Ito isometry for bounded Ito integral Let $(W_t)_{t \in [0, T]}$ be a Brownian motion and $T$ be a finite time.
If $\int^T_0 \beta_t d W_t$ is bounded and $\{ \beta_t \}_{t \in [0，T]}$ is locally integrable, I am curious whether the following assertions are true or not:


*

*${\mbox E} [ \int^T_0 \beta_t d W_t ] = 0$;

*${\mbox E} [ \int^T_0 |\beta_t|^2 d t ] < \infty$;

*${\mbox E} [ \int^T_0 |\beta_t|^2 d t ] = {\mbox E} [ |\int^T_0 \beta_t d W_t|^2 ]$.


I understand that the Ito isometry, i.e. Assertion 3, holds if $(\beta_t)_{t \in [0, T]}$ is square integrable. I am not sure whether the Ito isometry still holds not or when we only have that $\int^T_0 \beta_t d W_t$ is bounded.
 A: All of these statements can be false.
What follows is more or less the standard counterexample of a local martingale that is not a martingale.  I've stolen the details from an MO post of mine which constructs something slightly different.  (To avoid search-and-replace I'm keeping my process called $Y$ instead of $\beta$.)
Set $T=1$.  Let $r(t)$ be any positive, continuous, strictly increasing function on $[0,1)$ with $\int_0^1 r(s)^2\,ds = +\infty$ (for example, $r(t) = 1/(1-t)$).  Set $q(t) = \int_0^t r(s)^2\,ds$ so that $q$ is continuous on $[0,1)$, strictly increasing, and $q(1-) = +\infty$.  If we let $Z_t = \int_0^t r(s)\,dW_s$ for $t \in [0,1)$, then $Z_t$ is a time-changed Brownian motion; specifically, $\{Z_{q^{-1}(u)}, 0 \le u < +\infty\}$ is a Brownian motion with respect to the filtration $\{\mathcal{F}_{q^{-1}(u)}, 0 \le u < +\infty\}$.  (Observe that $Z_{q^{-1}(u)}$ is continuous with independent Gaussian increments having the correct variances.)  Let
$$\tau = \inf \left\{t : Z_t = 42\right\}.$$ 
Since Brownian motions are recurrent, almost surely $Z_{q^{-1}(u)}$ hits 42 for some $u < \infty$, thus $Z_t$ hits 42 for some $t < 1$.  So $\tau < 1$ almost surely.  Finally set
$$Y(t,\omega) = \begin{cases} r(t), & 0 \le t \le \tau(\omega)  \\ 0, & \tau(\omega) < t < \infty. \end{cases}$$
If we set $\sigma_n = \inf\{t : Y_t \ge n\}$ then we have $\sigma_n = \infty$ on the event $r(\tau) \le n$.  Since $\tau < 1$ a.s. we have $r(\tau) < \infty$ a.s., and so $\sigma_n \uparrow +\infty$ a.s.  Hence $Y_t$ is locally bounded (in particular, locally square integrable).
On the other hand, we have $\int_0^1 Y_s^2\,ds = \int_0^\tau r(s)^2\,ds = q(\tau)$.  Since $q(\tau)$ is the first passage time to 42 of the Brownian motion $Z_{q^{-1}(u)}$, and Brownian motions are null recurrent, we have $E \int_0^1 Y_s^2\,ds = E q(\tau) = \infty$.
Now $\int_0^t Y_s\,dW_s = Z_{t \wedge \tau}$.  In particular,  $\int_0^1 Y_s\,dW_s = Z_\tau = 42$ almost surely, which is certainly bounded.  So you can see that your hypotheses are satisfied but all three of your assertions fail.
