# Lebesgue–Stieltjes integral from 0 to $\infty$ on $\mathbb{R}^+$

In the Stochastic analysis course we encountered the following integral

$\int_0^\infty H^2_sd[M,M]_s$,

where $H_s$ is a predictable process, $M_s$ is a uniformly integrable martingale in $L^2$, $[M,M]_s$ is the quadratic variation.

We were working on the measure space $(\mathbb{R}^+ \times \Omega, \mathcal{B}(\mathbb{R}^+) \otimes \mathcal{F})$. As far as I understand, we consider the integral $w$-wise ($\omega \in \Omega$) as a Lebesgue–Stieltjes integral, with a stochastic process being a function $s \mapsto X_s(\omega), \ X_s(w): \mathbb{R}^+ \rightarrow \mathbb{R}$ for any $\omega$.

The question is: how the integral is defined (taking into account we are in $\mathbb{R}^+$ and the integral is to $\infty$)?

My guess was: we define it via a limit as $s \rightarrow \infty$ of the integral from 0 to $s$. However, my professor declined it (saying "there is no problem here as with a Riemann integral").

Let $\omega\in\Omega$ be fixed and let $F\!:\mathbb{R}_+\to \mathbb{R}_+$ be defined as $F(s)=[M,M]_s(\omega)$. Then $F$ is a non-negative, right-continuous and non-decreasing function with $F(0)=0$, and hence $$\mu_F((a,b])=F(b)-F(a),\quad0\leq a<b<\infty$$ defines a Borel-measure on $(\mathbb{R}_+,\mathcal{B}(\mathbb{R}_+))$. If $g\!:\mathbb{R}_+\to \mathbb{R}_+$ is a measurable function, then integrating with respect to $F$ is just ordinary Lebesgue-integration with respect to $\mu_F$, i.e. $$\int_0^\infty g(s)\,\mathrm dF(s):=\int_{\mathbb{R}_+}g(s)\mu_F(\mathrm ds).\tag{1}$$ Now, the integral $$\int_0^\infty H_s^2\mathrm d[M,M]_s$$ is just the $\omega$-wise construction of the integral in $(1)$.