$dt$ terms have zero quadratic variation Why does $ds$ integral have zero quadratic variation? Even if I have a integral of the form
$$\int X_s ds$$
where $X$ is a stochastic process? I know that a continuous process of finite variation has zero quadratic variation, but I do not see why this should be the case here.
thanks for your help
 A: In general if $X$ is a semimartingale and $H$ is a locally bounded predictable process, then $$
\Delta \left(\int_0^{\cdot} H_s\,\mathrm dX_s\right)_t=H_t \Delta X_t,\quad t\geq 0,
$$
so if $X$ is continuous, then so is any integral with respect to $X$. Now, integration with respect to the Lebesgue measure is just integration with respect to the semimartingale $X_t=t$ (which actually is of finite variation). Since $\Delta X_t=t-t=0$ we have that this $X$ is continuous and hence the integral is as well.
The integral is also of finite variation because the following holds for the stochastic integral:

If $X$ is a semimartingale and $H$ is locally bounded and predictable, then $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$ is also a semimartingale. If $X$ is a local martingale then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$, and lastly, if $X$ is of finite variation, then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$.

This can be seen in Jacod and Shiryaev's Limit Theorems for Stochastic Processes for example.
