Let $f:[0,\infty)\to\mathbb R$ be right-continuous and $$v(t):=\operatorname{Var}_{[0,\:t]}f$$ denote the variation of $f$ on $[0,t]$ for $t\ge0$. Assume $v(t)<\infty$ for all $t\ge0$ and let $$f^\pm:=\frac12(v\pm f).$$ It's easy to see that $f^\pm$ is nondecreasing and right-continuous. Hence, there's a unique $\sigma$-finite measure $\mu^\pm$ on $\mathcal B([0,\infty))$ with $$\mu^\pm([a,b])=f^\pm(b)-f^\pm(a)\;\;\;\text{for all }0\le a\le b<\infty\tag1.$$
Question 1: I've read that $$\mu:=\mu^+-\mu^-$$ is a well-defined (extended) signed measure. Shouldn't we need to assume that at least one of $\mu^+$ and $\mu^-$ is finite (if $B\in\mathcal B([0,\infty))$ with $\mu^\pm(B)=\infty$, then $\mu(B)=\infty-\infty$ is undefined)? If so, is there an easy to check criterion on $f$ which ensures this?
Question 2: How do we need to argue that $\mu$ is the unique signed measure on $\mathcal B([0,\infty))$ with $$\mu([a,b])=f(b)-f(a)\;\;\;\text{for all }0\le a\le b<\infty\tag2?$$
Remark: There might be a problem with the type of intervals in $(1)$. Thinking about it, I guess I should have chosen intervals of the form $(a,b]$ instead. The problem is that, unless the functions are left-continuous, the measure of $[a,b]$ might not differ from the measure of $(a,b]$.