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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!

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1 Answer 1

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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 $\mathcal A:= \{ F \times (a,b], F \in \mathcal F_t, 0\leq a<b\leq t \} \cup \{F \times \{ 0\}, F\in \mathcal F_t \}$ 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.

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