Let $H$ be a $\mathbb R$-Hilbert space. If $(\mathcal D(A),A_i)$ is a symmetric linear operator on $H$, write $A_1\le A_2$ if $$\langle A_1x,x\rangle_H\le\langle A_2x,x\rangle_H\;\;\;\text{for all }x\in\mathcal D(A)\tag1.$$ Let $(H_\lambda)_{\lambda\ge0}$ with
- $H_\lambda$ is a closed subspace of $H$ for all $\lambda\ge0$
- $(H_\lambda)_{\lambda\ge0}$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}H_\mu=H_\lambda\tag2$$
- $\bigcup_{\lambda\ge0}H_\lambda$ is dense
Now, let $\pi_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$ for $\lambda\ge0$. We can show that $$[0,\infty)\ni\lambda\mapsto\pi_\lambda\tag3$$ is nondecreasing (in the sense of $(1)$) and $$[0,\infty)\ni\lambda\mapsto\pi_\lambda x\tag4$$ is right-continuous for all $x\in H$.
Let $x,y\in H$. Note that $$g(\lambda):=\langle\pi_\lambda x,y\rangle_H=\frac12\left(\langle\pi_\lambda(x+y),x+y\rangle_H-\langle\pi_\lambda x,x\rangle_H-\langle\pi_\lambda y,y\rangle_H\right)\tag5$$ for all $\lambda\ge0$. $g$ is right-continuous. Moreover, by $(5)$, $g$ is the difference of nondecreasing functions and hence of bounded variation on any bounded interval. Does that suffice to conclude that there is a unique signed measure $\mu_g$ on $\mathcal B([0,\infty))$ with $$\mu((a,b])=g(b)-g(a)\;\;\;\text{for all }0\le a\le b\tag6?$$ If so, any function Borel measurable $f:\mathbb R\to\mathbb R$ is by definiton $\mu_g$-integrable if it is integrable with respect to the variation $|\mu_g|$ of $\mu_g$. Are we able to give an explicit formula for $|\mu_g|$ such that we can give a concrete characterization of $|\mu_g|$-integrability?
Basically, your question is about functions of bounded variation on $[0,\infty)$, unless I am misunderstanding your question.
Suppose $g : [0,\infty)\rightarrow\mathbb{R}$ is a function that is of bounded variation on every finite interval $[0,a]$. Then you can form the function $V_g(x)$ that is the variation of $g$ on $[0,x]$. This function is defined as $$ V_g(x)=\sup_{\mathscr{P}}\sum_{n=1}^{N}|g(x_{n-1})-g(x_n)| $$ where the $\sup$ is taken over all finite partitions $$\mathscr{P}=\{0=x_0 < x_1 < x_2 < \cdots x_N=x\}.$$ This allows you decompose $g$ into the difference $h-k$ of non-decreasing functions $$ h(x)=\frac{1}{2}(V_g(x)+g(x)),\;\; k(x)=\frac{1}{2}(V_g(x)-g(x)). $$ So $g=(h-k)/2$ has left- and right-hand limits at every point on $(0,\infty)$ and has a right limit at $0$. Suppose $g$ is normalized to be continuous from the right. Then so is $V_g$, which forces $h,k$ to be continuous from the right as well. From this you can find unique positive Borel measures $\mu_h$, $\mu_k$ on $[0,\infty)$ such that $h(x)=\mu_h[0,x]$, $k(x)=\mu_k[0,x]$. Automatically, $$ g(x)=h(x)-k(x),\;\; h(x)+k(x)=V_g(x). $$