measure $0$.