Let $F:\mathbb{R}\longrightarrow \mathbb{R}$ be increasing and right continuous. Then one can define $\mu_F$ the Lebesgue-Stieltjes measure induced by $F$, by setting $μ_F(a,b]=F(b)−F(a)$ for half-open intervals $(a,b]$.
I know that every Lebesgue-Stieltjes measure is finite over bounded sets. I am wondering if this allows me to conclude that $μ_F$ is $\sigma-$finite.
Motivation behind my question: I am wondering this because I know that every Lebesgue-Stieltjes measure is also a pre-measure, and I know that if I've got a pre-measure $\mu_0$ defined in the algebra $\mathcal{B}$, and $\mathcal{A}$ is the smallest $\sigma-$algebra containing $\mathcal{B}$, then one can extend $\mu_0$ to a measure over $\mathcal{A}$. If $\mu_0$ is a $\sigma-$finite measure, there is one only way to perform this extension.
I also know that every Lebesgue-Stieltjes measure $\mu_F$ defined over the algebra $\mathcal{B}_0$ can be extended in only one way to a Borel-measure, since $\mathcal{B}_{\mathbb{R}}$ is the smallest $\sigma-$algebra containing $\mathcal{B}_0$. This is why I was wondering if the oneness of this extension is due to the fact that every Lebesgue-Stieltjes measure is $\sigma-$finite.
In case I cannot asume that $μ_F$ is $\sigma-$finite, I would appreciate if someone could explain to me why can the extension I talk about in the motivation be performed in only one way.