If $\varphi$ is progressively measurable, is $\int_0^t\varphi_s\:{\rm d}s$ adapted?

Question

Let

• $T>0$
• $\lambda$ denote the Lebesgue measure on $[0,T]$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration on $(\Omega,\mathcal A)$
• $(\varphi_t)_{t\in[0,\:T]}$ be a real-valued $\mathcal F$-progressively measurable stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $\operatorname P\left[\varphi\in\mathcal L^1(\lambda)\right]=1$ and $$A_t:=\int1_{(0,\:t]}(s)\varphi(s)\:{\rm d}\lambda(s)\;\;\;\text{for }t\in[0,T]$$

How can we show that $A$ is $\mathcal F$-adapted?

It's clear that $A$ is $\operatorname P$-almost surely continuous and has $\operatorname P$-almost surely bounded variation.

Answer 1

By progressive measurability, the restriction of $\varphi$ to $\Omega\times[0,t]$ is $\mathcal F_t\otimes\mathcal B([0,t])$-measurable. From this and Fubini it follows that for each positive integer $n$, $$ A^{(n)}_t:=\int_0^t 1_{\{|\varphi(s)|\le n\}}\varphi(s)\,ds $$ is well-defined and $\mathcal F_t$-measurable. Finally, define $A_t:=\limsup_n A^{(n)}_t$. This random variable is also $\mathcal F_t$-measurable, and $A_t(\omega)=\int_0^t \varphi(\omega,s)\,ds$ for each $\omega$ for which $s\mapsto \varphi(\omega,s)\in L^1(\lambda)$, hence almost surely.