Let $g: \mathbb{R}\rightarrow \mathbb{R}$ given by $g(x)=x-\left \lfloor x \right \rfloor$, $x\in\mathbb{R}$
Show that $\left \{ g\geq a\right \} \in \mathcal{B}(\mathbb{R})$ for all $a\in \mathbb{R}$ and that $g \in\mathcal{M}^+(\mathcal{B}(\mathbb{R}))$. The notation $\left \{ g\geq a\right \} $ means $ \left \{ x \in \mathbb{R} : g(x) \geq a \right \}$.
If both the range and domain are measurable spaces, then a function is called measurable if the induced $σ$-algebra is a subset of the original $σ$-algebra. We usually show this by showing Borel-measurable when the inverse image $f^{-1}(U)$ is a Borel set for every open set $U$ in the target space.
The sum is throwing me off and the fact that the set is not easily for me to handle.
The graph of $g$ is an "inifnite sawtooth wave" and $g$ is clearly bounded by $0$ from below and $1$ from above. Then, the inverse image of a set of the form $[a,\infty)$ is easy to describe: $$ g^{-1}([a,\infty))=\begin{cases} \mathbb{R}, & a\le 0, \\ \bigcup_{i\in \mathbb{Z}}[i+a,i+1), & 0<a< 1, \\ \emptyset, & \ge 1. \end{cases} $$