If $a,b\in\mathbb R$ with $a<b$ and $f:[a,b]\to\mathbb R$ is of bounded variation, then $f$ is bounded and there are nondecreasing and nonnegative $f^\pm:[a,b]\to\mathbb R$ with $$f-f(a)=f^+-f^-\tag1$$ such that the measure $\mu_f^\pm$ induced by $f^\pm$ on $\mathcal B((a,b])$ is finite and $\mu^\pm$ is singular in respect to $\mu_f^\mp$, i.e. $(\mu_f^+,\mu_f^-)$ is the Jordan decomposition of the finite signed measure $$\mu_f:=\mu_f^+-\mu_f^-$$ on $\mathcal B((a,b])$. If a Borel measurable $u:(a,b]\to\mathbb R$ is Lebesgue integrable with respect to the variation $|\mu_f|:=\mu_f^+-\mu_f^-$ of $\mu_f$, then the Lebesgue-Stieltjes integral of $u$ with respect to $f$ is defined by $$\int u\:{\rm d}f:=\int u\:{\rm d}\mu_f=\int u\:{\rm d}\mu_f^+-\int u\:{\rm d}\mu_f^-\;.$$
Now, if $g:[0,\infty)\to\mathbb R$ is only locally of bounded variation, i.e. $\left.g\right|_{[0,\:t]}$ is of bounded variation for all $t>0$, then there is a decomposition of $(g^+,g^-)$ of $g$ as for $f$ in $(1)$. However, the measure $\mu_g^\pm$ induced by $g^\pm$ on $\mathcal B((0,\infty))$ is only known to be $\sigma$-finite and hence $\mu_g^+-\mu_g^-$ is not well-defined (since it may yield an $\infty-\infty$ expression).
In order to overcome this issue, $g^+$ or $g^-$ must be bounded.
Remark: If there is any other way to overcome this issue, please let me know.
In my scenario, I want to integrate with respect to the quadratic covariation $[M,N]$ of continuous local martingales $(M_t)_{t\ge0},(N_t)_{t\ge 0}$ on a complete filtered probability space.
The problem is that $[M,N]$ is only locally of bounded variation. Maybe the usual (Jordan) decomposition of $[M,N]$ yields at least one bounded part (If so, how can we prove that?). If not, how is $\int{\rm d}[M,N]$ defined?