Let $\left(f\left(t\right)\right)_{t\geq\,0}$ be a stochastic process with a.s. continuous paths adapted to a filtration $\left(\mathcal{F}_t\right)_{t\geq\,0}$.
Is the process $\left(\int_0^t f\left(s\right) ds\right)_{t\geq\,0}$ also adapted with respect to the same filtration? In other words, if $t>0$, is the random variable $\int_0^t f\left(s\right) ds$ $\mathcal{F}_t$-measurable?
Intuitively I would say yes since the integral can be seen as the limit of a Riemann sum involving terms that are all $\mathcal{F}_t$-measurable, but I have no clue about the technical details needed to assert that that measurability is preserved when taking the limit.
If the answer is no, however, what are some conditions one might add so that the result holds?
The key fact in measure theory is that pointwise limits of measurable functions are measurable. In our case, $\int_{0}^{t} f(s) \, ds$ is the pointwise limit of the sequence $F_{N} = \frac{1}{N} \sum_{j = 1}^{\lfloor tN \rfloor} f \left(\frac{j}{N}\right)$. For each $N$, $F_{N}$ is $\mathcal{F}_{t}$-measurable. Therefore, $\int_{0}^{t} f(s) \, ds = \lim_{N \to \infty} F_{N}$ is $\mathcal{F}_{t}$-measurable. From this, we see that the process $t \mapsto \int_{0}^{t} f(s) \, ds$ is adapted.