Consider a sub-martingale (e.g., a counting process) $N_t$ adapted to a filtration $\mathcal{F}_t$. According to the Doob-Meyer decomposition, we have:
$$N_t = M_t + A_t,$$
where $M_t$ is a martingale and $A_t$ is the compensator. Throughout, we will assume $A_t$ is absolutely continuous and predictable so that the intensity process $\lambda_t$ with $A_t = \int_0^t \lambda_sds$ exists.
Now, consider another filtration $\mathcal{G}_t$. My question is what are the relations between the following:
(a) $\lambda_t$ is adapted to $\mathcal{G}_t$ (i.e., $\lambda_t$ is $\mathcal{G}$-measuable).
(b) $A_t$ is adapted to $\mathcal{G}_t$.
(c) $M_t$ is adapted to $\mathcal{G}_t$.
(d) $N_t$ is adapted to $\mathcal{G}_t$
Are they equivalent? Given (a), can we dirive (b,c,d)?
After some searching, I find a partial answer to my question: yes, in some sense, (a),(b),(d) are equivalent. Whether (c) is equivalent to (a,b,d) remains unknown. I post my results below.
(a) $\Leftrightarrow$ (b). This is prove that the Lebesgue integral (as pointed out by M.N.) of a measurable function is measurable, and that the derivative of a measurable function is measurable.
The first one can be proved with the fact that "conditional expectation and integral can be exchanged", while the second one can be proved with the property that "The point limit of measurable function (if exists) is measurable".
Specifically, we have:
$$E(A(t)|\mathcal{G})=E(\int_0^t \lambda(s)ds|\mathcal{G})=\int_0^t E(\lambda(s)|\mathcal{G})ds=\int_0^t \lambda(s)ds = A(t),$$
therefore, $\Rightarrow$ is true.
We also have:
$$\lambda(t)=\lim_{n\to \infty} n(A(t-\frac{1}{n})-A(t)),$$
therefore, $\Leftarrow$ is true.
(a) $\Leftrightarrow$ (d). This requires the property:
$$\lambda_t=E({\frac{dN_t}{dt}}|\mathcal{F}_{t-}),$$
i.e., the intensity process describe the (average) changing rate of $N_t$. Therefore, $\lambda_t$ is measurable iff $dN_t$ (or $\frac{dN_t}{dt}$) is measurable.
To prove the property, note that:
$$E(M_t|\mathcal{F}_{t-})=M_{t-}$$
because $M_t$ is a martingale.
The LHS has $E(M_t|\mathcal{F}_{t-})=E(N_t-A_t|\mathcal{F}_{t-})=E(N_t|\mathcal{F}_{t-})-E(A_t|\mathcal{F}_{t-}) = E(N_t|\mathcal{F}_{t-}) - A_{t}$ because $A_t$ is a predictable process (i.e., measurable to $\mathcal{F}_{t-}$).
The RHS has $M_{t-}=N_{t-}-A_{t-}=E(N_{t-}|\mathcal{F}_{t-})$ because $N_{t-}$ is $\mathcal{F}_{t-}$-measurable by definition.
Therefore, combining the LHS and RHS, we have:
$$E(dN_t|\mathcal{F}_{t-})=E(N_t-N_{t-}|\mathcal{F}_{t-})=dA_t=\lambda_tdt.$$