Is $B(t-1)$ an Ito process? Let $I(\cdot)$ be an indicator, and $B_{t}$ be an 1-dim standard Brownian motion in a nice filtered probability space
$(\Omega, \mathcal{F}, P, \mathcal{F}_{t})$. We consider a random process
$$Y_{t} = I(t\ge 1) B_{t-1}.$$
Obviously, $Y$ is an $\mathcal F_{t}$-adapted process with finite quadratic variation 
$\langle Y \rangle_{t} = (t-1) I(t\ge 1)$.
[Q1.] Is it a Ito process, i.e. there exists a representation of the form
$$Y_{t} = \int_{0}^{t} b_{s}ds + \sigma_{s}d B_{s}, \quad \forall t\ge 0$$
for some $\mathcal F_t$-adapted processes $b$ and $\sigma$?
[Q2.] If No for [Q1.], then is it semi-martingale?
[Q3.] If the answers for the above are both NO, then it gives an example of finite QV which is not semi-martingale. Is there any other such an example which belongs to QV but not in semi-martingale?
 A: Note that for fixed $t_0$, we have by the martingale representation theorem that
$$Y_{t_0} = \int_0^{t_0} I\bigl( s \in [0,t_0-1) \bigr) \, dB_s.$$
In particular that for $t_0\leq 1$ the indicator yields $0$ trivially. 
However, if we want to consider the process $(Y_t)$, then the integrand $I\bigl( s \in [0,t-1) \bigr)$ is not progressively measurable in the Brownian filtration (i.e. not measurable w.r.t. the $\sigma$-field $\mathcal{B}([0,s]) \otimes \mathcal{F}_s$). Thus it is not an Ito process and also not a semi-martingale as the martingale representation is unique. This answers [Q1] and [Q2].
As for [Q3], one classical example for a process with finite quadratic variation that is not a semimartingale is fractional Brownian motion with Hurst-parameter $h<1/2$. (It has even quadratic variation constant $0$).
A: Let $\sigma_t=I(t\ge 1)$ and $b_t=0$ for all $t$. So $\sigma_t$ and $b_t$ are deterministic and certainly predictable, adapted, and integrable. Then the Ito process
$$
\int_0^t b_d\,ds + \sigma_s\,dB_s = \int_0^t I(s\ge 1)\,dB_s = I(t\ge 1)(B_t-B_1)
$$
is identically distributed (but not identical) with your process $I(t\ge 1)B_{t-1}$.
