Assume you have a filtered probability space with a Brownian motion $B_t$. Let $X_t$ be a progressively measurable process with $P(\int_0^T X_s^2 ds <\infty)=1.$ Assume that $\int_0^TX_sdB_s=0$ a.s. Can it be that $X_s$ is not equal to zero $Leb[0,T]\times P$ a.e.?
attempt:
If $E[\int_0^TX_s^2dt]<\infty$ this can not happen because then we have from the Itô-isometry $E[\int_0^TX_s^2dt]=E[(\int_0^TX_sdB_s)^2]=E[0]=0.$
But what if $E[\int_0^TX_s^2dt]=\infty$? In this case we know from the construction of the stochastic integral that $\int_0^tX_sdB_s$ is a local martingale, denote it by $Y_t$. We can stop it to get a martingale with a sequence of stopping times, $Y_{\min(t,\tau_n)}=\int_0^{\min(t,\tau_n)}X_s dB_s$. We then get
$$E\left[\left(\int_0^{\min(t,\tau_n)}X_s dB_s\right)^2\right]=E\left[\int_0^t1_{\{s\le \tau_n\}}X_s^2ds\right].$$ We also have
$$\lim\limits_{n \rightarrow \infty}E\left[\int_0^t1_{\{s\le \tau_n\}}X_s^2ds\right]=E\left[\lim\limits_{n \rightarrow \infty}\int_0^t1_{\{s\le \tau_n\}}X_s^2ds\right]=E\left[\int_0^tX_s^2ds\right]=\infty.$$
But I am not able to arrive to a contradiction because I do not know if
$$\lim\limits_{n \rightarrow \infty}E\left[\left(\int_0^{\min(t,\tau_n)}X_s dB_s\right)^2\right]{}^?=^?E\left[\lim\limits_{n \rightarrow \infty}\left(\int_0^{\min(t,\tau_n)}X_s dB_s\right)^2\right].$$
I can't use the monotone or the dominated convergence theorem in the last case as far as I see. Do you see how to solve this?
I also tried using Doobs inequality like this:
$$E\left[\int_0^T1_{\{s\le \tau_n\}}X_s^2ds\right]=E\left[\left(\int_0^{\min(T,\tau_n)}X_s dB_s\right)^2\right]\\ \le E\left[\sup\limits_{0\le t \le T}\left(\int_0^{\min(t,\tau_n)}X_s dB_s\right)^2\right]\le 4 E\left[\left(\int_0^{\min(T,\tau_n)}X_s dB_s\right)^2\right]=4E\left[\int_0^T1_{\{s\le \tau_n\}}X_s^2ds\right].$$
But again I am not able to arrive at a contradiction, because if I let $n$ go to infinity I am not allowed to move the limit inside the fourth term as for as I see.
Update:
I am able to prove it with the additional assumption that $\int_0^t X_s dB_s = 0$ a.s, for every $t \in [0,T]$. But I am specifically interested in the case where we only assume that $\int_0^T X_s dB_s = 0$ a.s..
The quesiton can there be solved if we can prove that if $\int_0^T X_s dB_s = 0$ a.s. then $\int_0^t X_s dB_s = 0$ a.s, for every $t \in [0,T]$.
Another way to prove it is if we can show that $\int_0^t X_s dB_s$ is a martingale. Because then $E[(\int_0^T X_s dB_s)^2]\ge E[(\int_0^t X_s dB_s)^2] $, because $x^2$ is convex and we use Jenssen. The problem is that since we do not know if $E[\int_0^T X_s^2 ds]<\infty$ we do not know that it is a martingale?
Best Answer
I suspect that the general statement is not true. Here is my attempt for constructing a counter-example.
Fix a function $\sigma : [0, T) \to (0, \infty)$ such that $\int_{0}^{t} \sigma(s)^2 \, \mathrm{d}s < \infty$ for all $t < T$ but $\int_{0}^{T} \sigma(s)^2 \, \mathrm{d}s = \infty$. Then $t \mapsto \int_{0}^{t} \sigma(s) \, \mathrm{d}B_s$ is a time-changed BM. So by the recurrence of BM, the stopping time
$$\tau =\inf\left\{ t \geq T/2: \int_{0}^{t} \sigma(s) \, \mathrm{d}B_s = 0 \right\}$$
satisfies $\tau < T$ a.s. Now define $(X_t)_{t\geq 0}$ by
$$ X_t = \sigma(t) \mathbf{1}_{\{t\leq \tau\}}.$$
Then $$ \int_{0}^{T} X_s^2 \, \mathrm{d}s = \int_{0}^{\tau} \sigma(s)^2 \, \mathrm{d}s < \infty \quad \text{a.s.} $$ and
$$ \int_{0}^{T} X_s \, \mathrm{d}B_s = \int_{0}^{\tau} \sigma(s) \, \mathrm{d}B_s = 0, $$
but certainly $X_t$ is not null.