Is the integral of an Ito process still an Ito process?

Question

Assume $r(t)$ is an Ito diffusion: $$dr_t = \mu_tdt + \sigma_tdW_t$$ Then, define the following process: $$X_t = \int_0^tr(s)ds$$ Is $X_t$ still an Ito diffusion?

Answer 1

$X_t$ is of finite variation. Per conventional definitions, it is not a diffusion process.

The integral defining $X_t$ is well defined in the classical sense path-by-path.