# Prove that $\int^t_0X_s dA_s$ is progressively measurable.

Let $$(\Omega, \mathcal F_\infty, \mathcal F= (\mathcal F_t)_{t\geq 0})$$ be a filtered probability space, let $$X = (X_t)_{t\geq 0}$$ be a progressively measurable process and $$A= (A_t)_{t\geq 0}$$ be a pathwise increasing, right-continuous, $$\mathcal F$$-adapted process and $$A_0=0$$ identically. Suppose we have

$$\quad \int_0^\infty |X_s(\omega)| dA_s(\omega) < \infty, \forall \omega$$

Prove that the process

$$Y_t(\omega) := \int_{0}^t X_s(\omega)dA_s(\omega), \forall \omega$$

is progressively measurable. Here $$\int^t_0 = \int_{[0,t]}$$ is the Lebesgue-Stieljes integral with the $$\sigma$$-finite measure $$\mu_{A(\omega)} (a,b] = A_b(\omega) - A_a(\omega)$$ on $$[0,\infty)$$.

What I have tried: I know that Lebesgue integrals are right-continuous so $$Y$$ is right-continuous, hence what is left is to prove $$Y_t \in \mathcal F_t$$. If $$X$$ is continuous, this is again doable for me since now I can write the integral as a Riemann limit. But for general progressively measurable process, I don't know what to do. My first thought is try (but fail) to approximate $$X_t(\omega)$$ by simple progressive processes, which are of the form

$$Z_t = Z_0 + \sum_{i=1}^{n-1} Z_i \mathbb 1_{[t_i,t_{i+1})}, \quad Z_i \in F_{t_i}$$

Any hints are highly appreciated!

You have to use the Monotone Class Theorem here. There are two ways to do this. First, let

$$\mathcal H := \left\{ \text{Processes} \ \ \Omega \times [0,t] \ni (\omega, s) \mapsto Z(\omega,s): \int^t_0 Z_s dA_s \quad \text{is} \ \mathcal F_t-\text{measurable} \right\}$$

and $$\mathcal A:= \{ F \times (a,b], F \in \mathcal F_t, 0\leq a so that $$\sigma(\mathcal A) = \mathcal F_t \times \mathcal B([0,t])$$. I let you check that $$\mathcal H$$ satisfies all the properties in the link (the last one 3. follows from Dominated convergence). Then conclude that $$\mathcal H$$ contains all the bounded processes measurable with respect to $$\mathcal F_t \times \mathcal B([0,t])$$.

Now assume your process $$X$$ is progressive and nonegative, so its restriction to $$\Omega \times [0,t]$$ is $$\mathcal F_t \times \mathcal B([0,t])$$-measurable and then you can find a sequence of bounded, $$\mathcal F_t \times \mathcal B([0,t])$$-measurable processes $$(X^n)_{n \geq 1}$$ such that $$X^n \uparrow X$$ ( this means $$(\omega,s)$$-wise in $$\Omega \times [0,t]$$ ). By monotone convergence theorem, one has

$$\omega \mapsto \int^t_0X_s(\omega) dA_s(\omega) = \lim_{n \rightarrow \infty}\int^t_0X^n_s(\omega) dA_s(\omega),$$ and so $$\int^t_0X_s dA_s$$ is $$\mathcal F_t$$-measurable.

Now with general progressive $$X$$ with $$\int^\infty_0 |X_s| dA_s <0$$ $$\omega$$-wise like yours, split it into positive and negative parts (they are still progressive) and use the result above to see

$$\omega \mapsto \int^t_0X_s(\omega) dA_s(\omega) = \int^t_0X^+_s(\omega) dA_s(\omega) - \int^t_0X^-_s(\omega) dA_s(\omega) \in \mathcal F_t.$$

Second approach is less direct (but I should say it's rather the same). You consider the set

$$\mathcal G:= \{ G \in \mathcal F_t \times \mathcal B([0,t]): \int^t_0 \mathbb 1_G(s) dA_s \in \mathcal F_t\}.$$ Show that this set is a $$\sigma$$-algebra containing $$\mathcal A$$ (above) and so $$\mathcal G =\mathcal F_t \times \mathcal B([0,t])$$. Now the restriction of your $$X$$ on $$\Omega \times [0,t]$$ is again $$\mathcal G$$-measurable. Follow the same trick above, use a sequence of simple processes to approximate $$X$$ (in $$\Omega \times [0,t]$$). You can fill in the details.